Get the biggest chronological drop, min and max from an array with O(n)
7
I have written a javascript function for analyzing the biggest drop in an array. But one little issue is still there. As the max value, I always get max value from my hole array and not from my drop.
Example:
Array: [100,90,80,120]
The biggest drop would be between 100 and 80. So max must be 100, and min 80. My function always returns the highest value from the whole array. in my case 120
function checkData(data) {
let max = 0
let min = 0
let drop = 0
for (let i = 0; i < data.length; i++) {
if (max < data[i]) {
max = data[i] //?
} else {
let tempDrop = max - data[i]
drop = Math.max(tempDrop, drop)
min = max - drop
}
}
return [max, min, drop]
}
I want to get the chronological correct biggest delta from left to right
javascript arrays algorithm calculation
asked Nov 26 at 12:36
Pommesloch
19310
|
show 11 more comments
7
I have written a javascript function for analyzing the biggest drop in an array. But one little issue is still there. As the max value, I always get max value from my hole array and not from my drop.
Example:
Array: [100,90,80,120]
The biggest drop would be between 100 and 80. So max must be 100, and min 80. My function always returns the highest value from the whole array. in my case 120
function checkData(data) {
let max = 0
let min = 0
let drop = 0
for (let i = 0; i < data.length; i++) {
if (max < data[i]) {
max = data[i] //?
} else {
let tempDrop = max - data[i]
drop = Math.max(tempDrop, drop)
min = max - drop
}
}
return [max, min, drop]
}
I want to get the chronological correct biggest delta from left to right
javascript arrays algorithm calculation
asked Nov 26 at 12:36
Pommesloch
19310
1
When you say drop, do you mean difference?
– cmprogram
Nov 26 at 12:40
@cmprogram, I mean the biggest delta between max an min.
– Pommesloch
Nov 26 at 12:41
1
is this array sorted? what do you mean by saying drop?
– yossico
Nov 26 at 12:41
return [max, min, drop]What are max, min and drop?
– Wais Kamal
Nov 26 at 12:42
what would be the drop output to [100,110,120,70,60,50,90,300,200]?
– yossico
Nov 26 at 12:42
|
show 11 more comments
7
7
7
1
I have written a javascript function for analyzing the biggest drop in an array. But one little issue is still there. As the max value, I always get max value from my hole array and not from my drop.
Example:
Array: [100,90,80,120]
The biggest drop would be between 100 and 80. So max must be 100, and min 80. My function always returns the highest value from the whole array. in my case 120
function checkData(data) {
let max = 0
let min = 0
let drop = 0
for (let i = 0; i < data.length; i++) {
if (max < data[i]) {
max = data[i] //?
} else {
let tempDrop = max - data[i]
drop = Math.max(tempDrop, drop)
min = max - drop
}
}
return [max, min, drop]
}
I want to get the chronological correct biggest delta from left to right
javascript arrays algorithm calculation
asked Nov 26 at 12:36
Pommesloch
19310
I have written a javascript function for analyzing the biggest drop in an array. But one little issue is still there. As the max value, I always get max value from my hole array and not from my drop.
Example:
Array: [100,90,80,120]
The biggest drop would be between 100 and 80. So max must be 100, and min 80. My function always returns the highest value from the whole array. in my case 120
function checkData(data) {
let max = 0
let min = 0
let drop = 0
for (let i = 0; i < data.length; i++) {
if (max < data[i]) {
max = data[i] //?
} else {
let tempDrop = max - data[i]
drop = Math.max(tempDrop, drop)
min = max - drop
}
}
return [max, min, drop]
}
I want to get the chronological correct biggest delta from left to right
javascript arrays algorithm calculation
javascript arrays algorithm calculation
asked Nov 26 at 12:36
Pommesloch
19310
asked Nov 26 at 12:36
Pommesloch
19310
edited Nov 26 at 13:20
asked Nov 26 at 12:36
Pommesloch
19310
asked Nov 26 at 12:36
Pommesloch
19310
asked Nov 26 at 12:36
Pommesloch
19310
19310
1
When you say drop, do you mean difference?
– cmprogram
Nov 26 at 12:40
@cmprogram, I mean the biggest delta between max an min.
– Pommesloch
Nov 26 at 12:41
1
is this array sorted? what do you mean by saying drop?
– yossico
Nov 26 at 12:41
return [max, min, drop]What are max, min and drop?
– Wais Kamal
Nov 26 at 12:42
what would be the drop output to [100,110,120,70,60,50,90,300,200]?
– yossico
Nov 26 at 12:42
|
show 11 more comments
1
When you say drop, do you mean difference?
– cmprogram
Nov 26 at 12:40
@cmprogram, I mean the biggest delta between max an min.
– Pommesloch
Nov 26 at 12:41
1
is this array sorted? what do you mean by saying drop?
– yossico
Nov 26 at 12:41
return [max, min, drop]What are max, min and drop?
– Wais Kamal
Nov 26 at 12:42
what would be the drop output to [100,110,120,70,60,50,90,300,200]?
– yossico
Nov 26 at 12:42
1
1
When you say drop, do you mean difference?
– cmprogram
Nov 26 at 12:40
When you say drop, do you mean difference?
– cmprogram
Nov 26 at 12:40
@cmprogram, I mean the biggest delta between max an min.
– Pommesloch
Nov 26 at 12:41
@cmprogram, I mean the biggest delta between max an min.
– Pommesloch
Nov 26 at 12:41
1
1
is this array sorted? what do you mean by saying drop?
– yossico
Nov 26 at 12:41
is this array sorted? what do you mean by saying drop?
– yossico
Nov 26 at 12:41
return [max, min, drop] What are max, min and drop?– Wais Kamal
Nov 26 at 12:42
return [max, min, drop] What are max, min and drop?– Wais Kamal
Nov 26 at 12:42
what would be the drop output to [100,110,120,70,60,50,90,300,200]?
– yossico
Nov 26 at 12:42
what would be the drop output to [100,110,120,70,60,50,90,300,200]?
– yossico
Nov 26 at 12:42
|
show 11 more comments
6 Answers
6
active
oldest
votes
6
Your loop should keep track of the current drop and compare it to the previous largest drop. You can do this by tracking indexes:
function checkData(data) {
let bestDropStart = 0
let bestDropEnd = 0
let bestDrop = 0
let currentDropStart = 0
let currentDropEnd = 0
let currentDrop = 0
for (let i = 1; i < data.length; i++) {
if (data[i] < data[i - 1]) {
// we are dropping
currentDropEnd = i
currentDrop = data[currentDropStart] - data[i]
} else {
// the current drop ended; check if it's better
if (currentDrop > bestDrop) {
bestDrop = currentDrop
bestDropStart = currentDropStart
bestDropEnd = currentDropEnd
}
// start a new drop
currentDropStart = currentDropEnd = i
currentDrop = 0
}
}
// check for a best drop at end of data
if (currentDrop > bestDrop) {
bestDrop = currentDrop
bestDropStart = currentDropStart
bestDropEnd = currentDropEnd
}
// return the best drop data
return [data[bestDropStart], data[bestDropEnd], bestDrop]
}
console.log(checkData([100, 90, 80, 120]))
console.log(checkData([100, 90, 80, 120, 30]))
console.log(checkData([70, 100, 90, 80]))
console.log(checkData([100, 90, 80, 120, 30, 50]))You can also do it by just keeping the start and end values for the current and best drops, but my preference would be to explicitly track the indexes. It just seems clearer (easier to debug and maintain) to me that way.
