# Triplet Sum in Array

QUESTION

Given an array A[] of n numbers and another number x, determine whether or not there exist three elements in A[] whose sum is exactly x.\n\nExpected time complexity is O(n^2).\n\nInput:\n\nThe first line of input contains an integer T denoting the number of test cases.\nThe first line of each test case is n and x, n is the size of array.\nThe second line of each test case contains n integers representing array elements C[i].\n\nOutput:\n\nPrint 1 if there exist three elements in A whose sum is exactly x, else 0.\n\nConstraints:\n\n1 T 100\n1 N 200\n1 A[i] 1000.

“TESTCASE_1”: “2\n6 13\n1 4 45 6 10 8\n5 10\n1 2 4 3 6\n###—###SEPERATOR—###—\n1\n1”, “TESTCASE_2”: “2\n6 13\n11 14 15 16 3 5\n8 1\n4 12 19 13 6\n###—###SEPERATOR—###—\n0\n0”, “TESTCASE_3”: “0\n###—###SEPERATOR—###—\n0”, “TESTCASE_4”: “0\n###—###SEPERATOR—###—\n0”, “TESTCASE_5”: “0\n###—###SEPERATOR—###—\n0

``````# include <stdio.h>
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
int n,sum,i,j,k;
scanf("%d",&n);
scanf("%d",&sum);
int arr[n];
for(i=0;i<n;i++)
scanf("%d",&arr[i]);
int flag=0;
for(i=0;i<n;i++)
{
for(j=0;j<n;j++)
{
if(i!=j)
{
for(k=0;k<n;k++)
{
if(k!=i && k!=j)
{
if(arr[i]+arr[j]+arr[k]==sum)
flag=1;
}
}
}
}
}
printf("%d",flag);
printf("\n");
}
return 0;
}``````