Short and simple C++ - O(n) time, O(1) space

  • 0

    Solution relies on knowing the highest possible sum when looking at the subarray to the right (iterating in reverse).

    int maxSubArray(int A[], int n) {
        int res = A[0], rightSum = 0;
        for(int i = n-1; i >= 0; --i){
            rightSum = max(A[i], A[i] + rightSum);
            res = max(res, rightSum);
        return res;

  • -1

    If A is a list that all members are negative, then your solution is wrong

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