answered Nov 26 at 13:56
Ted Hopp
199k41312421
add a comment |
2
It sounds like you'd like the returned max and min to be the same ones used in calculating the largest drop rather than the overall max and min. In that case just add an additional variable to store those when drop is updated. Replace
drop = Math.max(tempDrop, drop)
with an if statement (pseudocode):
if current_drop is greater than stored_drop:
stored_drop = current_drop
stored_max_min = current max and min
then return the additional stored max and min.
answered Nov 26 at 13:39
גלעד ברקן
12.2k21439
add a comment |
2
You need to iterate the array once (O(n)) and keep a running count of min and max values and:
- Reset it every time the value increases w.r.t. previous value
- But note them down if the difference is greater than previously noted values
function checkData(array) {
var currentmax = array[0];
var currentmin = array[0];
var overallmax = array[0];
var overallmin = array[0];
var i;
for (i = 1; i < array.length; i++) {
if (array[i] <= array[i - 1]) {
// value is same or decreased
currentmin = array[i];
} else {
// check if previous iteration was the largest
if (currentmax - currentmin > overallmax - overallmin) {
overallmax = currentmax;
overallmin = currentmin;
}
// restart
currentmax = array[i];
currentmin = array[i];
}
}
// check if last iteration was the largest
if (currentmax - currentmin > overallmax - overallmin) {
overallmax = currentmax;
overallmin = currentmin;
}
return [overallmax, overallmin, overallmax - overallmin];
}
console.log(checkData([100, 90, 80, 120]));
console.log(checkData([100, 90, 80, 120, 200, 100, 90]));
console.log(checkData([10, 100, 50, 50, 10, 100, 50, 50, 1]));answered Nov 26 at 13:36
Salman A
175k66336424
This fails for [100, 90, 80, 120, 200, 100 ,90]. The return must be max = 200, min = 90 and drop = 110
– Pommesloch
Nov 26 at 13:50
1
the logic is sound (I think) and i just need to update the max min everytime the values change. See revised answer.
– Salman A
Nov 26 at 13:55
add a comment |
0
You need to keep track of two things:
- The maximum value you have up to an element
i. - The minimum value you have from an element
ito the end of the list.
Then you just need to find the drop for each index and select which one is biggest.
Here is the code:
function maxDrop (data) {
if (!data || data.length === 0) {
return;
}
const max = ;
const min = ;
for (let i = 0; i < data.length; i++) {
const left = data.slice(0, i);
const right = data.slice(i, data.length);
max.push(Math.max.apply(null, left));
min.push(Math.min.apply(null, right));
}
let maxDrop = max[0] - min[0];
let maxValue = max[0];
let minValue = min[0];
for (let j = 0; j < data.length; j++) {
const drop = max[j] - min[j];
if (maxDrop < drop) {
maxDrop = drop;
maxValue = max[j];
minValue = min[j];
}
}
return [maxValue, minValue, maxDrop];
}
console.log(maxDrop([100,90,80,120]))
console.log(maxDrop([100,110,120,300,200,70,60,50,90,400]))
console.log(maxDrop([100,90,80,120,30]))answered Nov 26 at 13:34
kakamg0
886412
This fails for[100, 90, 80, 120, 30]. The output is[100, 80, 20]but it should be[120, 30, 90].
– Ted Hopp
Nov 26 at 13:37
@TedHopp Fixed it, I was excluding the last element when searching for the minimum value. Thanks!
– kakamg0
Nov 26 at 13:56
add a comment |
0
To add one more to the mix..
A possible approach is to create all the drop - ranges first, than get the one with the largest drop
A version with reduce:
function checkData(data) {
let dropInd = 1,
mx = data.reduce((arr,val, ind) => {
if(ind && val > data[ind-1])dropInd++;
arr[dropInd] = arr[dropInd] || {max:val};
arr[dropInd].min = val;
arr[dropInd].drop = arr[dropInd].max - val;
return arr;
}, ) //after reduce, islands are created based on 'dropInd'
.sort((v1,v2)=>v2.drop - v1.drop) //sort by drop descending
[0]; //the first value now contains the maximum drop
return [mx.max,mx.min,mx.drop];
}
console.log(checkData([100, 90, 80, 120]));
console.log(checkData([100, 90, 80, 120, 200, 100 ,90]));
console.log(checkData([200,100, 90, 80, 120, 100 ,90]));
console.log(checkData([200,120,190, 90, 80, 120, 100 ,90]));answered Nov 26 at 14:12
Me.Name
10.1k22039
add a comment |
-1
This works exactly as intended (as per your expected output), as shown in the following code snippet.
The issue must be in how you are sending your data as an array.
alert(checkData([100, 90, 80, 120]));
function checkData(data) {
let max = 0
let min = 0
let drop = 0
for (let i = 0; i < data.length; i++) {
if (max < data[i]) {
max = data[i] //?
} else {
let tempDrop = max - data[i]
drop = Math.max(tempDrop)
min = max - drop
}
}
return [max, min, drop]
}// Output : 120, 80, 20
However, if what you are looking for is a Max, a Min, and then the biggest difference, i.e. between the the max and min, which in your given example would be 40, then you can simply do this.
alert(checkData([100, 90, 80, 120]));
function checkData(data) {
let max = data.reduce(function(a, b) {
return Math.max(a, b);
});
let min = data.reduce(function(a, b) {
return Math.min(a, b);
});
let drop = max-min;
return [max, min, drop];
}// Output : 120, 80, 40
I only mention this as I don't know why you've said you expect the biggest difference to be 20, given 100 and 80, when you have 120 and 80 in the same array, meaning the biggest difference is 40.
answered Nov 26 at 12:50
cmprogram
1,083519
Thanks, but the correct return must be max = 100, min = 80 and drop = 20. Look at the graphic that I added to my post
– Pommesloch
Nov 26 at 13:11
But 100 isn't the max... If you don't want the 120 to be included, just don't include it in the array...
– cmprogram
Nov 26 at 13:13
This data comes from a shares API. That is why I need this chronological from left to right delta. The shares course is crashed here from 100 to 80. This is a -20% drop. In my attempt, I miss the chronological aspect
– Pommesloch
Nov 26 at 13:18
1
OP wants the max and min associated with the largest drop, where a drop is a sequence of descending data values. The largest drop is from 100 to 80 (indexes 0 through 2).
