Count of Triplets With Sum Less than Given Value

Problem Statement

We have given an array containing N number of elements. In the given array, Count the number of triplets with a sum less than the given value.

Example

Input

a[] = {1, 2, 3, 4, 5, 6, 7, 8}
Sum = 10

Output

7
Possible triplets are : (1,2,3), (1,2,4), (1,2,5), (1,2,6), (1,3,4), (1,3,5), (2,3,4)

Input

a[]  = {3, 7, 9, 1, 2, 5, 11, 4}
Sum = 10

Output

6
Possible triplets are : (3,1,2), (3,1,5), (3,1,4), (3,2,4), (1,2,5), (1,2,4)

Approach 1

Algorithm

1. Run 3 nested loops, selecting all triplets possible.

2. Initialize result = 0.

3. If the Sum of elements is less than Sum, increment count.

4. Print results as a count of triplets satisfying the condition.

Implementation

C++ Program for Count of Triplets With Sum Less than Given Value

#include <bits/stdc++.h>
using namespace std;
 
//Main function
int main()
{
    int array[] = {3, 7, 9, 1, 2, 5, 11, 4};
    int N = sizeof(array)/sizeof(array[0]);
    int sum = 10;
    int result = 0;
    //First for loop for selecting first element
    //Second loop for second element
    //Third loop for third element.
    //check for triplets satisfing the condition, and increment result
    for (int i = 0; i < N-2; i++)
    {
       for (int j = i+1; j < N-1; j++)
       {
           for (int k = j+1; k < N; k++)
           {
               if(array[i]+array[j]+array[k] < sum)
                   {
                      result++;
                   }
           }
       }
    }
    cout<<"Number of Triplets found = ";
    cout<<result;
    return 0;
}

Number of Triplets found = 6

import static java.lang.Math.pow;
import java.util.Scanner;
class sum
{
    public static void main(String[] args)  
    { 
        Scanner sr = new Scanner(System.in);
        int n = sr.nextInt();
        int sum = sr.nextInt();
        int a[] = new int[n];
        for(int i=0;i<n;i++)
        {
            a[i] = sr.nextInt();
        }
        //First for loop for selecting first element
        //Second loop for second element
        //Third loop for third element.
        //check for triplets satisfing the condition, and increment result
        int result=0;
        for (int i = 0; i < n-2; i++)
        {
           for (int j = i+1; j < n-1; j++)
           {
               for (int k = j+1; k < n; k++)
               {
                   if(a[i]+a[j]+a[k] < sum)
                       {
                          result++;
                       }
               }
           }
        }
        System.out.print("Number of Triplets found = ");
        System.out.println(result);
    }
}

5 6
1 2 3 2 1

Number of Triplets found = 5

Approach 2

Algorithm

1. Sort the given array, initialize result = 0

2. Iterate from i to N -2 (N is the size of the array), and take array[i] and the first element of the triplet.

3. Initialize the other two elements as corner elements. array[i+1] and array[N-1]

4. move j and k until they meet to do,

1. If the sum is greater than the given sum, then move the pointer of the last element. (array[N-1]).
2. Else if the sum is less than the given sum, then there can k -j possible tird elements satisfying, so add (k-j) to result. And move array[i+1] pointer.

Step 5: Print the result after ending the loop.

Implementation

C++ Program for Count of Triplets With Sum Less than Given Value

#include <bits/stdc++.h>
using namespace std;
 
// Main function
int main()
{
    int array[] ={1, 2,3,5,6, 3, 2, 5, 7};//input array
    int N = sizeof(array)/sizeof(array[0]);//size of the input array(N)
    int Sum = 9;//input Sum
    int result = 0;//initilizing count of triplets
    sort(array,array+N);//sorting the input array
    //selecting first element
    for(int i = 0; i < N - 2; i++)
    {
        int j=i+1,k=N-1;//second and third elements as last two elements
        while(j<k)
        {
            // Sum of triplet is greater than or equalto given sum, move to find smaller values
            if(array[i] + array[j] + array[k] >= Sum)
            {
                k--;
            }
            // Sum of triplet is less than given sum
            else
            {
                //for current i and j there can be k-j elements, move j pointer
                result += (k - j);
                j++;
            }
        }
    }
    cout<<"Number of Triplets found = ";
    cout<<result;
}

Number of Triplets found = 14

import static java.lang.Math.pow;
import java.util.Arrays;
import java.util.Scanner;
class sum
{
    public static void main(String[] args)  
    { 
        Scanner sr = new Scanner(System.in);
        int n = sr.nextInt();
        int sum = sr.nextInt();
        int a[] = new int[n];
        for(int i=0;i<n;i++)
        {
            a[i] = sr.nextInt();
        }
        //First for loop for selecting first element
        //Second loop for second element
        //Third loop for third element.
        //check for triplets satisfing the condition, and increment result
        int result=0;
        Arrays. sort(a);//sorting the input array
        //selecting first element
        for(int i = 0; i < n - 2; i++)
        {
            int j=i+1,k=n-1;//second and third elements as last two elements
            while(j<k)
            {
                // Sum of triplet is greater than or equalto given sum, move to find smaller values
                if(a[i] + a[j] + a[k] >= sum)
                {
                    k--;
                }
                // Sum of triplet is less than given sum
                else
                {
                    //for current i and j there can be k-j elements, move j pointer
                    result += (k - j);
                    j++;
                }
            }
        }
        System.out.print("Number of Triplets found = ");
        System.out.println(result);
    }
}

9 9
1 2 3 5 6 3 2 5 7

Number of Triplets found = 14