Skip to content
Related Articles

Related Articles

Improve Article

Print triplets with sum less than k

  • Difficulty Level : Easy
  • Last Updated : 24 May, 2021

Given an array of distinct integers and a sum value. Print all triplets with sum smaller than given sum value. Expected Time Complexity is O(n2).

Examples

Input : arr[] = {-2, 0, 1, 3}
        sum = 2.
Output : (-2, 0, 1)
         (-2, 0, 3) 
Explanation :  The two triplets have sum less than 2.

Input : arr[] = {5, 1, 3, 4, 7}
        sum = 12.
Output : (1, 3, 4)
         (1, 3, 5)
         (1, 3, 7) 
         (1, 4, 5)
               

A Simple Solution is to run three loops to consider all triplets one by one. For every triplet, compare the sums and print current triplet if its sum is smaller than given sum. 

C++




// A Simple C++ program to count triplets with sum
// smaller than a given value
#include<bits/stdc++.h>
using namespace std;
 
int printTriplets(int arr[], int n, int sum)
{
    // Fix the first element as A[i]
    for (int i = 0; i < n-2; i++)
    {
       // Fix the second element as A[j]
       for (int j = i+1; j < n-1; j++)
       {
           // Now look for the third number
           for (int k = j+1; k < n; k++)
               if (arr[i] + arr[j] + arr[k] < sum)
                  cout << arr[i] << ", " << arr[j]
                       << ", " << arr[k] << endl;
       }
    }
}
 
// Driver program
int main()
{
    int arr[] = {5, 1, 3, 4, 7};
    int n = sizeof arr / sizeof arr[0];
    int sum = 12;
    printTriplets(arr, n, sum);
    return 0;
}

Java




// A Simple Java program to
// count triplets with sum
// smaller than a given value
import java.io.*;
 
class GFG
{
static int printTriplets(int arr[],
                         int n, int sum)
{
    // Fix the first
    // element as A[i]
    for (int i = 0; i < n - 2; i++)
    {
         
    // Fix the second
    // element as A[j]
    for (int j = i + 1;
             j < n - 1; j++)
    {
        // Now look for
        // the third number
        for (int k = j + 1; k < n; k++)
            if (arr[i] + arr[j] + arr[k] < sum)
                System.out.println(arr[i] + ", " +
                                   arr[j] + ", " +
                                   arr[k]);
    }
    }
    return 0;
}
 
// Driver Code
public static void main (String[] args)
{
    int arr[] = {5, 1, 3, 4, 7};
    int n = arr.length;
    int sum = 12;
    printTriplets(arr, n, sum);
}
}
 
// This code is contributed
// by anuj_67.

Python3




# A Simple python 3 program to count
# triplets with sum smaller than a
# given value
 
def printTriplets(arr, n, sum):
     
    # Fix the first element as A[i]
    for i in range(0, n - 2, 1):
         
        # Fix the second element as A[j]
        for j in range(i + 1, n - 1, 1):
             
            # Now look for the third number
            for k in range(j + 1, n, 1):
                if (arr[i] + arr[j] + arr[k] < sum):
                    print(arr[i], ",", arr[j], ",", arr[k])
# Driver Code
if __name__ == '__main__':
    arr =[5, 1, 3, 4, 7]
    n = len(arr)
    sum = 12
    printTriplets(arr, n, sum)
     
# This code is contributed by
# Sahil_Shelangia

C#




// A Simple C# program to
// count triplets with sum
// smaller than a given value
using System;
class GFG
{
static int printTriplets(int[] arr,
                        int n, int sum)
{
    // Fix the first
    // element as A[i]
    for (int i = 0; i < n - 2; i++)
    {
         
    // Fix the second
    // element as A[j]
    for (int j = i + 1;
            j < n - 1; j++)
    {
        // Now look for
        // the third number
        for (int k = j + 1; k < n; k++)
            if (arr[i] + arr[j] + arr[k] < sum)
                Console.WriteLine(arr[i] + ", " +
                                arr[j] + ", " +
                                arr[k]);
    }
    }
    return 0;
}
 
// Driver Code
public static void Main ()
{
    int[] arr = {5, 1, 3, 4, 7};
    int n = arr.Length;
    int sum = 12;
    printTriplets(arr, n, sum);
}
}
 
// This code is contributed
// by Mukul Singh.

