From 3c005c0254993983579e453c39eee51504ab7fa8 Mon Sep 17 00:00:00 2001 From: Jack Sangdahl Date: Mon, 29 Sep 2025 20:20:34 -0600 Subject: 2221. find triangular sum of an array --- find_triangular_sum_arr.c | 26 ++++++++++++++++++++++++++ 1 file changed, 26 insertions(+) create mode 100644 find_triangular_sum_arr.c (limited to 'find_triangular_sum_arr.c') diff --git a/find_triangular_sum_arr.c b/find_triangular_sum_arr.c new file mode 100644 index 0000000..0033f7c --- /dev/null +++ b/find_triangular_sum_arr.c @@ -0,0 +1,26 @@ +// 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 +} -- cgit v1.2.3