– Ted Hopp
Nov 26 at 13:34
add a comment |
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6 Answers
6
active
oldest
votes
6 Answers
6
active
oldest
votes
active
oldest
votes
active
oldest
votes
6
Your loop should keep track of the current drop and compare it to the previous largest drop. You can do this by tracking indexes:
function checkData(data) {
let bestDropStart = 0
let bestDropEnd = 0
let bestDrop = 0
let currentDropStart = 0
let currentDropEnd = 0
let currentDrop = 0
for (let i = 1; i < data.length; i++) {
if (data[i] < data[i - 1]) {
// we are dropping
currentDropEnd = i
currentDrop = data[currentDropStart] - data[i]
} else {
// the current drop ended; check if it's better
if (currentDrop > bestDrop) {
bestDrop = currentDrop
bestDropStart = currentDropStart
bestDropEnd = currentDropEnd
}
// start a new drop
currentDropStart = currentDropEnd = i
currentDrop = 0
}
}
// check for a best drop at end of data
if (currentDrop > bestDrop) {
bestDrop = currentDrop
bestDropStart = currentDropStart
bestDropEnd = currentDropEnd
}
// return the best drop data
return [data[bestDropStart], data[bestDropEnd], bestDrop]
}
console.log(checkData([100, 90, 80, 120]))
console.log(checkData([100, 90, 80, 120, 30]))
console.log(checkData([70, 100, 90, 80]))
console.log(checkData([100, 90, 80, 120, 30, 50]))You can also do it by just keeping the start and end values for the current and best drops, but my preference would be to explicitly track the indexes. It just seems clearer (easier to debug and maintain) to me that way.
answered Nov 26 at 13:56
Ted Hopp
199k41312421
add a comment |
6
Your loop should keep track of the current drop and compare it to the previous largest drop. You can do this by tracking indexes:
function checkData(data) {
let bestDropStart = 0
let bestDropEnd = 0
let bestDrop = 0
let currentDropStart = 0
let currentDropEnd = 0
let currentDrop = 0
for (let i = 1; i < data.length; i++) {
if (data[i] < data[i - 1]) {
// we are dropping
currentDropEnd = i
currentDrop = data[currentDropStart] - data[i]
} else {
// the current drop ended; check if it's better
if (currentDrop > bestDrop) {
bestDrop = currentDrop
bestDropStart = currentDropStart
bestDropEnd = currentDropEnd
}
// start a new drop
currentDropStart = currentDropEnd = i
currentDrop = 0
}
}
// check for a best drop at end of data
if (currentDrop > bestDrop) {
bestDrop = currentDrop
bestDropStart = currentDropStart
bestDropEnd = currentDropEnd
}
// return the best drop data
return [data[bestDropStart], data[bestDropEnd], bestDrop]
}
console.log(checkData([100, 90, 80, 120]))
console.log(checkData([100, 90, 80, 120, 30]))
console.log(checkData([70, 100, 90, 80]))
console.log(checkData([100, 90, 80, 120, 30, 50]))You can also do it by just keeping the start and end values for the current and best drops, but my preference would be to explicitly track the indexes. It just seems clearer (easier to debug and maintain) to me that way.
answered Nov 26 at 13:56
Ted Hopp
199k41312421
add a comment |
6
6
6
Your loop should keep track of the current drop and compare it to the previous largest drop. You can do this by tracking indexes:
function checkData(data) {
let bestDropStart = 0
let bestDropEnd = 0
let bestDrop = 0
let currentDropStart = 0
let currentDropEnd = 0
let currentDrop = 0
for (let i = 1; i < data.length; i++) {
if (data[i] < data[i - 1]) {
// we are dropping
currentDropEnd = i
currentDrop = data[currentDropStart] - data[i]
} else {
// the current drop ended; check if it's better
if (currentDrop > bestDrop) {
bestDrop = currentDrop
bestDropStart = currentDropStart
bestDropEnd = currentDropEnd
}
// start a new drop
currentDropStart = currentDropEnd = i
currentDrop = 0
}
}
// check for a best drop at end of data
if (currentDrop > bestDrop) {
bestDrop = currentDrop
bestDropStart = currentDropStart
bestDropEnd = currentDropEnd
}
// return the best drop data
return [data[bestDropStart], data[bestDropEnd], bestDrop]
}
console.log(checkData([100, 90, 80, 120]))
console.log(checkData([100, 90, 80, 120, 30]))
console.log(checkData([70, 100, 90, 80]))
console.log(checkData([100, 90, 80, 120, 30, 50]))You can also do it by just keeping the start and end values for the current and best drops, but my preference would be to explicitly track the indexes. It just seems clearer (easier to debug and maintain) to me that way.
answered Nov 26 at 13:56
Ted Hopp
199k41312421
Your loop should keep track of the current drop and compare it to the previous largest drop. You can do this by tracking indexes:
function checkData(data) {
let bestDropStart = 0
let bestDropEnd = 0
let bestDrop = 0
let currentDropStart = 0
let currentDropEnd = 0
let currentDrop = 0
for (let i = 1; i < data.length; i++) {
if (data[i] < data[i - 1]) {
// we are dropping
currentDropEnd = i
currentDrop = data[currentDropStart] - data[i]
} else {
// the current drop ended; check if it's better
if (currentDrop > bestDrop) {
bestDrop = currentDrop
bestDropStart = currentDropStart
bestDropEnd = currentDropEnd
}
// start a new drop
currentDropStart = currentDropEnd = i
currentDrop = 0
}
}
// check for a best drop at end of data
if (currentDrop > bestDrop) {
bestDrop = currentDrop
bestDropStart = currentDropStart
bestDropEnd = currentDropEnd
}
// return the best drop data
return [data[bestDropStart], data[bestDropEnd], bestDrop]
}
console.log(checkData([100, 90, 80, 120]))
console.log(checkData([100, 90, 80, 120, 30]))
console.log(checkData([70, 100, 90, 80]))
console.log(checkData([100, 90, 80, 120, 30, 50]))You can also do it by just keeping the start and end values for the current and best drops, but my preference would be to explicitly track the indexes. It just seems clearer (easier to debug and maintain) to me that way.
function checkData(data) {
let bestDropStart = 0
let bestDropEnd = 0
let bestDrop = 0
let currentDropStart = 0
let currentDropEnd = 0
let currentDrop = 0
for (let i = 1; i < data.length; i++) {
if (data[i] < data[i - 1]) {
// we are dropping
currentDropEnd = i
currentDrop = data[currentDropStart] - data[i]
} else {
// the current drop ended; check if it's better
if (currentDrop > bestDrop) {
bestDrop = currentDrop
bestDropStart = currentDropStart
bestDropEnd = currentDropEnd
}
// start a new drop
currentDropStart = currentDropEnd = i
currentDrop = 0
}
}
// check for a best drop at end of data
if (currentDrop > bestDrop) {
bestDrop = currentDrop
bestDropStart = currentDropStart
bestDropEnd = currentDropEnd
}
// return the best drop data
return [data[bestDropStart], data[bestDropEnd], bestDrop]
}
console.log(checkData([100, 90, 80, 120]))
console.log(checkData([100, 90, 80, 120, 30]))
console.log(checkData([70, 100, 90, 80]))
console.log(checkData([100, 90, 80, 120, 30, 50]))function checkData(data) {
let bestDropStart = 0
let bestDropEnd = 0
let bestDrop = 0
let currentDropStart = 0
let currentDropEnd = 0
let currentDrop = 0
for (let i = 1; i < data.length; i++) {
if (data[i] < data[i - 1]) {
// we are dropping
currentDropEnd = i
currentDrop = data[currentDropStart] - data[i]
} else {
// the current drop ended; check if it's better
if (currentDrop > bestDrop) {
bestDrop = currentDrop
bestDropStart = currentDropStart
bestDropEnd = currentDropEnd
}
// start a new drop
currentDropStart = currentDropEnd = i
currentDrop = 0
}
}
// check for a best drop at end of data
if (currentDrop > bestDrop) {
bestDrop = currentDrop
bestDropStart = currentDropStart
bestDropEnd = currentDropEnd
}
// return the best drop data
return [data[bestDropStart], data[bestDropEnd], bestDrop]
}
console.log(checkData([100, 90, 80, 120]))
console.log(checkData([100, 90, 80, 120, 30]))
console.log(checkData([70, 100, 90, 80]))
console.log(checkData([100, 90, 80, 120, 30, 50]))answered Nov 26 at 13:56
Ted Hopp
199k41312421
edited Nov 26 at 15:39
answered Nov 26 at 13:56
Ted Hopp
199k41312421
answered Nov 26 at 13:56
Ted Hopp
199k41312421
answered Nov 26 at 13:56
Ted Hopp
199k41312421
199k41312421
add a comment |
add a comment |
2
It sounds like you'd like the returned max and min to be the same ones used in calculating the largest drop rather than the overall max and min. In that case just add an additional variable to store those when drop is updated. Replace
drop = Math.max(tempDrop, drop)
with an if statement (pseudocode):
if current_drop is greater than stored_drop:
stored_drop = current_drop
stored_max_min = current max and min
then return the additional stored max and min.