PHP




<?php
// A Simple PHP program to count triplets
// with sum smaller than a given value
 
function printTriplets(&$arr, $n, $sum)
{
    // Fix the first element as A[i]
    for ($i = 0; $i < $n - 2; $i++)
    {
    // Fix the second element as A[j]
    for ($j = $i + 1; $j < $n - 1; $j++)
    {
        // Now look for the third number
        for ($k = $j + 1; $k < $n; $k++)
            if ($arr[$i] + $arr[$j] +
                           $arr[$k] < $sum)
            {
                echo($arr[$i]);
                echo(", " );
                echo($arr[$j]);
                echo(", ");
                echo($arr[$k]);
                echo("\n");
            }
    }
    }
}
 
// Driver Code
$arr = array(5, 1, 3, 4, 7);
$n = sizeof($arr);
$sum = 12;
printTriplets($arr, $n, $sum);
 
// This code is contributed
// by Shivi_Aggarwal
?>

Javascript




<script>
 
// A Simple JavaScript program to
// count triplets with sum
// smaller than a given value
 
function printTriplets(arr, n, sum)
{
    // Fix the first element as A[i]
    for (let i = 0; i < n-2; i++)
    {
    // Fix the second element as A[j]
    for (let j = i+1; j < n-1; j++)
    {
        // Now look for the third number
        for (let k = j+1; k < n; k++)
            if (arr[i] + arr[j] + arr[k] < sum)
                document.write(arr[i] + ", " + arr[j]
                    + ", " + arr[k] + "<br>");
    }
    }
}
 
// Driver program
    let arr = [5, 1, 3, 4, 7];
    let n = arr.length;
    let sum = 12;
    printTriplets(arr, n, sum);
 
 
// This code is contributed by Surbhi Tyagi.
 
</script>
Output: 
5, 1, 3
5, 1, 4
1, 3, 4
1, 3, 7

 

Time complexity of above solution is O(n3).



An Efficient Solution can print triplets in O(n2) by sorting the array first, and then using method 1 of this post in a loop.

1) Sort the input array in increasing order.
2) Initialize result as 0.
3) Run a loop from i = 0 to n-2.  An iteration of this loop 
   finds all triplets with arr[i] as first element.
     a) Initialize other two elements as corner elements
        of subarray
        arr[i+1..n-1], i.e., j = i+1 and k = n-1
     b) Move j and k toward each other until they meet,
        i.e., while (j = sum), then do k--

        // Else for current i and j, there are (k-j) possible  
        // third elements that satisfy the constraint.
        (ii) Else print elements from j to k

Below is the implementation of above idea. 

C++




// C++ program to print triplets with sum smaller
// than a given value
#include <bits/stdc++.h>
using namespace std;
 
int printTriplets(int arr[], int n, int sum)
{
    // Sort input array
    sort(arr, arr + n);
 
    // Every iteration of loop counts triplet with
    // first element as arr[i].
    for (int i = 0; i < n - 2; i++) {
 
        // Initialize other two elements as corner
        // elements of subarray arr[j+1..k]
        int j = i + 1, k = n - 1;
 
        // Use Meet in the Middle concept
        while (j < k) {
 
            // If sum of current triplet is more or equal,
            // move right corner to look for smaller values
            if (arr[i] + arr[j] + arr[k] >= sum)
                k--;
 
            // Else move left corner
            else {
 
                // This is important. For current i and j,
                // there are total k-j third elements.
                for (int x = j + 1; x <= k; x++)
                    cout << arr[i] << ", " << arr[j]
                         << ", " << arr[x] << endl;
                j++;
            }
        }
    }
}
 
// Driver program
int main()
{
    int arr[] = { 5, 1, 3, 4, 7 };
    int n = sizeof arr / sizeof arr[0];
    int sum = 12;
    printTriplets(arr, n, sum);
    return 0;
}

Java




// Java program to print
// triplets with sum smaller
// than a given value
import java.util.*;
import java.lang.*;
import java.io.*;
 
class GFG
{
static void printTriplets(int arr[],
                          int n, int sum)
{
    // Sort input array
    Arrays.sort(arr);
 
    // Every iteration of loop
    // counts triplet with
    // first element as arr[i].
    for (int i = 0; i < n - 2; i++)
    {
 
        // Initialize other two elements
        //  as corner elements of subarray
        // arr[j+1..k]
        int j = i + 1, k = n - 1;
 
        // Use Meet in the
        // Middle concept
        while (j < k)
        {
 
            // If sum of current triplet
            // is more or equal, move right
            // corner to look for smaller values
            if (arr[i] + arr[j] + arr[k] >= sum)
                k--;
 
            // Else move left corner
            else
            {
 
                // This is important. For
                // current i and j, there
                // are total k-j third elements.
                for (int x = j + 1; x <= k; x++)
                    System.out.println(arr[i] + ", " +
                                       arr[j] + ", " +
                                       arr[x]);
                j++;
            }
        }
    }
}
 
// Driver Code
public static void main(String args[])
{
    int arr[] = { 5, 1, 3, 4, 7 };
    int n = arr.length;
    int sum = 12;
    printTriplets(arr, n, sum);
}
}
 
// This code is contributed
// by Subhadeep

Python3




# Python3 program to print
# triplets with sum smaller
# than a given value
def printTriplets(arr, n, sum):
     
    # Sort input array
    arr.sort()
  
    # Every iteration of loop
    # counts triplet with
    # first element as arr[i].
    for i in range(n - 2):
 
        # Initialize other two elements
        # as corner elements of subarray
        # arr[j+1..k]
        (j, k) = (i + 1, n - 1)
  
