Find the smallest element greater than current element
$begingroup$
my question is that given a list of integers (lets say A), for each element A[i], find the smallest element A[j] which could satisfy A[i] < A[j] and i < j. Return -1 if there is no such element.
For example, given [4,2,1,9,3], return [9,3,3,-1,-1]
I have come up with a brute force solution which costs $O(n^2)$, but I'm wondering if there could be a more efficient solution.
algorithms arrays
edited Dec 20 '18 at 2:41
xskxzr
3,93121033
asked Dec 19 '18 at 22:01
AustinAustin
182
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add a comment |
$begingroup$
my question is that given a list of integers (lets say A), for each element A[i], find the smallest element A[j] which could satisfy A[i] < A[j] and i < j. Return -1 if there is no such element.
For example, given [4,2,1,9,3], return [9,3,3,-1,-1]
I have come up with a brute force solution which costs $O(n^2)$, but I'm wondering if there could be a more efficient solution.
algorithms arrays
edited Dec 20 '18 at 2:41
xskxzr
3,93121033
asked Dec 19 '18 at 22:01
AustinAustin
182
$endgroup$
$begingroup$
How difficult would this problem be if you had an array of integers, sorted in ascending order?
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– gnasher729
Dec 19 '18 at 22:05
$begingroup$
I believe that this could be done in as little as O(n)! I'll elaborate when I have access to a computer.
$endgroup$
– xuq01
Dec 19 '18 at 23:20
$begingroup$
I think my solution is wrong; the minimum time complexity must be O(n log n) I think.
$endgroup$
– xuq01
Dec 20 '18 at 3:11
add a comment |
$begingroup$
my question is that given a list of integers (lets say A), for each element A[i], find the smallest element A[j] which could satisfy A[i] < A[j] and i < j. Return -1 if there is no such element.
For example, given [4,2,1,9,3], return [9,3,3,-1,-1]
I have come up with a brute force solution which costs $O(n^2)$, but I'm wondering if there could be a more efficient solution.
algorithms arrays
edited Dec 20 '18 at 2:41
xskxzr
3,93121033
asked Dec 19 '18 at 22:01
AustinAustin
182
$endgroup$
my question is that given a list of integers (lets say A), for each element A[i], find the smallest element A[j] which could satisfy A[i] < A[j] and i < j. Return -1 if there is no such element.
For example, given [4,2,1,9,3], return [9,3,3,-1,-1]
I have come up with a brute force solution which costs $O(n^2)$, but I'm wondering if there could be a more efficient solution.
algorithms arrays
algorithms arrays
edited Dec 20 '18 at 2:41
xskxzr
3,93121033
asked Dec 19 '18 at 22:01
AustinAustin
182
edited Dec 20 '18 at 2:41
xskxzr
3,93121033
asked Dec 19 '18 at 22:01
AustinAustin
182
edited Dec 20 '18 at 2:41
xskxzr
3,93121033
edited Dec 20 '18 at 2:41
xskxzr
3,93121033
edited Dec 20 '18 at 2:41
xskxzr
3,93121033
3,93121033
asked Dec 19 '18 at 22:01
AustinAustin
182
asked Dec 19 '18 at 22:01
AustinAustin
182
asked Dec 19 '18 at 22:01
AustinAustin
182
182
$begingroup$
How difficult would this problem be if you had an array of integers, sorted in ascending order?
$endgroup$
– gnasher729
Dec 19 '18 at 22:05
$begingroup$
I believe that this could be done in as little as O(n)! I'll elaborate when I have access to a computer.
$endgroup$
– xuq01
Dec 19 '18 at 23:20
$begingroup$
I think my solution is wrong; the minimum time complexity must be O(n log n) I think.
$endgroup$
– xuq01
Dec 20 '18 at 3:11
add a comment |
$begingroup$
How difficult would this problem be if you had an array of integers, sorted in ascending order?
$endgroup$
– gnasher729
Dec 19 '18 at 22:05
$begingroup$
I believe that this could be done in as little as O(n)! I'll elaborate when I have access to a computer.
$endgroup$
– xuq01
Dec 19 '18 at 23:20
$begingroup$
I think my solution is wrong; the minimum time complexity must be O(n log n) I think.