answered Nov 26 at 13:39
גלעד ברקן
12.2k21439
add a comment |
2
It sounds like you'd like the returned max and min to be the same ones used in calculating the largest drop rather than the overall max and min. In that case just add an additional variable to store those when drop is updated. Replace
drop = Math.max(tempDrop, drop)
with an if statement (pseudocode):
if current_drop is greater than stored_drop:
stored_drop = current_drop
stored_max_min = current max and min
then return the additional stored max and min.
answered Nov 26 at 13:39
גלעד ברקן
12.2k21439
add a comment |
2
2
2
It sounds like you'd like the returned max and min to be the same ones used in calculating the largest drop rather than the overall max and min. In that case just add an additional variable to store those when drop is updated. Replace
drop = Math.max(tempDrop, drop)
with an if statement (pseudocode):
if current_drop is greater than stored_drop:
stored_drop = current_drop
stored_max_min = current max and min
then return the additional stored max and min.
answered Nov 26 at 13:39
גלעד ברקן
12.2k21439
It sounds like you'd like the returned max and min to be the same ones used in calculating the largest drop rather than the overall max and min. In that case just add an additional variable to store those when drop is updated. Replace
drop = Math.max(tempDrop, drop)
with an if statement (pseudocode):
if current_drop is greater than stored_drop:
stored_drop = current_drop
stored_max_min = current max and min
then return the additional stored max and min.
answered Nov 26 at 13:39
גלעד ברקן
12.2k21439
answered Nov 26 at 13:39
גלעד ברקן
12.2k21439
answered Nov 26 at 13:39
גלעד ברקן
12.2k21439
answered Nov 26 at 13:39
גלעד ברקן
12.2k21439
12.2k21439
add a comment |
add a comment |
2
You need to iterate the array once (O(n)) and keep a running count of min and max values and:
- Reset it every time the value increases w.r.t. previous value
- But note them down if the difference is greater than previously noted values
function checkData(array) {
var currentmax = array[0];
var currentmin = array[0];
var overallmax = array[0];
var overallmin = array[0];
var i;
for (i = 1; i < array.length; i++) {
if (array[i] <= array[i - 1]) {
// value is same or decreased
currentmin = array[i];
} else {
// check if previous iteration was the largest
if (currentmax - currentmin > overallmax - overallmin) {
overallmax = currentmax;
overallmin = currentmin;
}
// restart
currentmax = array[i];
currentmin = array[i];
}
}
// check if last iteration was the largest
if (currentmax - currentmin > overallmax - overallmin) {
overallmax = currentmax;
overallmin = currentmin;
}
return [overallmax, overallmin, overallmax - overallmin];
}
console.log(checkData([100, 90, 80, 120]));
console.log(checkData([100, 90, 80, 120, 200, 100, 90]));
console.log(checkData([10, 100, 50, 50, 10, 100, 50, 50, 1]));answered Nov 26 at 13:36
Salman A
175k66336424
This fails for [100, 90, 80, 120, 200, 100 ,90]. The return must be max = 200, min = 90 and drop = 110
– Pommesloch
Nov 26 at 13:50
1
the logic is sound (I think) and i just need to update the max min everytime the values change. See revised answer.
– Salman A
Nov 26 at 13:55
add a comment |
2
You need to iterate the array once (O(n)) and keep a running count of min and max values and:
- Reset it every time the value increases w.r.t. previous value
- But note them down if the difference is greater than previously noted values
function checkData(array) {
var currentmax = array[0];
var currentmin = array[0];
var overallmax = array[0];
var overallmin = array[0];
var i;
for (i = 1; i < array.length; i++) {
if (array[i] <= array[i - 1]) {
// value is same or decreased
currentmin = array[i];
} else {
// check if previous iteration was the largest
if (currentmax - currentmin > overallmax - overallmin) {
overallmax = currentmax;
overallmin = currentmin;
}
// restart
currentmax = array[i];
currentmin = array[i];
}
}
// check if last iteration was the largest
if (currentmax - currentmin > overallmax - overallmin) {
overallmax = currentmax;
overallmin = currentmin;
}
return [overallmax, overallmin, overallmax - overallmin];
}
console.log(checkData([100, 90, 80, 120]));
console.log(checkData([100, 90, 80, 120, 200, 100, 90]));
console.log(checkData([10, 100, 50, 50, 10, 100, 50, 50, 1]));answered Nov 26 at 13:36
Salman A
175k66336424
This fails for [100, 90, 80, 120, 200, 100 ,90]. The return must be max = 200, min = 90 and drop = 110
– Pommesloch
Nov 26 at 13:50
1
the logic is sound (I think) and i just need to update the max min everytime the values change. See revised answer.
