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918. Maximum Sum Circular Subarray (Medium)

发布时间 2023-07-25 12:21:54作者: zwyyy456

Description

918. Maximum Sum Circular Subarray (Medium)

Given a circular integer array nums of length n, return the maximum possible sum of a non-empty subarray of nums.

A circular array means the end of the array connects to the beginning of the array. Formally, the next element of nums[i] is nums[(i + 1) % n] and the previous element of nums[i] is nums[(i - 1 + n) % n].

A subarray may only include each element of the fixed buffer nums at most once. Formally, for a subarray nums[i], nums[i + 1], ..., nums[j], there does not exist i <= k1, k2 <= j with k1 % n == k2 % n.

 

Example 1:

Input: nums = [1,-2,3,-2]
Output: 3
Explanation: Subarray [3] has maximum sum 3.

Example 2:

Input: nums = [5,-3,5]
Output: 10
Explanation: Subarray [5,5] has maximum sum 5 + 5 = 10.

Example 3:

Input: nums = [-3,-2,-3]
Output: -2
Explanation: Subarray [-2] has maximum sum -2.

 

Constraints:

  • n == nums.length
  • 1 <= n <= 3 * 104
  • -3 * 104 <= nums[i] <= 3 * 104

Solution

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

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]);
    }
};