Triplet sum

 Triplet sum

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You have been given a random integer array/list(ARR) and a number X. Find and return the triplet(s) in the array/list which sum to X.

Note :

Given array/list can contain duplicate elements.

Input format :

The first line contains an Integer 't' which denotes the number of test cases or queries to be run. Then the test cases follow.

First line of each test case or query contains an integer 'N' representing the size of the first array/list.

Second line contains 'N' single space separated integers representing the elements in the array/list.

Third line contains an integer 'X'.

Output format :

For each test case, print the total number of triplets present in the array/list.

Output for every test case will be printed in a separate line.

Constraints :

1 <= t <= 10^2
0 <= N <= 10^3
0 <= X <= 10^9

Time Limit: 1 sec

Sample Input 1:

1
7
1 2 3 4 5 6 7 
12

Sample Output 1:

5

Sample Input 2:

2
7
1 2 3 4 5 6 7 
19
9
2 -5 8 -6 0 5 10 11 -3
10

Sample Output 2:

0
5

 Explanation for Input 2:

Since there doesn't exist any triplet with sum equal to 19 for the first query, we print 0.

For the second query, we have 5 triplets in total that sum up to 10. They are, (2, 8, 0), (2, 11, -3), (-5, 5, 10), (8, 5, -3) and (-6, 5, 11)

// int tripletSum(int *arr, int n, int num)
// {
//  //Write your code here
//     int cnt=0;
//     sort(arr,arr+n);
//     for(int i=0;i<n;i++){
//         for(int j=i+1;j<n;j++){
//             for(int k=j+1;k<n;k++){
//                 if(arr[i]+arr[j]+arr[k]==num) cnt++;
//             }
//         }
//     }
//     return cnt;
    
// }
#include<bits/stdc++.h>
using namespace std;
int pairSum(int *arr, int startIndex, int endIndex, int num) 
{ 
    int numPair = 0; 
    while (startIndex < endIndex) 
    { 
        if (arr[startIndex] + arr[endIndex] < num) 
        { 
            startIndex++; 
            
        } 
        else if (arr[startIndex] + arr[endIndex] > num) 
        { 
            endIndex--; 
            
        } 
        else 
        { 
            int elementAtStart = arr[startIndex];
            int elementAtEnd = arr[endIndex]; 
            if (elementAtStart == elementAtEnd) 
            { 
                int totalElementsFromStartToEnd = (endIndex - startIndex) + 1; 
                numPair += (totalElementsFromStartToEnd * (totalElementsFromStartToEnd - 1) / 2); 
                return numPair; 
                
            }
            int tempStartIndex = startIndex + 1; 
            int tempEndIndex = endIndex - 1; 
            while (tempStartIndex <= tempEndIndex && arr[tempStartIndex] == elementAtStart) 
            { 
                tempStartIndex += 1; 
                
            } 
            while (tempEndIndex >= tempStartIndex && arr[tempEndIndex] == elementAtEnd) 
            { 
                tempEndIndex -= 1; 
            } 
            int totalElementsFromStart = (tempStartIndex - startIndex); 
            int totalElementsFromEnd = (endIndex - tempEndIndex); 
            numPair += (totalElementsFromStart * totalElementsFromEnd); 
            startIndex = tempStartIndex; endIndex = tempEndIndex; 
        }
    } 
    return numPair; 
} 
int tripletSum(int *arr, int n, int num) 
{ 
    sort(arr, arr + n); 
    int numTriplets = 0; 
    for (int i = 0; i < n; i++) 
    {
        int pairSumFor = num - arr[i]; 
        int numPairs = pairSum(arr, (i + 1), (n - 1), pairSumFor); 
        numTriplets += numPairs;
    }
    return numTriplets;
}

#include <iostream>
#include <algorithm>
using namespace std;
#include "solution.h"
int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    int t;
    cin >> t;
    while (t--)
    {
        int size;
        int x;
        cin >> size;
        
        int *input = new int[size];
        for (int i = 0; i < size; i++)
        {
            cin >> input[i];
        }
        cin >> x;
        cout << tripletSum(input, size, x) << endl;
        delete[] input;
    }
    return 0;
}