Maximum Subarray

It's obvious an DP problem.

Let's define:

$dp[i] :=$ the maximum sum of subarray in the first $i$ elements.

 

From the base point, the optimal subarray may contain element $A[i]$ or not.

1. optimal subarray without $A[i]$, so $dp[i] = dp[i-1]$.
2. optimal subarray ending with $A[i]$.

There're some subtle issues about the second situation. In order to get the second case, we need to know the optimal solution of array ending with $A[i-1]$.

 

So, we have to define another dp table:

$End[i] :=$ the maximum sum of subarray ending with $A[i]$.

Obviously, the $End$ array satisfies:

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1	End[i] = max(End[i-1] + A[i], A[i]);

 In fact, we can simplify as

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1	End[i] = (End[i-1] > 0) ? End[i-1] + A[i] : A[i];

 

As both $dp[i]$ and $End[i]$ only relate the previous step, we can use a variable instead of an array to store the result.

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1 2	end = (end > 0) ? end + A[i] : A[i]; dp = max(dp, end);

 

Of course, at the initial stage, we consider only one element $A[0]$, thus, we can initialize $dp, end$ as: 

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1	dp = end = A[0];

 

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The complete code is:

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class Solution { public:     int maxSubArray(int A[], int n) {         int dp = A[0];         int end = dp;                   for(int i = 1; i < n; ++i){             end = end > 0 ? end + A[i] : A[i];             dp = dp > end ? dp : end;         }                   return dp;     } };