– Salman A
Nov 26 at 13:55
add a comment |
2
2
2
You need to iterate the array once (O(n)) and keep a running count of min and max values and:
- Reset it every time the value increases w.r.t. previous value
- But note them down if the difference is greater than previously noted values
function checkData(array) {
var currentmax = array[0];
var currentmin = array[0];
var overallmax = array[0];
var overallmin = array[0];
var i;
for (i = 1; i < array.length; i++) {
if (array[i] <= array[i - 1]) {
// value is same or decreased
currentmin = array[i];
} else {
// check if previous iteration was the largest
if (currentmax - currentmin > overallmax - overallmin) {
overallmax = currentmax;
overallmin = currentmin;
}
// restart
currentmax = array[i];
currentmin = array[i];
}
}
// check if last iteration was the largest
if (currentmax - currentmin > overallmax - overallmin) {
overallmax = currentmax;
overallmin = currentmin;
}
return [overallmax, overallmin, overallmax - overallmin];
}
console.log(checkData([100, 90, 80, 120]));
console.log(checkData([100, 90, 80, 120, 200, 100, 90]));
console.log(checkData([10, 100, 50, 50, 10, 100, 50, 50, 1]));answered Nov 26 at 13:36
Salman A
175k66336424
You need to iterate the array once (O(n)) and keep a running count of min and max values and:
- Reset it every time the value increases w.r.t. previous value
- But note them down if the difference is greater than previously noted values
function checkData(array) {
var currentmax = array[0];
var currentmin = array[0];
var overallmax = array[0];
var overallmin = array[0];
var i;
for (i = 1; i < array.length; i++) {
if (array[i] <= array[i - 1]) {
// value is same or decreased
currentmin = array[i];
} else {
// check if previous iteration was the largest
if (currentmax - currentmin > overallmax - overallmin) {
overallmax = currentmax;
overallmin = currentmin;
}
// restart
currentmax = array[i];
currentmin = array[i];
}
}
// check if last iteration was the largest
if (currentmax - currentmin > overallmax - overallmin) {
overallmax = currentmax;
overallmin = currentmin;
}
return [overallmax, overallmin, overallmax - overallmin];
}
console.log(checkData([100, 90, 80, 120]));
console.log(checkData([100, 90, 80, 120, 200, 100, 90]));
console.log(checkData([10, 100, 50, 50, 10, 100, 50, 50, 1]));function checkData(array) {
var currentmax = array[0];
var currentmin = array[0];
var overallmax = array[0];
var overallmin = array[0];
var i;
for (i = 1; i < array.length; i++) {
if (array[i] <= array[i - 1]) {
// value is same or decreased
currentmin = array[i];
} else {
// check if previous iteration was the largest
if (currentmax - currentmin > overallmax - overallmin) {
overallmax = currentmax;
overallmin = currentmin;
}
// restart
currentmax = array[i];
currentmin = array[i];
}
}
// check if last iteration was the largest
if (currentmax - currentmin > overallmax - overallmin) {
overallmax = currentmax;
overallmin = currentmin;
}
return [overallmax, overallmin, overallmax - overallmin];
}
console.log(checkData([100, 90, 80, 120]));
console.log(checkData([100, 90, 80, 120, 200, 100, 90]));
console.log(checkData([10, 100, 50, 50, 10, 100, 50, 50, 1]));function checkData(array) {
var currentmax = array[0];
var currentmin = array[0];
var overallmax = array[0];
var overallmin = array[0];
var i;
for (i = 1; i < array.length; i++) {
if (array[i] <= array[i - 1]) {
// value is same or decreased
currentmin = array[i];
} else {
// check if previous iteration was the largest
if (currentmax - currentmin > overallmax - overallmin) {
overallmax = currentmax;
overallmin = currentmin;
}
// restart
currentmax = array[i];
currentmin = array[i];
}
}
// check if last iteration was the largest
if (currentmax - currentmin > overallmax - overallmin) {
overallmax = currentmax;
overallmin = currentmin;
}
return [overallmax, overallmin, overallmax - overallmin];
}
console.log(checkData([100, 90, 80, 120]));
console.log(checkData([100, 90, 80, 120, 200, 100, 90]));
console.log(checkData([10, 100, 50, 50, 10, 100, 50, 50, 1]));answered Nov 26 at 13:36
Salman A
175k66336424
edited Nov 26 at 15:16
answered Nov 26 at 13:36
Salman A
175k66336424
answered Nov 26 at 13:36
Salman A
175k66336424
answered Nov 26 at 13:36
Salman A
175k66336424
175k66336424
This fails for [100, 90, 80, 120, 200, 100 ,90]. The return must be max = 200, min = 90 and drop = 110
– Pommesloch
Nov 26 at 13:50
1
the logic is sound (I think) and i just need to update the max min everytime the values change. See revised answer.
– Salman A
Nov 26 at 13:55
add a comment |
This fails for [100, 90, 80, 120, 200, 100 ,90]. The return must be max = 200, min = 90 and drop = 110
– Pommesloch
Nov 26 at 13:50
1
the logic is sound (I think) and i just need to update the max min everytime the values change. See revised answer.
– Salman A
Nov 26 at 13:55
This fails for [100, 90, 80, 120, 200, 100 ,90]. The return must be max = 200, min = 90 and drop = 110
– Pommesloch
Nov 26 at 13:50
This fails for [100, 90, 80, 120, 200, 100 ,90]. The return must be max = 200, min = 90 and drop = 110
– Pommesloch
Nov 26 at 13:50
1
1
the logic is sound (I think) and i just need to update the max min everytime the values change. See revised answer.
– Salman A
Nov 26 at 13:55
the logic is sound (I think) and i just need to update the max min everytime the values change. See revised answer.
– Salman A
Nov 26 at 13:55
add a comment |
0
You need to keep track of two things:
- The maximum value you have up to an element
i. - The minimum value you have from an element
ito the end of the list.
Then you just need to find the drop for each index and select which one is biggest.
Here is the code:
function maxDrop (data) {
if (!data || data.length === 0) {
return;
}
const max = ;
const min = ;
for (let i = 0; i < data.length; i++) {
const left = data.slice(0, i);
const right = data.slice(i, data.length);
max.push(Math.max.apply(null, left));
min.push(Math.min.apply(null, right));
}
let maxDrop = max[0] - min[0];
let maxValue = max[0];
let minValue = min[0];
for (let j = 0; j < data.length; j++) {
const drop = max[j] - min[j];
if (maxDrop < drop) {
maxDrop = drop;
maxValue = max[j];
minValue = min[j];
}
}
return [maxValue, minValue, maxDrop];
}
console.log(maxDrop([100,90,80,120]))
console.log(maxDrop([100,110,120,300,200,70,60,50,90,400]))
console.log(maxDrop([100,90,80,120,30]))answered Nov 26 at 13:34
kakamg0
886412
This fails for[100, 90, 80, 120, 30]. The output is[100, 80, 20]but it should be[120, 30, 90].
– Ted Hopp
Nov 26 at 13:37
@TedHopp Fixed it, I was excluding the last element when searching for the minimum value. Thanks!
– kakamg0
Nov 26 at 13:56
add a comment |
0
You need to keep track of two things:
- The maximum value you have up to an element
i. - The minimum value you have from an element
ito the end of the list.
Then you just need to find the drop for each index and select which one is biggest.
Here is the code:
function maxDrop (data) {
if (!data || data.length === 0) {
return;
}
const max = ;
const min = ;
for (let i = 0; i < data.length; i++) {
const left = data.slice(0, i);
const right = data.slice(i, data.length);
max.push(Math.max.apply(null, left));
min.push(Math.min.apply(null, right));
}
let maxDrop = max[0] - min[0];
let maxValue = max[0];
let minValue = min[0];
for (let j = 0; j < data.length; j++) {
const drop = max[j] - min[j];
if (maxDrop < drop) {
maxDrop = drop;
maxValue = max[j];
minValue = min[j];
}
}
return [maxValue, minValue, maxDrop];
}
console.log(maxDrop([100,90,80,120]))
console.log(maxDrop([100,110,120,300,200,70,60,50,90,400]))
console.log(maxDrop([100,90,80,120,30]))answered Nov 26 at 13:34
kakamg0
886412
This fails for[100, 90, 80, 120, 30]. The output is[100, 80, 20]but it should be[120, 30, 90].
– Ted Hopp
Nov 26 at 13:37
@TedHopp Fixed it, I was excluding the last element when searching for the minimum value. Thanks!
– kakamg0
Nov 26 at 13:56
add a comment |
0
0
0
You need to keep track of two things:
- The maximum value you have up to an element
i. - The minimum value you have from an element
ito the end of the list.
Then you just need to find the drop for each index and select which one is biggest.
