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Medium
Queue
Array
Divide and Conquer
Dynamic Programming
Monotonic Queue

918. Maximum Sum Circular Subarray

中文文档

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

Solution 1: Maintain Prefix Maximum

The maximum sum of a circular subarray can be divided into two cases:

  • Case 1: The subarray with the maximum sum does not include the circular part, which is the ordinary maximum subarray sum;
  • Case 2: The subarray with the maximum sum includes the circular part, which can be transformed into: the total sum of the array minus the minimum subarray sum.

Therefore, we maintain the following variables:

  • The minimum prefix sum $pmi$, initially $0$;
  • The maximum prefix sum $pmx$, initially $-\infty$;
  • The prefix sum $s$, initially $0$;
  • The minimum subarray sum $smi$, initially $\infty$;
  • The answer $ans$, initially $-\infty$.

Next, we only need to traverse the array $nums$. For the current element $x$ we are traversing, we perform the following update operations:

  • Update the prefix sum $s = s + x$;
  • Update the answer $ans = \max(ans, s - pmi)$, which is the answer for Case 1 (the prefix sum $s$ minus the minimum prefix sum $pmi$ can give the maximum subarray sum);
  • Update $smi = \min(smi, s - pmx)$, which is the minimum subarray sum for Case 2;
  • Update $pmi = \min(pmi, s)$, which is the minimum prefix sum;
  • Update $pmx = \max(pmx, s)$, which is the maximum prefix sum.

After the traversal, we return the maximum value of $ans$ and $s - smi$ as the answer.

The time complexity is $O(n)$, where $n$ is the length of the array. The space complexity is $O(1)$.

Python3

class Solution:
    def maxSubarraySumCircular(self, nums: List[int]) -> int:
        pmi, pmx = 0, -inf
        ans, s, smi = -inf, 0, inf
        for x in nums:
            s += x
            ans = max(ans, s - pmi)
            smi = min(smi, s - pmx)
            pmi = min(pmi, s)
            pmx = max(pmx, s)
        return max(ans, s - smi)

Java

class Solution {
    public int maxSubarraySumCircular(int[] nums) {
        final int inf = 1 << 30;
        int pmi = 0, pmx = -inf;
        int ans = -inf, s = 0, smi = inf;
        for (int x : nums) {
            s += x;
            ans = Math.max(ans, s - pmi);
            smi = Math.min(smi, s - pmx);
            pmi = Math.min(pmi, s);
            pmx = Math.max(pmx, s);
        }
        return Math.max(ans, s - smi);
    }
}

C++

class Solution {
public:
    int maxSubarraySumCircular(vector<int>& nums) {
        const int inf = 1 << 30;
        int pmi = 0, pmx = -inf;
        int ans = -inf, s = 0, smi = inf;
        for (int x : nums) {
            s += x;
            ans = max(ans, s - pmi);
            smi = min(smi, s - pmx);
            pmi = min(pmi, s);
            pmx = max(pmx, s);
        }
        return max(ans, s - smi);
    }
};

Go

func maxSubarraySumCircular(nums []int) int {
	const inf = 1 << 30
	pmi, pmx := 0, -inf
	ans, s, smi := -inf, 0, inf
	for _, x := range nums {
		s += x
		ans = max(ans, s-pmi)
		smi = min(smi, s-pmx)
		pmi = min(pmi, s)
		pmx = max(pmx, s)
	}
	return max(ans, s-smi)
}

TypeScript

function maxSubarraySumCircular(nums: number[]): number {
    let [pmi, pmx] = [0, -Infinity];
    let [ans, s, smi] = [-Infinity, 0, Infinity];
    for (const x of nums) {
        s += x;
        ans = Math.max(ans, s - pmi);
        smi = Math.min(smi, s - pmx);
        pmi = Math.min(pmi, s);
        pmx = Math.max(pmx, s);
    }
    return Math.max(ans, s - smi);
}