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] <= 300001 <= A.length <= 30000
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)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;
}
}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;
};class Solution {
public:
int maxSubarraySumCircular(vector<int>& nums) {
int s1 = nums[0], s2 = nums[0], f1 = nums[0], f2 = nums[0], total = nums[0];
for (int i = 1; i < nums.size(); ++i) {
total += nums[i];
f1 = nums[i] + max(f1, 0);
f2 = nums[i] + min(f2, 0);
s1 = max(s1, f1);
s2 = min(s2, f2);
}
return s1 > 0 ? max(s1, total - s2) : s1;
}
};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
}