Formatted question description: https://leetcode.ca/all/918.html

918. Maximum Sum Circular Subarray (Medium)

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

Related Topics:
Array

Solution 1.

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

// Time: O(N)
// Space: O(N)
class Solution {
public:
    int maxSubarraySumCircular(vector<int>& A) {
        int N = A.size(), sum = 0, ans = INT_MIN;
        vector<int> p(2 * N + 1);
        for (int i = 0; i < 2 * N; ++i) {
            p[i + 1] = (sum += A[i % N]);
        }
        deque<int> q;
        for (int i = 0; i < 2 * N + 1; ++i) {
            if (i >= N && q.front() == i - N - 1) q.pop_front();
            if (q.size()) ans = max(ans, p[i] - p[q.front()]);
            while (q.size() && p[q.back()] >= p[i]) q.pop_back();
            q.push_back(i);
        }
        return ans;
    }
};

Java

class Solution {
    public int maxSubarraySumCircular(int[] A) {
        if (A == null || A.length == 0)
            return 0;
        int sum = A[0];
        int maxNum = A[0];
        int newMaxSum = A[0], maxSum = A[0];
        int newMinSum = A[0], minSum = A[0];
        int length = A.length;
        for (int i = 1; i < length; i++) {
            sum += A[i];
            maxNum = Math.max(maxNum, A[i]);
            newMaxSum = Math.max(newMaxSum + A[i], A[i]);
            maxSum = Math.max(maxSum, newMaxSum);
            newMinSum = Math.min(newMinSum + A[i], A[i]);
            minSum = Math.min(minSum, newMinSum);
        }
        if (maxNum <= 0)
            return maxNum;
        else
            return Math.max(maxSum, sum - minSum);
    }
}

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