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

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.

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>

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

?>

Output

Yes

Time Complexity: O(1)

Auxiliary Space: O(1)

Approach 2: Dynamic Programming:

Here’s the dynamic programming (DP) approach to solve the same problem:

• Define a 2D array dp of size (n+1) x (k+1).
• Initialize dp[i][j] to false if either i is 0 or j is 0, and to true if j is 1.
• For i from 1 to n and j from 2 to k, do the following steps:
• a. If i >= j, then set dp[i][j] to dp[i-1][j] || dp[i-j][j-1].
• b. If i < j, then set dp[i][j] to dp[i-1][j].
• If dp[n][k] is true, return true, else return false.
• Here’s the C++ code for the above DP approach:

C++

#include <iostream>
#include <vector>
using namespace std;

bool canSumToDistinctIntegers(int n, int k) {
    vector<vector<bool>> dp(n+1, vector<bool>(k+1, false));
    for (int i = 0; i <= n; i++) {
        dp[i][0] = false;
    }
    for (int j = 0; j <= k; j++) {
        dp[0][j] = false;
    }
    for (int j = 1; j <= k; j++) {
        dp[1][j] = true;
    }
    for (int i = 1; i <= n; i++) {
        for (int j = 2; j <= k; j++) {
            if (i >= j) {
                dp[i][j] = dp[i-1][j] || dp[i-j][j-1];
            }
            else {
                dp[i][j] = dp[i-1][j];
            }
        }
    }
    return dp[n][k];
}

int main() {
    int n = 12, k = 4;
    if (canSumToDistinctIntegers(n, k)) {
        cout << "Yes" << endl;
    }
    else {
        cout << "No" << endl;
    }
    return 0;
}

Java

public class GFG {
    public static boolean canSumToDistinctIntegers(int n, int k) {
        // Create a 2D DP array to store the results of subproblems
        boolean[][] dp = new boolean[n + 1][k + 1];

        // Initialize base cases
        for (int i = 0; i <= n; i++) {
            dp[i][0] = false;
        }

        for (int j = 0; j <= k; j++) {
            dp[0][j] = false;
        }

        for (int j = 1; j <= k; j++) {
            dp[1][j] = true;
        }

        // Fill the DP array bottom-up
        for (int i = 1; i <= n; i++) {
            for (int j = 2; j <= k; j++) {
                if (i >= j) {
                    dp[i][j] = dp[i - 1][j] || dp[i - j][j - 1];
                } else {
                    dp[i][j] = dp[i - 1][j];
                }
            }
        }

        return dp[n][k];
    }

    public static void main(String[] args) {
        int n = 12, k = 4;

        // Check if it is possible to form a sum of k distinct integers
        if (canSumToDistinctIntegers(n, k)) {
            System.out.println("Yes");
        } else {
            System.out.println("No");
        }
    }
}

Python3

def can_sum_to_distinct_integers(n, k):
    # Create a 2D DP table with n+1 rows and k+1 columns, initialized with False values
    dp = [[False] * (k+1) for _ in range(n+1)]

    # Base case: If k is 0, it is always possible to form an empty set, so set dp[i][0] to True for all i.
    for i in range(n+1):
        dp[i][0] = False

    # Base case: If n is 0, it is not possible to form a set with any sum, so set dp[0][j] to False for all j.
    for j in range(k+1):
        dp[0][j] = False

    # Base case: If k is 1, it is always possible to form a set with just one element equal to n,
    # so set dp[1][j] to True for all j.
    for j in range(1, k+1):
        dp[1][j] = True

    # Fill the DP table using bottom-up approach
    for i in range(1, n+1):
        for j in range(2, k+1):
            if i >= j:
                # If the current number i is greater than or equal to j,
                # we have two options: include i or exclude i to form the sum j.
                # dp[i-1][j] represents excluding i, and dp[i-j][j-1] represents including i.
                dp[i][j] = dp[i-1][j] or dp[i-j][j-1]
            else:
                # If the current number i is less than j, we can only exclude i to form the sum j.
                dp[i][j] = dp[i-1][j]

    # The final result is stored in dp[n][k], which represents whether it is possible to form a set of k distinct integers
    # whose sum is n.
    return dp[n][k]

# Driver code
n = 12
k = 4

if can_sum_to_distinct_integers(n, k):
    print("YES")
else:
    print("No")

C#

using System;

public class GFG
{
    public static bool CanSumToDistinctIntegers(int n, int k)
    {
        // Create a 2D DP array to store the results of subproblems
        bool[,] dp = new bool[n + 1, k + 1];

        // Initialize base cases
        for (int i = 0; i <= n; i++)
        {
            dp[i, 0] = false;
        }

        for (int j = 0; j <= k; j++)
        {
            dp[0, j] = false;
        }

        for (int j = 1; j <= k; j++)
        {
            dp[1, j] = true;
        }

        // Fill the DP array bottom-up
        for (int i = 1; i <= n; i++)
        {
            for (int j = 2; j <= k; j++)
            {
                if (i >= j)
                {
                    dp[i, j] = dp[i - 1, j] || dp[i - j, j - 1];
                }
                else
                {
                    dp[i, j] = dp[i - 1, j];
                }
            }
        }

        return dp[n, k];
    }

    public static void Main()
    {
        int n = 12, k = 4;

        // Check if it is possible to form a sum of k distinct integers
        if (CanSumToDistinctIntegers(n, k))
        {
            Console.WriteLine("Yes");
        }
        else
        {
            Console.WriteLine("No");
        }

        // To pause the console before exiting
        Console.ReadLine();
    }
}

Javascript

// This function determines if it is possible to represent a given integer n
// as the sum of k distinct positive integers
function canSumToDistinctIntegers(n, k) {

    // Create a 2D array to store the DP table, with n+1 rows and k+1 columns
    let dp = new Array(n+1);
    for (let i = 0; i <= n; i++) {
        dp[i] = new Array(k+1).fill(false);
    }

    // Set base cases
    for (let i = 0; i <= n; i++) {
        dp[i][0] = false;
    }
    for (let j = 0; j <= k; j++) {
        dp[0][j] = false;
    }
    for (let j = 1; j <= k; j++) {
        dp[1][j] = true;
    }

    // Fill in the DP table using a nested loop
    for (let i = 1; i <= n; i++) {
        for (let j = 2; j <= k; j++) {
            if (i >= j) {
                dp[i][j] = dp[i-1][j] || dp[i-j][j-1];
            }
            else {
                dp[i][j] = dp[i-1][j];
            }
        }
    }

    // Return the result, which is stored in the last cell of the DP table
    return dp[n][k];
}

// Test the function with some sample input
let n = 12, k = 4;
if (canSumToDistinctIntegers(n, k)) {
    console.log("Yes");
}
else {
    console.log("No");
}

Output: 

Yes

Time Complexity:  O(nk), where n is the maximum possible value of n (the input number), and k is the maximum possible value of k

Auxiliary Space: O(nk) because we need to store the intermediate results in a 2D array.