Here is the code:
function maxDrop (data) {
if (!data || data.length === 0) {
return;
}
const max = ;
const min = ;
for (let i = 0; i < data.length; i++) {
const left = data.slice(0, i);
const right = data.slice(i, data.length);
max.push(Math.max.apply(null, left));
min.push(Math.min.apply(null, right));
}
let maxDrop = max[0] - min[0];
let maxValue = max[0];
let minValue = min[0];
for (let j = 0; j < data.length; j++) {
const drop = max[j] - min[j];
if (maxDrop < drop) {
maxDrop = drop;
maxValue = max[j];
minValue = min[j];
}
}
return [maxValue, minValue, maxDrop];
}
console.log(maxDrop([100,90,80,120]))
console.log(maxDrop([100,110,120,300,200,70,60,50,90,400]))
console.log(maxDrop([100,90,80,120,30]))answered Nov 26 at 13:34
kakamg0
886412
You need to keep track of two things:
- The maximum value you have up to an element
i. - The minimum value you have from an element
ito the end of the list.
Then you just need to find the drop for each index and select which one is biggest.
Here is the code:
function maxDrop (data) {
if (!data || data.length === 0) {
return;
}
const max = ;
const min = ;
for (let i = 0; i < data.length; i++) {
const left = data.slice(0, i);
const right = data.slice(i, data.length);
max.push(Math.max.apply(null, left));
min.push(Math.min.apply(null, right));
}
let maxDrop = max[0] - min[0];
let maxValue = max[0];
let minValue = min[0];
for (let j = 0; j < data.length; j++) {
const drop = max[j] - min[j];
if (maxDrop < drop) {
maxDrop = drop;
maxValue = max[j];
minValue = min[j];
}
}
return [maxValue, minValue, maxDrop];
}
console.log(maxDrop([100,90,80,120]))
console.log(maxDrop([100,110,120,300,200,70,60,50,90,400]))
console.log(maxDrop([100,90,80,120,30]))function maxDrop (data) {
if (!data || data.length === 0) {
return;
}
const max = ;
const min = ;
for (let i = 0; i < data.length; i++) {
const left = data.slice(0, i);
const right = data.slice(i, data.length);
max.push(Math.max.apply(null, left));
min.push(Math.min.apply(null, right));
}
let maxDrop = max[0] - min[0];
let maxValue = max[0];
let minValue = min[0];
for (let j = 0; j < data.length; j++) {
const drop = max[j] - min[j];
if (maxDrop < drop) {
maxDrop = drop;
maxValue = max[j];
minValue = min[j];
}
}
return [maxValue, minValue, maxDrop];
}
console.log(maxDrop([100,90,80,120]))
console.log(maxDrop([100,110,120,300,200,70,60,50,90,400]))
console.log(maxDrop([100,90,80,120,30]))function maxDrop (data) {
if (!data || data.length === 0) {
return;
}
const max = ;
const min = ;
for (let i = 0; i < data.length; i++) {
const left = data.slice(0, i);
const right = data.slice(i, data.length);
max.push(Math.max.apply(null, left));
min.push(Math.min.apply(null, right));
}
let maxDrop = max[0] - min[0];
let maxValue = max[0];
let minValue = min[0];
for (let j = 0; j < data.length; j++) {
const drop = max[j] - min[j];
if (maxDrop < drop) {
maxDrop = drop;
maxValue = max[j];
minValue = min[j];
}
}
return [maxValue, minValue, maxDrop];
}
console.log(maxDrop([100,90,80,120]))
console.log(maxDrop([100,110,120,300,200,70,60,50,90,400]))
console.log(maxDrop([100,90,80,120,30]))answered Nov 26 at 13:34
kakamg0
886412
edited Nov 26 at 13:55
answered Nov 26 at 13:34
kakamg0
886412
answered Nov 26 at 13:34
kakamg0
886412
answered Nov 26 at 13:34
kakamg0
886412
886412
This fails for[100, 90, 80, 120, 30]. The output is[100, 80, 20]but it should be[120, 30, 90].
– Ted Hopp
Nov 26 at 13:37
@TedHopp Fixed it, I was excluding the last element when searching for the minimum value. Thanks!
– kakamg0
Nov 26 at 13:56
add a comment |
This fails for[100, 90, 80, 120, 30]. The output is[100, 80, 20]but it should be[120, 30, 90].
– Ted Hopp
Nov 26 at 13:37
@TedHopp Fixed it, I was excluding the last element when searching for the minimum value. Thanks!
– kakamg0
Nov 26 at 13:56
This fails for
[100, 90, 80, 120, 30]. The output is [100, 80, 20] but it should be [120, 30, 90].– Ted Hopp
Nov 26 at 13:37
This fails for
[100, 90, 80, 120, 30]. The output is [100, 80, 20] but it should be [120, 30, 90].– Ted Hopp
Nov 26 at 13:37
@TedHopp Fixed it, I was excluding the last element when searching for the minimum value. Thanks!
– kakamg0
Nov 26 at 13:56
@TedHopp Fixed it, I was excluding the last element when searching for the minimum value. Thanks!
– kakamg0
Nov 26 at 13:56
add a comment |
0
To add one more to the mix..
A possible approach is to create all the drop - ranges first, than get the one with the largest drop
A version with reduce:
function checkData(data) {
let dropInd = 1,
mx = data.reduce((arr,val, ind) => {
if(ind && val > data[ind-1])dropInd++;
arr[dropInd] = arr[dropInd] || {max:val};
arr[dropInd].min = val;
arr[dropInd].drop = arr[dropInd].max - val;
return arr;
}, ) //after reduce, islands are created based on 'dropInd'
.sort((v1,v2)=>v2.drop - v1.drop) //sort by drop descending
[0]; //the first value now contains the maximum drop
return [mx.max,mx.min,mx.drop];
}
console.log(checkData([100, 90, 80, 120]));
console.log(checkData([100, 90, 80, 120, 200, 100 ,90]));
console.log(checkData([200,100, 90, 80, 120, 100 ,90]));
console.log(checkData([200,120,190, 90, 80, 120, 100 ,90]));answered Nov 26 at 14:12
Me.Name
10.1k22039
add a comment |
0
To add one more to the mix..
A possible approach is to create all the drop - ranges first, than get the one with the largest drop
A version with reduce:
function checkData(data) {
let dropInd = 1,
mx = data.reduce((arr,val, ind) => {
if(ind && val > data[ind-1])dropInd++;
arr[dropInd] = arr[dropInd] || {max:val};
arr[dropInd].min = val;
arr[dropInd].drop = arr[dropInd].max - val;
return arr;
}, ) //after reduce, islands are created based on 'dropInd'
.sort((v1,v2)=>v2.drop - v1.drop) //sort by drop descending
[0]; //the first value now contains the maximum drop
return [mx.max,mx.min,mx.drop];
}
console.log(checkData([100, 90, 80, 120]));
console.log(checkData([100, 90, 80, 120, 200, 100 ,90]));
console.log(checkData([200,100, 90, 80, 120, 100 ,90]));
console.log(checkData([200,120,190, 90, 80, 120, 100 ,90]));answered Nov 26 at 14:12
Me.Name
10.1k22039
add a comment |
0
0
0
To add one more to the mix..
