---
description: https://leetcode.com/problems/maximum-sum-circular-subarray/
---

# 918. Maximum Sum Circular Subarray

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`

## Thoughts

给定数组看作首尾相连的循环数组，问循环数组内最大子数组和是多少。

![](https://2251957238-files.gitbook.io/~/files/v0/b/gitbook-legacy-files/o/assets%2F-LcOoIFG2VRxGaJDySQN%2F-LcOoJHd7TzQh5w1SKcJ%2F-LcOothcLaRVoB5acBEF%2F918_1.png?generation=1555211464380933\&alt=media)

![](https://2251957238-files.gitbook.io/~/files/v0/b/gitbook-legacy-files/o/assets%2F-LcOoIFG2VRxGaJDySQN%2F-LcOoJHd7TzQh5w1SKcJ%2F-LcOothefdTQwytLRNTX%2F918_2.png?generation=1555211465202637\&alt=media)

1. 如果不是环，subarray sum标准解法就是遍历每个元素算出当前sum, 拿它减去之前的min\_sum得出的sum就是max\_sum候选, 遍历时不断更新max\_sum候选，最后剩下的就是最大的。
2. 成环后最直接的想法就是把它看成原数组x2. 这时max\_sum可能是如图二中那样拼接而成。这时我们与其找max\_sum, 不如找无环情况下的min\_sum, 最后max\_sum = total\_sum - min\_sum。找min\_sum的流程和找max\_sum是一致的。
3. 最后比较无环时的max\_sum和有环时total - min\_sum哪个大。 但如果全是负的时候，min\_sum会变成整个数组，导致total - min\_sum = 0, 因此全是负的时候，应当选最小的那个负数，即无环时的max\_sum。

## Code

```python
class Solution:
    def maxSubarraySumCircular(self, A: List[int]) -> int:
        N = len(A)
        dp_mx, dp_mi, s = [-math.inf] * (N + 1), [math.inf] * (N + 1), 0
        for i, a in enumerate(A):
            s += a
            dp_mx[i + 1] = max(dp_mx[i], 0) + a
            dp_mi[i + 1] = min(dp_mi[i], 0) + a
        if dp_mi[-1] == s: return max(dp_mx[1:])
        return max(max(dp_mx[1:]), s - min(dp_mi[1:]) )
        
```

```cpp
class Solution {
public:
    int maxSubarraySumCircular(vector<int>& A) {
        int max_res = INT_MIN, min_sum = 0, sum = 0;
        for (int a : A) {
            sum += a;
            max_res = max(max_res, sum - min_sum);
            min_sum = min(min_sum, sum);
        }

        int min_res = INT_MAX, max_sum = 0;
        sum = 0;
        for (int a : A) {
            sum += a;
            min_res = min(min_res, sum - max_sum);
            max_sum = max(max_sum, sum);
        }
        if (min_res == sum) return max_res;

        return max(max_res, sum - min_res);
    }
};
```

## Analysis

时间复杂度O(N)
