Maximum Sum Circular Subarray Problem
Description
LeetCode Problem 918.
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:
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Input: nums = [1,-2,3,-2]
Output: 3
Explanation: Subarray [3] has maximum sum 3
Example 2:
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Input: nums = [5,-3,5]
Output: 10
Explanation: Subarray [5,5] has maximum sum 5 + 5 = 10
Example 3:
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Input: nums = [3,-1,2,-1]
Output: 4
Explanation: Subarray [2,-1,3] has maximum sum 2 + (-1) + 3 = 4
Example 4:
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Input: nums = [3,-2,2,-3]
Output: 3
Explanation: Subarray [3] and [3,-2,2] both have maximum sum 3
Example 5:
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Input: nums = [-2,-3,-1]
Output: -1
Explanation: Subarray [-1] has maximum sum -1
Constraints:
- n == nums.length
- 1 <= n <= 3 * 10^4
- -3 * 10^4 <= nums[i] <= 3 * 10^4
Sample C++ Code
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class Solution {
public:
int maxSubarraySumCircular(vector<int>& nums) {
int total = 0, minSum = nums[0], maxSum = nums[0], curMax = 0, curMin = 0;
for (auto &e : nums) {
total += e;
curMax = max(e, curMax+e);
maxSum = max(maxSum, curMax);
curMin = min(e, curMin+e);
minSum = min(minSum, curMin);
}
return maxSum < 0 ? maxSum : max(maxSum, total-minSum);
}
};