Count the triplets in C

Count the triplets in C

Problem statement: Given an Array of distinct integers, count all the triplets such that the sum of two elements equals the third element.

#include <stdio.h>

int countTriplet = 0;

void triplet(int a, int b, int c)
{
    if (a + b == c)
    {
        printf("%d\t%d\t%d\n", a, b, c);
        countTriplet++;
    }
}

void combination(int arr[], int end)
{

    for (int i = 0; i < end - 2; i++)
        triplet(arr[i], arr[i + 1], arr[i + 2]);
}

void swap(int *xp, int *yp)
{
    int temp = *xp;
    *xp = *yp;
    *yp = temp;
}

void sort(int arr[], int n)
{
    int i, j;
    for (i = 0; i < n - 1; i++)
        for (j = 0; j < n - i - 1; j++)
            if (arr[j] > arr[j + 1])
                swap(&arr[j], &arr[j + 1]);
}

void printArray(int arr[], int n)
{
    for (int i = 0; i < n; i++)
    {
        printf("%d\t", arr[i]);
    }
    printf("\n*******************\n");
}

int main()
{
    int arr[] = {1, 2, 3, 5, 8, 13, 21};
    int data[3];
    int n = sizeof(arr) / sizeof(arr[0]);
    printf("Array is:");
    printArray(arr, n);
    sort(arr, n);
    combination(arr, n);
    if (!countTriplet)
        printf("No triplet in array!");
    return 0;
}

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