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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:
-30000 <= A[i] <= 30000
1 <= A.length <= 30000
Related Topics:
Array
Solution 1.
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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); } } ############ class Solution { public int maxSubarraySumCircular(int[] nums) { int s1 = nums[0], s2 = nums[0], f1 = nums[0], f2 = nums[0], total = nums[0]; for (int i = 1; i < nums.length; ++i) { total += nums[i]; f1 = nums[i] + Math.max(f1, 0); f2 = nums[i] + Math.min(f2, 0); s1 = Math.max(s1, f1); s2 = Math.min(s2, f2); } return s1 > 0 ? Math.max(s1, total - s2) : s1; } }
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// 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; } };
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class Solution: def maxSubarraySumCircular(self, nums: List[int]) -> int: s1 = s2 = f1 = f2 = nums[0] for num in nums[1:]: f1 = num + max(f1, 0) f2 = num + min(f2, 0) s1 = max(s1, f1) s2 = min(s2, f2) return s1 if s1 <= 0 else max(s1, sum(nums) - s2)
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func maxSubarraySumCircular(nums []int) int { s1, s2, f1, f2, total := nums[0], nums[0], nums[0], nums[0], nums[0] for i := 1; i < len(nums); i++ { total += nums[i] f1 = nums[i] + max(f1, 0) f2 = nums[i] + min(f2, 0) s1 = max(s1, f1) s2 = min(s2, f2) } if s1 <= 0 { return s1 } return max(s1, total-s2) } func max(a, b int) int { if a > b { return a } return b } func min(a, b int) int { if a < b { return a } return b }
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function maxSubarraySumCircular(nums: number[]): number { let pre1 = nums[0], pre2 = nums[0]; let ans1 = nums[0], ans2 = nums[0]; let sum = nums[0]; for (let i = 1; i < nums.length; ++i) { let cur = nums[i]; sum += cur; pre1 = Math.max(pre1 + cur, cur); ans1 = Math.max(pre1, ans1); pre2 = Math.min(pre2 + cur, cur); ans2 = Math.min(pre2, ans2); } return ans1 > 0 ? Math.max(ans1, sum - ans2) : ans1; }