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918. Maximum Sum Circular Subarray

Description

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

Solutions

• Java
• C++
• Python
• Go
• TypeScript

• class Solution {
      public int maxSubarraySumCircular(int[] nums) {
          int s1 = nums[0], s2 = nums[0], f1 = nums[0], f2 = nums[0], total = nums[0];
          for (int i = 1; i < nums.length; ++i) {
              total += nums[i];
              f1 = nums[i] + Math.max(f1, 0);
              f2 = nums[i] + Math.min(f2, 0);
              s1 = Math.max(s1, f1);
              s2 = Math.min(s2, f2);
          }
          return s1 > 0 ? Math.max(s1, total - s2) : s1;
      }
  }
  

• class Solution {
  public:
      int maxSubarraySumCircular(vector<int>& nums) {
          int s1 = nums[0], s2 = nums[0], f1 = nums[0], f2 = nums[0], total = nums[0];
          for (int i = 1; i < nums.size(); ++i) {
              total += nums[i];
              f1 = nums[i] + max(f1, 0);
              f2 = nums[i] + min(f2, 0);
              s1 = max(s1, f1);
              s2 = min(s2, f2);
          }
          return s1 > 0 ? max(s1, total - s2) : s1;
      }
  };
  

• class Solution:
      def maxSubarraySumCircular(self, nums: List[int]) -> int:
          s1 = s2 = f1 = f2 = nums[0]
          for num in nums[1:]:
              f1 = num + max(f1, 0)
              f2 = num + min(f2, 0)
              s1 = max(s1, f1)
              s2 = min(s2, f2)
          return s1 if s1 <= 0 else max(s1, sum(nums) - s2)
  
  

• func maxSubarraySumCircular(nums []int) int {
  	s1, s2, f1, f2, total := nums[0], nums[0], nums[0], nums[0], nums[0]
  	for i := 1; i < len(nums); i++ {
  		total += nums[i]
  		f1 = nums[i] + max(f1, 0)
  		f2 = nums[i] + min(f2, 0)
  		s1 = max(s1, f1)
  		s2 = min(s2, f2)
  	}
  	if s1 <= 0 {
  		return s1
  	}
  	return max(s1, total-s2)
  }
  

• function maxSubarraySumCircular(nums: number[]): number {
      let pre1 = nums[0],
          pre2 = nums[0];
      let ans1 = nums[0],
          ans2 = nums[0];
      let sum = nums[0];
  
      for (let i = 1; i < nums.length; ++i) {
          let cur = nums[i];
          sum += cur;
          pre1 = Math.max(pre1 + cur, cur);
          ans1 = Math.max(pre1, ans1);
  
          pre2 = Math.min(pre2 + cur, cur);
          ans2 = Math.min(pre2, ans2);
      }
      return ans1 > 0 ? Math.max(ans1, sum - ans2) : ans1;
  }
  
  

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