A possible approach is to create all the drop - ranges first, than get the one with the largest drop
A version with reduce:
function checkData(data) {
let dropInd = 1,
mx = data.reduce((arr,val, ind) => {
if(ind && val > data[ind-1])dropInd++;
arr[dropInd] = arr[dropInd] || {max:val};
arr[dropInd].min = val;
arr[dropInd].drop = arr[dropInd].max - val;
return arr;
}, ) //after reduce, islands are created based on 'dropInd'
.sort((v1,v2)=>v2.drop - v1.drop) //sort by drop descending
[0]; //the first value now contains the maximum drop
return [mx.max,mx.min,mx.drop];
}
console.log(checkData([100, 90, 80, 120]));
console.log(checkData([100, 90, 80, 120, 200, 100 ,90]));
console.log(checkData([200,100, 90, 80, 120, 100 ,90]));
console.log(checkData([200,120,190, 90, 80, 120, 100 ,90]));answered Nov 26 at 14:12
Me.Name
10.1k22039
To add one more to the mix..
A possible approach is to create all the drop - ranges first, than get the one with the largest drop
A version with reduce:
function checkData(data) {
let dropInd = 1,
mx = data.reduce((arr,val, ind) => {
if(ind && val > data[ind-1])dropInd++;
arr[dropInd] = arr[dropInd] || {max:val};
arr[dropInd].min = val;
arr[dropInd].drop = arr[dropInd].max - val;
return arr;
}, ) //after reduce, islands are created based on 'dropInd'
.sort((v1,v2)=>v2.drop - v1.drop) //sort by drop descending
[0]; //the first value now contains the maximum drop
return [mx.max,mx.min,mx.drop];
}
console.log(checkData([100, 90, 80, 120]));
console.log(checkData([100, 90, 80, 120, 200, 100 ,90]));
console.log(checkData([200,100, 90, 80, 120, 100 ,90]));
console.log(checkData([200,120,190, 90, 80, 120, 100 ,90]));function checkData(data) {
let dropInd = 1,
mx = data.reduce((arr,val, ind) => {
if(ind && val > data[ind-1])dropInd++;
arr[dropInd] = arr[dropInd] || {max:val};
arr[dropInd].min = val;
arr[dropInd].drop = arr[dropInd].max - val;
return arr;
}, ) //after reduce, islands are created based on 'dropInd'
.sort((v1,v2)=>v2.drop - v1.drop) //sort by drop descending
[0]; //the first value now contains the maximum drop
return [mx.max,mx.min,mx.drop];
}
console.log(checkData([100, 90, 80, 120]));
console.log(checkData([100, 90, 80, 120, 200, 100 ,90]));
console.log(checkData([200,100, 90, 80, 120, 100 ,90]));
console.log(checkData([200,120,190, 90, 80, 120, 100 ,90]));function checkData(data) {
let dropInd = 1,
mx = data.reduce((arr,val, ind) => {
if(ind && val > data[ind-1])dropInd++;
arr[dropInd] = arr[dropInd] || {max:val};
arr[dropInd].min = val;
arr[dropInd].drop = arr[dropInd].max - val;
return arr;
}, ) //after reduce, islands are created based on 'dropInd'
.sort((v1,v2)=>v2.drop - v1.drop) //sort by drop descending
[0]; //the first value now contains the maximum drop
return [mx.max,mx.min,mx.drop];
}
console.log(checkData([100, 90, 80, 120]));
console.log(checkData([100, 90, 80, 120, 200, 100 ,90]));
console.log(checkData([200,100, 90, 80, 120, 100 ,90]));
console.log(checkData([200,120,190, 90, 80, 120, 100 ,90]));answered Nov 26 at 14:12
Me.Name
10.1k22039
edited Nov 26 at 16:17
answered Nov 26 at 14:12
Me.Name
10.1k22039
answered Nov 26 at 14:12
Me.Name
10.1k22039
answered Nov 26 at 14:12
Me.Name
10.1k22039
10.1k22039
add a comment |
add a comment |
-1
This works exactly as intended (as per your expected output), as shown in the following code snippet.
The issue must be in how you are sending your data as an array.
alert(checkData([100, 90, 80, 120]));
function checkData(data) {
let max = 0
let min = 0
let drop = 0
for (let i = 0; i < data.length; i++) {
if (max < data[i]) {
max = data[i] //?
} else {
let tempDrop = max - data[i]
drop = Math.max(tempDrop)
min = max - drop
}
}
return [max, min, drop]
}// Output : 120, 80, 20
However, if what you are looking for is a Max, a Min, and then the biggest difference, i.e. between the the max and min, which in your given example would be 40, then you can simply do this.
alert(checkData([100, 90, 80, 120]));
function checkData(data) {
let max = data.reduce(function(a, b) {
return Math.max(a, b);
});
let min = data.reduce(function(a, b) {
return Math.min(a, b);
});
let drop = max-min;
return [max, min, drop];
}// Output : 120, 80, 40
I only mention this as I don't know why you've said you expect the biggest difference to be 20, given 100 and 80, when you have 120 and 80 in the same array, meaning the biggest difference is 40.
answered Nov 26 at 12:50
cmprogram
1,083519
Thanks, but the correct return must be max = 100, min = 80 and drop = 20. Look at the graphic that I added to my post
– Pommesloch
Nov 26 at 13:11
But 100 isn't the max... If you don't want the 120 to be included, just don't include it in the array...
– cmprogram
Nov 26 at 13:13
This data comes from a shares API. That is why I need this chronological from left to right delta. The shares course is crashed here from 100 to 80. This is a -20% drop. In my attempt, I miss the chronological aspect
– Pommesloch
Nov 26 at 13:18
1
OP wants the max and min associated with the largest drop, where a drop is a sequence of descending data values. The largest drop is from 100 to 80 (indexes 0 through 2).
– Ted Hopp
Nov 26 at 13:34
add a comment |
-1
This works exactly as intended (as per your expected output), as shown in the following code snippet.
The issue must be in how you are sending your data as an array.
alert(checkData([100, 90, 80, 120]));
function checkData(data) {
let max = 0
let min = 0
let drop = 0
for (let i = 0; i < data.length; i++) {
if (max < data[i]) {
max = data[i] //?
} else {
let tempDrop = max - data[i]
drop = Math.max(tempDrop)
min = max - drop
}
}
return [max, min, drop]
}// Output : 120, 80, 20
However, if what you are looking for is a Max, a Min, and then the biggest difference, i.e. between the the max and min, which in your given example would be 40, then you can simply do this.
alert(checkData([100, 90, 80, 120]));
function checkData(data) {
let max = data.reduce(function(a, b) {
return Math.max(a, b);
});
let min = data.reduce(function(a, b) {
return Math.min(a, b);
});
let drop = max-min;
return [max, min, drop];
}// Output : 120, 80, 40
I only mention this as I don't know why you've said you expect the biggest difference to be 20, given 100 and 80, when you have 120 and 80 in the same array, meaning the biggest difference is 40.
answered Nov 26 at 12:50
cmprogram
1,083519
Thanks, but the correct return must be max = 100, min = 80 and drop = 20. Look at the graphic that I added to my post
– Pommesloch
Nov 26 at 13:11
But 100 isn't the max... If you don't want the 120 to be included, just don't include it in the array...
– cmprogram
Nov 26 at 13:13
This data comes from a shares API. That is why I need this chronological from left to right delta. The shares course is crashed here from 100 to 80. This is a -20% drop. In my attempt, I miss the chronological aspect
– Pommesloch
Nov 26 at 13:18
1
OP wants the max and min associated with the largest drop, where a drop is a sequence of descending data values. The largest drop is from 100 to 80 (indexes 0 through 2).
