918. Maximum Sum Circular Subarray

Medium (Trung bình) C++ Python 🔗 Xem trên LeetCode

📋 Đề Bài

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

💻 Lời Giải

C++ 0918-maximum-sum-circular-subarray.cpp
class Solution {
public:
    int maxSubarraySumCircular(vector<int>& nums) {
        const int oo = 1e5;
        int minSub = oo;
        int ansMin = oo;
        int maxSub = -oo;
        int ansMax = -oo;
        int s = 0;
        for (int num : nums) {
            minSub = min(num, minSub + num);
            ansMin = min(ansMin, minSub);
            maxSub = max(num, maxSub + num);
            ansMax = max(ansMax, maxSub);
            s += num;
        }
        if (ansMax < 0) {
            return ansMax;
        }
        return max(s - ansMin, ansMax);
    }
};
Python 0918-maximum-sum-circular-subarray.py
class Solution:
    def maxSubarraySumCircular(self, nums: List[int]) -> int:
        maxSub = -10**5
        maxSum = -10**5
        minSub = 10**5
        minSum = 10**5
        total = 0
        
        for num in nums:
            maxSum = max(num, maxSum + num)
            maxSub = max(maxSub, maxSum)
            minSum = min(num, minSum + num)
            minSub = min(minSub, minSum)
            total += num
            
        return max(maxSub, total - minSub) if maxSub > 0 else maxSub