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

ANSWER

# 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;
}
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