        # Use Meet in the
        # Middle concept
        while (j < k):
         
            # If sum of current triplet
            # is more or equal, move right
            # corner to look for smaller values
            if (arr[i] + arr[j] + arr[k] >= sum):
                k -= 1
  
            # Else move left corner
            else:
  
                # This is important. For
                # current i and j, there
                # are total k-j third elements.
                for x in range(j + 1, k + 1):
                    print(str(arr[i]) + ", " +
                          str(arr[j]) + ", " +
                          str(arr[x]))
        
                j += 1
 
# Driver code
if __name__=="__main__":
     
    arr = [ 5, 1, 3, 4, 7 ]
    n = len(arr)
    sum = 12
     
    printTriplets(arr, n, sum);
 
# This code is contributed by rutvik_56

C#




// C# program to print
// triplets with sum smaller
// than a given value
using System;
 
class GFG
{
static void printTriplets(int[] arr,
                        int n, int sum)
{
    // Sort input array
    Array.Sort(arr);
 
    // Every iteration of loop
    // counts triplet with
    // first element as arr[i].
    for (int i = 0; i < n - 2; i++)
    {
 
        // Initialize other two elements
        // as corner elements of subarray
        // arr[j+1..k]
        int j = i + 1, k = n - 1;
 
        // Use Meet in the
        // Middle concept
        while (j < k)
        {
 
            // If sum of current triplet
            // is more or equal, move right
            // corner to look for smaller values
            if (arr[i] + arr[j] + arr[k] >= sum)
                k--;
 
            // Else move left corner
            else
            {
 
                // This is important. For
                // current i and j, there
                // are total k-j third elements.
                for (int x = j + 1; x <= k; x++)
                    Console.WriteLine(arr[i] + ", " +
                                    arr[j] + ", " +
                                    arr[x]);
                j++;
            }
        }
    }
}
 
// Driver Code
public static void Main()
{
    int[] arr = { 5, 1, 3, 4, 7 };
    int n = arr.Length;
    int sum = 12;
    printTriplets(arr, n, sum);
}
}
 
// This code is contributed
// by Akanksha Rai

PHP




<?php
// PHP program to print triplets with
// sum smaller than a given value
 
function printTriplets($arr, $n, $sum)
{
    // Sort input array
    sort($arr, 0);
 
    // Every iteration of loop counts
    // triplet with first element as arr[i].
    for ($i = 0; $i < $n - 2; $i++)
    {
 
        // Initialize other two elements as corner
        // elements of subarray arr[j+1..k]
        $j = $i + 1; $k = $n - 1;
 
        // Use Meet in the Middle concept
        while ($j < $k)
        {
 
            // If sum of current triplet is more
            // or equal, move right corner to
            // look for smaller values
            if ($arr[$i] + $arr[$j] +
                $arr[$k] >= $sum)
                $k--;
 
            // Else move left corner
            else
            {
 
                // This is important. For current i and j,
                // there are total k-j third elements.
                for ($x = $j + 1; $x <= $k; $x++)
                    echo $arr[$i] . ", " . $arr[$j] .
                              ", " . $arr[$x] . "\n";
                $j++;
            }
        }
    }
}
 
// Driver Code
$arr = array(5, 1, 3, 4, 7);
$n = sizeof($arr);
$sum = 12;
printTriplets($arr, $n, $sum);
 
// This code is contributed
// by Akanksha Rai
?>

Javascript




<script>
 
    // JavaScript program to print
    // triplets with sum smaller
    // than a given value
     
    function printTriplets(arr, n, sum)
    {
        // Sort input array
        arr.sort(function(a, b){return a - b});
 
        // Every iteration of loop
        // counts triplet with
        // first element as arr[i].
        for (let i = 0; i < n - 2; i++)
        {
 
            // Initialize other two elements
            // as corner elements of subarray
            // arr[j+1..k]
            let j = i + 1, k = n - 1;
 
            // Use Meet in the
            // Middle concept
            while (j < k)
            {
 
                // If sum of current triplet
                // is more or equal, move right
                // corner to look for smaller values
                if (arr[i] + arr[j] + arr[k] >= sum)
                    k--;
 
                // Else move left corner
                else
                {
 
                    // This is important. For
                    // current i and j, there
                    // are total k-j third elements.
                    for (let x = j + 1; x <= k; x++)
                        document.write(arr[i] + ", " +
                                        arr[j] + ", " +
                                        arr[x] + "</br>");
                    j++;
                }
            }
        }
    }
     
    let arr = [ 5, 1, 3, 4, 7 ];
    let n = arr.length;
    let sum = 12;
    printTriplets(arr, n, sum);
     
</script>
Output: 
1, 3, 4
1, 3, 5
1, 3, 7
1, 4, 5

 

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.




My Personal Notes arrow_drop_up
Recommended Articles
Page :