Skip to content
Related Articles

Related Articles

Improve Article

Check whether a number can be represented as sum of K distinct positive integers

  • Difficulty Level : Medium
  • Last Updated : 08 Apr, 2021

Given two integers N and K, the task is to check whether N can be represented as sum of K distinct positive integers.

Examples: 

Input: N = 12, K = 4 
Output: Yes 
N = 1 + 2 + 4 + 5 = 12 (12 as sum of 4 distinct integers)

Input: N = 8, K = 4 
Output: No 
 

Approach: Consider the series 1 + 2 + 3 + … + K which has exactly K distinct integers with minimum possible sum i.e. Sum = (K * (K – 1)) / 2. Now, if N < Sum then it is not possible to represent N as the sum of K distinct positive integers but if N ≥ Sum then any integer say X ≥ 0 can be added to Sum to generate the sum equal to N i.e. 1 + 2 + 3 + … + (K – 1) + (K + X) ensuring that there are exactly K distinct positive integers.



Below is the implementation of the above approach:  

C++




// C++ implementation of the approach
#include <iostream>
using namespace std;
 
// Function that returns true if n
// can be represented as the sum of
// exactly k distinct positive integers
bool solve(int n, int k)
{
    // If n can be represented as
    // 1 + 2 + 3 + ... + (k - 1) + (k + x)
    if (n >= (k * (k + 1)) / 2) {
        return true;
    }
 
    return false;
}
 
// Driver code
int main()
{
    int n = 12, k = 4;
 
    if (solve(n, k))
        cout << "Yes";
    else
        cout << "No";
 
    return 0;
}

Java




// Java implementation of the approach
import java.io.*;
 
class GFG {
 
    // Function that returns true if n
    // can be represented as the sum of
    // exactly k distinct positive integers
    static boolean solve(int n, int k)
    {
        // If n can be represented as
        // 1 + 2 + 3 + ... + (k - 1) + (k + x)
        if (n >= (k * (k + 1)) / 2) {
            return true;
        }
 
        return false;
    }
 
    // Driver code
    public static void main(String[] args)
    {
        int n = 12, k = 4;
 
        if (solve(n, k))
            System.out.println("Yes");
        else
            System.out.println("No");
    }
}
 
// This code is contributed by anuj_67..

Python3




# Python 3 implementation of the approach
 
# Function that returns true if n
# can be represented as the sum of
# exactly k distinct positive integers
def solve(n,k):
    # If n can be represented as
    # 1 + 2 + 3 + ... + (k - 1) + (k + x)
    if (n >= (k * (k + 1)) // 2):
        return True
 
    return False
 
# Driver code
if __name__ == '__main__':
    n = 12
    k = 4
 
    if (solve(n, k)):
        print("Yes")
    else:
        print("No")
 
# This code is contributed by
# Surendra_Gangwar

C#




// C# implementation of the approach
using System;
 
class GFG
{
    // Function that returns true if n
    // can be represented as the sum of
    // exactly k distinct positive integers
    static bool solve(int n, int k)
    {
        // If n can be represented as
        // 1 + 2 + 3 + ... + (k - 1) + (k + x)
        if (n >= (k * (k + 1)) / 2) {
            return true;
        }
 
        return false;
    }
 
    // Driver code
    static public void Main ()
    {
        int n = 12, k = 4;
 
        if (solve(n, k))
            Console.WriteLine("Yes");
        else
            Console.WriteLine("No");
    }
}
 
// This code is contributed by ajit.

PHP




<?php
 
// PHP implementation of the approach
 
// Function that returns true if n
// can be represented as the sum of
// exactly k distinct positive integers
function solve($n, $k)
{
    // If n can be represented as
    // 1 + 2 + 3 + ... + (k - 1) + (k + x)
    if ($n >= ($k * ($k + 1)) / 2) {
        return true;
    }
 
    return false;
}
 
// Driver code
 
$n = 12;
$k = 4;
 
if (solve($n, $k))
    echo  "Yes";
else
    echo  "No";
 
// This code is contributed by ihritik
 
?>

Javascript




<script>
 
// Javascript implementation of the approach
 
// Function that returns true if n
// can be represented as the sum of
// exactly k distinct positive integers
function solve(n, k)
{
     
    // If n can be represented as
    // 1 + 2 + 3 + ... + (k - 1) + (k + x)
    if (n >= (k * (k + 1)) / 2)
    {
        return true;
    }
    return false;
}
 
// Driver code
var n = 12, k = 4;
 
if (solve(n, k))
    document.write("Yes");
else
    document.write("No");
 
// This code is contributed by todaysgaurav
 
</script>
Output: 
Yes

 

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 :