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Maximum Subarray Sum with One Deletion
Bookmark
Brute Force
Forward-Backward
Single-Pass DP
Input
Classic
Delete the Dip
All Positive
All Negative
Custom
arr
=
[1, -2, 0, 3]
0
1
2
3
1
-2
0
3
0
1
2
3
1
-2
0
3
subarraySum =
-1
- arr[0] =
-2
best =
1
0
1
2
3
1
-2
0
3
j
i
[0..1] sum
(-1)
- arr[0] →
(-2)
subarraySum =
-1
- arr[0] =
-2
best =
1
0
1
2
3
1
-2
0
3
j
i
[0..2] sum
(-1)
- arr[0] →
(-2)
subarraySum =
2
best =
2
0
1
2
3
1
-2
0
3
j
i
[0..3] sum
2
new best
subarraySum =
2
- arr[2] =
2
best =
4
0
1
2
3
1
-2
0
3
j
i
[0..3] sum
2
- arr[2] →
2
subarraySum =
-2
best =
4
0
1
2
3
1
-2
0
3
j
i
[1..2] sum
(-2)
subarraySum =
1
best =
4
0
1
2
3
1
-2
0
3
j
i
[1..3] sum
1
subarraySum =
1
- arr[3] =
-2
best =
4
0
1
2
3
1
-2
0
3
j
i
[1..3] sum
1
- arr[3] →
(-2)
subarraySum =
3
- arr[2] =
3
best =
4
0
1
2
3
1
-2
0
3
j
i
[2..3] sum
3
- arr[2] →
3
subarraySum =
-
best =
4
0
1
2
3
1
-2
0
3
Max sum (one deletion) =
4
algo
master
.
io
Step:
For every subarray, score it as-is, then score it again with each of its elements removed.
0 / 27
Input
Classic
Delete the Dip
All Positive
All Negative
Custom
arr
=
[1, -2, 0, 3]
0 / 27
algo
master
.
io
Step:
For every subarray, score it as-is, then score it again with each of its elements removed.
