Maximum Sum Circular Subarray

Problem Statement - link #

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.

Examples #

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 * 10^4
• -3 * 10^4 <= nums[i] <= 3 * 10^4

Solutions #

class Solution {
public:
    int maxSubarraySum(vector<int>& nums) {
        int curSum=nums[0], resSum=nums[0];
        for( int i=1; i<nums.size(); i++ ) {
            curSum = max( nums[i], curSum+nums[i] );
            resSum = max( resSum, curSum );
        }
        return resSum;
    }
    int maxSubarraySumCircular(vector<int>& nums) {
        int linearRes = maxSubarraySum(nums);
        if( linearRes < 0 )
            return linearRes;
        int totalSum = 0;
        for(int &n: nums){
            totalSum += n;
            n *= -1;
        }
        int circularRes = totalSum + maxSubarraySum(nums);
        return max( linearRes, circularRes );
    }
};