README_EN.md

918. Maximum Sum Circular Subarray

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Description

Given a circular array C of integers represented by A, find the maximum possible sum of a non-empty subarray of C.

Here, a circular array means the end of the array connects to the beginning of the array.  (Formally, C[i] = A[i] when 0 <= i < A.length, and C[i+A.length] = C[i] when i >= 0.)

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

 

Example 1:

Input: [1,-2,3,-2]

Output: 3

Explanation: Subarray [3] has maximum sum 3

Example 2:

Input: [5,-3,5]

Output: 10

Explanation: Subarray [5,5] has maximum sum 5 + 5 = 10

Example 3:

Input: [3,-1,2,-1]

Output: 4

Explanation: Subarray [2,-1,3] has maximum sum 2 + (-1) + 3 = 4

Example 4:

Input: [3,-2,2,-3]

Output: 3

Explanation: Subarray [3] and [3,-2,2] both have maximum sum 3

Example 5:

Input: [-2,-3,-1]

Output: -1

Explanation: Subarray [-1] has maximum sum -1

 

Note:

1. -30000 <= A[i] <= 30000
2. 1 <= A.length <= 30000

Solutions

Python3

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)

Java

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;
    }
}

TypeScript

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;
};

C++

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;
    }
};

Go

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)
}

func max(a, b int) int {
	if a > b {
		return a
	}
	return b
}

func min(a, b int) int {
	if a < b {
		return a
	}
	return b
}

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