Description Link to heading

918. Maximum Sum Circular Subarray (Medium)

Solution Link to heading

We define $dp[i]$ to represent the maximum sum of subarrays that end at $nums[i]$. Then, we can discuss two cases:

  1. The subarray is a continuous segment, i.e., $tail \geq head$.
  2. The subarray is divided into two segments, i.e., $tail < head$.

In each case, we can update the $dp[i]$ value accordingly to find the maximum sum of subarrays that end at each element $nums[i]$.

Code Link to heading 

class Solution {
  public:
    int maxSubarraySumCircular(vector<int> &nums) {
        int n = nums.size();
        if (n == 1) {
            return nums[0];
        }
        vector<int> dp(n, INT_MIN);
        int sum = accumulate(nums.begin(), nums.end(), 0);
        vector<int> normal(n, INT_MIN);
        dp[0] = nums[0];
        for (int i = 1; i < n; ++i) {
            dp[i] = max(nums[i], nums[i] + dp[i - 1]);
        }
        vector<int> min_sum(n, 0);
        min_sum[n - 1] = nums[n - 1];
        for (int i = n - 2; i >= 0; --i) {
            min_sum[i] = min(min_sum[i + 1] + nums[i], nums[i]);
        }
        int res = INT_MIN;
        for (int i = 0; i < n - 1; ++i) {
            dp[i] = max(dp[i], sum - min_sum[i + 1]);
            res = max(res, dp[i]);
        }
        return max(res, dp[n - 1]);
    }
};