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// 918. Maximum Sum Circular Subarray
#include <stdio.h>
#define MAX(a, b) (((a) > (b)) ? (a) : (b))
/*
* given: circular int array 'nums' of length 'n', return the max possible
* sum of non-empty subarray of 'nums'
*
* a circular array means the end of the array connects to the beginning.
* formally, the previous element of 'nums[i]' is 'nums[(i -/+ 1) % 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
*/
int helper(int *a, int n) {
// kadane's dp algorithm
// [https://medium.com/@rsinghal757/kadanes-algorithm-dynamic-programming-how-and-why-does-it-work-3fd8849ed73d]
int currMax, sum;
sum = currMax = a[0];
for (int i = 1; i < n; i++) {
currMax = MAX(a[i], currMax + a[i]);
sum = MAX(currMax, sum);
}
return sum;
}
int maxSubarraySumCircular(int *nums, int numsSize) {
int k1, k2, total = 0;
k1 = helper(nums, numsSize);
for (int i = 0; i < numsSize; i++) {
total += nums[i];
nums[i] *= -1;
}
k2 = helper(nums, numsSize);
int ans = total + k2;
return (ans > k1 && ans != 0) ? ans : k1;
}
int main() {
int nums1[] = {1, -2, 3, -2}, nums2[] = {5, -3, 5}, nums3[] = {-3, -2, -3};
printf("%d\n", maxSubarraySumCircular(nums1, 4)); // expect: 3
printf("%d\n", maxSubarraySumCircular(nums2, 3)); // expect: 3
printf("%d\n", maxSubarraySumCircular(nums3, 3)); // expect: 3
}
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