$endgroup$
– xuq01
Dec 20 '18 at 3:11
$begingroup$
How difficult would this problem be if you had an array of integers, sorted in ascending order?
$endgroup$
– gnasher729
Dec 19 '18 at 22:05
$begingroup$
How difficult would this problem be if you had an array of integers, sorted in ascending order?
$endgroup$
– gnasher729
Dec 19 '18 at 22:05
$begingroup$
I believe that this could be done in as little as O(n)! I'll elaborate when I have access to a computer.
$endgroup$
– xuq01
Dec 19 '18 at 23:20
$begingroup$
I believe that this could be done in as little as O(n)! I'll elaborate when I have access to a computer.
$endgroup$
– xuq01
Dec 19 '18 at 23:20
$begingroup$
I think my solution is wrong; the minimum time complexity must be O(n log n) I think.
$endgroup$
– xuq01
Dec 20 '18 at 3:11
$begingroup$
I think my solution is wrong; the minimum time complexity must be O(n log n) I think.
$endgroup$
– xuq01
Dec 20 '18 at 3:11
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
A better solution would be using a balanced binary search tree.
You can process the elements from the right to left and for each element you find the vertex in the tree with the smallest key greater than this element (a search query on the tree suffice) and then you add the element to the tree.
Total complexity is $O(nlog(n))$ since we are iterating over the elements once and having one query and one insert operations in each iteration.
answered Dec 19 '18 at 22:10
narek Bojikiannarek Bojikian
44317
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add a comment |
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
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active
oldest
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active
oldest
votes
$begingroup$
A better solution would be using a balanced binary search tree.
You can process the elements from the right to left and for each element you find the vertex in the tree with the smallest key greater than this element (a search query on the tree suffice) and then you add the element to the tree.
Total complexity is $O(nlog(n))$ since we are iterating over the elements once and having one query and one insert operations in each iteration.
answered Dec 19 '18 at 22:10
narek Bojikiannarek Bojikian
44317
$endgroup$
add a comment |
$begingroup$
A better solution would be using a balanced binary search tree.
You can process the elements from the right to left and for each element you find the vertex in the tree with the smallest key greater than this element (a search query on the tree suffice) and then you add the element to the tree.
Total complexity is $O(nlog(n))$ since we are iterating over the elements once and having one query and one insert operations in each iteration.
answered Dec 19 '18 at 22:10
narek Bojikiannarek Bojikian
44317
$endgroup$
add a comment |
$begingroup$
A better solution would be using a balanced binary search tree.
You can process the elements from the right to left and for each element you find the vertex in the tree with the smallest key greater than this element (a search query on the tree suffice) and then you add the element to the tree.
Total complexity is $O(nlog(n))$ since we are iterating over the elements once and having one query and one insert operations in each iteration.
answered Dec 19 '18 at 22:10
narek Bojikiannarek Bojikian
44317
$endgroup$
A better solution would be using a balanced binary search tree.
You can process the elements from the right to left and for each element you find the vertex in the tree with the smallest key greater than this element (a search query on the tree suffice) and then you add the element to the tree.
Total complexity is $O(nlog(n))$ since we are iterating over the elements once and having one query and one insert operations in each iteration.
answered Dec 19 '18 at 22:10
narek Bojikiannarek Bojikian
44317
edited Dec 19 '18 at 22:59
answered Dec 19 '18 at 22:10
narek Bojikiannarek Bojikian
44317
answered Dec 19 '18 at 22:10
narek Bojikiannarek Bojikian
44317
answered Dec 19 '18 at 22:10
narek Bojikiannarek Bojikian
44317
44317
add a comment |
add a comment |
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$begingroup$
How difficult would this problem be if you had an array of integers, sorted in ascending order?
$endgroup$
– gnasher729
Dec 19 '18 at 22:05
$begingroup$
I believe that this could be done in as little as O(n)! I'll elaborate when I have access to a computer.
$endgroup$
– xuq01
Dec 19 '18 at 23:20
$begingroup$
I think my solution is wrong; the minimum time complexity must be O(n log n) I think.
$endgroup$
– xuq01
Dec 20 '18 at 3:11