tanishq2507's blog

How to prove this?
By tanishq2507, history, 17 months ago, In English

Maximum absolute sum of a subarray =Max prefix sum — Min prefix sum(prefix sum includes zero).

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17 months ago, # |
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Well this statement isn't exactly true, let me show you why.
take the following array {1,2,-10}
the maximum subarray sum is 1+2=3
but according to your formula its max prefix — min prefix = 3-(-7)=11
which obviously isn't true.
What you're looking for is the maximum value of a prefix sums minus the min prefix sum that ends before it
.

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    17 months ago, # ^ |
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    It's the maximum absolute sum of a subarray, so for your array, that would be abs(sum({-10})) = 10, and indeed 3-(-7) = 10.

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17 months ago, # |
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Notation: $$$n$$$ length of array, $$$a_i$$$ ($$$0 \le i < n$$$) $$$i$$$th element, $$$s_{lr} = \sum_{l\le j<r}a_j$$$ ($$$0 \le l \le r \le n$$$) sum of elements from $$$l$$$ to $$$r$$$, $$$p_i = s_{0i}$$$ prefix sum

Part 1: maximum absolute sum of a subarray $$$\ge$$$ max prefix sum $$$-$$$ min prefix sum

Let $$$p_M$$$ and $$$p_m$$$ be the maximum and minimum prefix sum, respectively. WLOG assume $$$M \le m$$$. Then $$$\max(|s_{lr}|) \ge |s_{Mm}| = |p_m - p_M| = p_M - p_m$$$.

Part 2: maximum absolute sum of a subarray $$$\le$$$ max prefix sum $$$-$$$ min prefix sum

Let $$$L$$$, $$$R$$$ ($$$L\le R$$$) be indices such that $$$\max(|s_{lr}|) = |s_{LR}|$$$. WLOG assume $$$p_L \le p_R$$$. Then, $$$\max(|s_{lr}|) = |s_{LR}| = |p_R - p_L| = p_R - p_L \le \max(p_i) - \min(p_i)$$$.

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