// 2221. Find Triangular Sum of an Array #include "leetcode.h" /* * you are given a 0-indexed integer array 'nums', where 'nums[i]' is a digit * between 0 and 9 inclusive. the triangular sum of 'nums' is the value of the * only element present in 'nums' after the following process concludes: let * 'nums' comprise of 'n' elements. if 'n == 1', end the process. otherwise, * create a new 0-indexed integer array 'new_nums' of length 'n - 1'. for each * index 'i' where '0 <= i < n - 1', assign the value of 'new_nums[i]' as * '(nums[i] + nums[i + 1]) % 10'. replace the array 'nums' with 'new_nums', and * repeat the entire process starting again. */ int triangularSum(int *nums, int numsSize) { for (int i = numsSize - 1; i >= 1; i--) for (int j = 0; j < i; j++) nums[j] = (nums[j] + nums[j + 1]) % 10; return nums[0]; } int main() { int n1[] = {1, 2, 3, 4, 5}, n2[] = {5}; printf("%d\n", triangularSum(n1, ARRAY_SIZE(n1))); // expect: 8 printf("%d\n", triangularSum(n2, ARRAY_SIZE(n2))); // expect: 5 }