– Ted Hopp
Nov 26 at 13:34
add a comment |
-1
-1
-1
This works exactly as intended (as per your expected output), as shown in the following code snippet.
The issue must be in how you are sending your data as an array.
alert(checkData([100, 90, 80, 120]));
function checkData(data) {
let max = 0
let min = 0
let drop = 0
for (let i = 0; i < data.length; i++) {
if (max < data[i]) {
max = data[i] //?
} else {
let tempDrop = max - data[i]
drop = Math.max(tempDrop)
min = max - drop
}
}
return [max, min, drop]
}// Output : 120, 80, 20
However, if what you are looking for is a Max, a Min, and then the biggest difference, i.e. between the the max and min, which in your given example would be 40, then you can simply do this.
alert(checkData([100, 90, 80, 120]));
function checkData(data) {
let max = data.reduce(function(a, b) {
return Math.max(a, b);
});
let min = data.reduce(function(a, b) {
return Math.min(a, b);
});
let drop = max-min;
return [max, min, drop];
}// Output : 120, 80, 40
I only mention this as I don't know why you've said you expect the biggest difference to be 20, given 100 and 80, when you have 120 and 80 in the same array, meaning the biggest difference is 40.
answered Nov 26 at 12:50
cmprogram
1,083519
This works exactly as intended (as per your expected output), as shown in the following code snippet.
The issue must be in how you are sending your data as an array.
alert(checkData([100, 90, 80, 120]));
function checkData(data) {
let max = 0
let min = 0
let drop = 0
for (let i = 0; i < data.length; i++) {
if (max < data[i]) {
max = data[i] //?
} else {
let tempDrop = max - data[i]
drop = Math.max(tempDrop)
min = max - drop
}
}
return [max, min, drop]
}// Output : 120, 80, 20
However, if what you are looking for is a Max, a Min, and then the biggest difference, i.e. between the the max and min, which in your given example would be 40, then you can simply do this.
alert(checkData([100, 90, 80, 120]));
function checkData(data) {
let max = data.reduce(function(a, b) {
return Math.max(a, b);
});
let min = data.reduce(function(a, b) {
return Math.min(a, b);
});
let drop = max-min;
return [max, min, drop];
}// Output : 120, 80, 40
I only mention this as I don't know why you've said you expect the biggest difference to be 20, given 100 and 80, when you have 120 and 80 in the same array, meaning the biggest difference is 40.
alert(checkData([100, 90, 80, 120]));
function checkData(data) {
let max = 0
let min = 0
let drop = 0
for (let i = 0; i < data.length; i++) {
if (max < data[i]) {
max = data[i] //?
} else {
let tempDrop = max - data[i]
drop = Math.max(tempDrop)
min = max - drop
}
}
return [max, min, drop]
}alert(checkData([100, 90, 80, 120]));
function checkData(data) {
let max = 0
let min = 0
let drop = 0
for (let i = 0; i < data.length; i++) {
if (max < data[i]) {
max = data[i] //?
} else {
let tempDrop = max - data[i]
drop = Math.max(tempDrop)
min = max - drop
}
}
return [max, min, drop]
}alert(checkData([100, 90, 80, 120]));
function checkData(data) {
let max = data.reduce(function(a, b) {
return Math.max(a, b);
});
let min = data.reduce(function(a, b) {
return Math.min(a, b);
});
let drop = max-min;
return [max, min, drop];
}alert(checkData([100, 90, 80, 120]));
function checkData(data) {
let max = data.reduce(function(a, b) {
return Math.max(a, b);
});
let min = data.reduce(function(a, b) {
return Math.min(a, b);
});
let drop = max-min;
return [max, min, drop];
}answered Nov 26 at 12:50
cmprogram
1,083519
edited Nov 26 at 13:01
answered Nov 26 at 12:50
cmprogram
1,083519
answered Nov 26 at 12:50
cmprogram
1,083519
answered Nov 26 at 12:50
cmprogram
1,083519
1,083519
Thanks, but the correct return must be max = 100, min = 80 and drop = 20. Look at the graphic that I added to my post
– Pommesloch
Nov 26 at 13:11
But 100 isn't the max... If you don't want the 120 to be included, just don't include it in the array...
– cmprogram
Nov 26 at 13:13
This data comes from a shares API. That is why I need this chronological from left to right delta. The shares course is crashed here from 100 to 80. This is a -20% drop. In my attempt, I miss the chronological aspect
– Pommesloch
Nov 26 at 13:18
1
OP wants the max and min associated with the largest drop, where a drop is a sequence of descending data values. The largest drop is from 100 to 80 (indexes 0 through 2).
– Ted Hopp
Nov 26 at 13:34
add a comment |
Thanks, but the correct return must be max = 100, min = 80 and drop = 20. Look at the graphic that I added to my post
– Pommesloch
Nov 26 at 13:11
But 100 isn't the max... If you don't want the 120 to be included, just don't include it in the array...
– cmprogram
Nov 26 at 13:13
This data comes from a shares API. That is why I need this chronological from left to right delta. The shares course is crashed here from 100 to 80. This is a -20% drop. In my attempt, I miss the chronological aspect
– Pommesloch
Nov 26 at 13:18
1
OP wants the max and min associated with the largest drop, where a drop is a sequence of descending data values. The largest drop is from 100 to 80 (indexes 0 through 2).
– Ted Hopp
Nov 26 at 13:34
Thanks, but the correct return must be max = 100, min = 80 and drop = 20. Look at the graphic that I added to my post
– Pommesloch
Nov 26 at 13:11
Thanks, but the correct return must be max = 100, min = 80 and drop = 20. Look at the graphic that I added to my post
– Pommesloch
Nov 26 at 13:11
But 100 isn't the max... If you don't want the 120 to be included, just don't include it in the array...
– cmprogram
Nov 26 at 13:13
But 100 isn't the max... If you don't want the 120 to be included, just don't include it in the array...
– cmprogram
Nov 26 at 13:13
This data comes from a shares API. That is why I need this chronological from left to right delta. The shares course is crashed here from 100 to 80. This is a -20% drop. In my attempt, I miss the chronological aspect
– Pommesloch
Nov 26 at 13:18
This data comes from a shares API. That is why I need this chronological from left to right delta. The shares course is crashed here from 100 to 80. This is a -20% drop. In my attempt, I miss the chronological aspect
– Pommesloch
Nov 26 at 13:18
1
1
OP wants the max and min associated with the largest drop, where a drop is a sequence of descending data values. The largest drop is from 100 to 80 (indexes 0 through 2).
– Ted Hopp
Nov 26 at 13:34
OP wants the max and min associated with the largest drop, where a drop is a sequence of descending data values. The largest drop is from 100 to 80 (indexes 0 through 2).
– Ted Hopp
Nov 26 at 13:34
add a comment |
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1
When you say drop, do you mean difference?
– cmprogram
Nov 26 at 12:40
@cmprogram, I mean the biggest delta between max an min.
– Pommesloch
Nov 26 at 12:41
1
is this array sorted? what do you mean by saying drop?
– yossico
Nov 26 at 12:41
return [max, min, drop]What are max, min and drop?– Wais Kamal
Nov 26 at 12:42
what would be the drop output to [100,110,120,70,60,50,90,300,200]?
– yossico
Nov 26 at 12:42