// 918. Maximum Sum Circular Subarray #include #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 }