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+// 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
+}