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.)
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| 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:
- -30000 <= A[i] <= 30000
- 1 <= A.length <= 30000
Dev notes:
Solution:
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| class Solution {
public int maxSubarraySumCircular(int[] A) {
int N = A.length;
int ans = A[0];
int curr = A[0];
for (int i = 1; i < N; i++) {
curr = Math.max(curr, 0) + A[i];
ans = Math.max(ans, curr);
}
int[] rightsums = new int[N];
rightsums[N - 1] = A[N - 1];
for (int i = N - 2; i >= 0; i--) {
rightsums[i] = rightsums[i + 1] + A[i];
}
int[] maxright = new int[N];
maxright[N-1] = A[N-1];
for (int i = N - 2; i >= 0; i--) {
maxright[i] = Math.max(maxright[i + 1], rightsums[i]);
}
int leftsum = 0;
for (int i = 0; i < N - 2; i++) {
leftsum +=A[i];
ans = Math.max(ans, leftsum + maxright[i + 2]);
}
return ans;
}
}
|