Data Modelling & AI Data Structure & Algorithm

Number of distinct ways to represent a number as sum of K unique primes

23 June 2025

0

Given an integer N, and an integer K, the task is to count the number of distinct ways to represent the number N as a sum of K unique primes.
Note: Distinct means, let N = 7 and K = 2, then the only way can be {2,5}, because {5,2} is same as {2,5}. So only 1 way.

Examples:

Input: N = 10, K = 2
Output: 1
Explanation:
The only way is {3, 7} or {7, 3}

Input: N = 100, K = 5
Output: 55

Approach: The problem can be solved using Dynamic Programming and Sieve of Eratosthenes.

Let dp[i][j][sum] be our 3D DP array, which stores the number of distinct ways to form a sum using j number of primes where the last index of prime selected is i in the prime vector.
The prime numbers can be efficiently computed using Sieve of Eratosthenes. So, we can get a check of prime in O(1) time.
Recurrence:

We can either include this current prime to our sum, or we can exclude it.
dp[i][j][sum] = solve(i+1, j+1, sum+prime[i]) + solve(i+1, j, sum)

Below is the implementation of the above approach :

C++

// C++ program to count the Number 
// of distinct ways to represent 
// a number as K different primes 
#include <bits/stdc++.h> 
using namespace std; 
 
// Prime vector 
vector<int> prime; 
 
// Sieve array of prime 
bool isprime[1000]; 
 
// DP array 
int dp[200][20][1000]; 
 
void sieve() 
{ 
    // Initialise all numbers 
    // as prime 
    memset(isprime, true, 
        sizeof(isprime)); 
 
    // Sieve of Eratosthenes. 
    for (int i = 2; i * i <= 1000; 
        i++) 
    { 
        if (isprime[i]) 
        { 
            for (int j = i * i; 
                j <= 1000; j += i) 
            { 
                isprime[j] = false; 
            } 
        } 
    } 
    // Push all the primes into 
    // prime vector 
    for (int i = 2; i <= 1000; i++) 
    { 
        if (isprime[i]) 
        { 
            prime.push_back(i); 
        } 
    } 
} 
 
// Function to get the number of 
// distinct ways to get sum 
// as K different primes 
int CountWays(int i, int j, int sum, 
        int n, int k) 
{ 
 
    // If index went out of prime 
    // array size or the sum became 
    // larger than n return 0 
    if (i > prime.size() || sum > n) 
    { 
        return 0; 
    } 
 
    // If sum becomes equal to n and 
    // j becomes exactly equal to k. 
    // Return 1, else if j is still 
    // not equal to k, return 0 
    if (sum == n) { 
        if (j == k) { 
            return 1; 
        } 
        return 0; 
    } 
 
    // If sum!=n and still j as 
    // exceeded, return 0 
    if (j == k) 
        return 0; 
 
    // If that state is already 
    // calculated, return directly 
    // the ans 
    if (dp[i][j][sum]) 
        return dp[i][j][sum]; 
 
    int inc = 0, exc = 0; 
    // Include the current prime 
    inc = CountWays(i + 1, j + 1, 
                sum + prime[i], 
                n, k); 
 
    // Exclude the current prime 
    exc = CountWays(i + 1, j, sum, 
                n, k); 
 
    // Return by memoizing the ans 
    return dp[i][j][sum] = inc + exc; 
} 
 
// Driver code 
int main() 
{ 
 
    // Precompute primes by sieve 
    sieve(); 
 
    int N = 100, K = 5; 
     
    cout << CountWays(0, 0, 0, N, K); 
} 

Java

// Java program to count the number
// of distinct ways to represent
// a number as K different primes
import java.io.*;
import java.util.*;
 
class GFG{
 
// Prime vector
static ArrayList<Integer> prime = new ArrayList<Integer>();
 
// Sieve array of prime
static boolean[] isprime = new boolean[1000];
 
// DP array
static int[][][] dp = new int[200][20][1000];
 
static void sieve()
{
     
    // Initialise all numbers
    // as prime
    for(int i = 0; i < 1000; i++)
        isprime[i] = true;
 
    //  Sieve of Eratosthenes.
    for(int i = 2; i * i < 1000; i++) 
    {
        if (isprime[i]) 
        {
            for(int j = i * i; 
                    j < 1000; j += i)
            {
                isprime[j] = false;
            }
        }
    }
     
    // Push all the primes into
    // prime vector
    for(int i = 2; i < 1000; i++)
    {
        if (isprime[i]) 
        {
            prime.add(i);
        }
    }
}
 
// Function to get the number of
// distinct ways to get sum
// as K different primes
static int CountWays(int i, int j, int sum,
                     int n, int k)
{
     
    // If index went out of prime
    // array size or the sum became
    // larger than n return 0
    if (i >= prime.size() - 1 || sum > n)
    {
        return 0;
    }
 
    // If sum becomes equal to n and
    // j becomes exactly equal to k.
    // Return 1, else if j is still
    // not equal to k, return 0
    if (sum == n) 
    {
        if (j == k) 
        {
            return 1;
        }
        return 0;
    }
 
    // If sum!=n and still j as
    // exceeded, return 0
    if (j == k)
        return 0;
 
    // If that state is already
    // calculated, return directly
    // the ans
    if (dp[i][j][sum] != 0)
        return dp[i][j][sum];
 
    int inc = 0, exc = 0;
    // Include the current prime
    inc = CountWays(i + 1, j + 1, 
                  sum + prime.get(i),
                  n, k);
 
    // Exclude the current prime
    exc = CountWays(i + 1, j, sum, n, k);
 
    // Return by memoizing the ans
    return dp[i][j][sum] = inc + exc;
}
 
// Driver code
public static void main(String[] args)
{
 
    // Precompute primes by sieve
    sieve();
 
    int N = 100, K = 5;
 
    System.out.println(CountWays(0, 0, 0, N, K));
}
}
 
// This code is contributed by akhilsaini

Python3

# Python3 program to count the number
# of distinct ways to represent
# a number as K different primes
 
# Prime list
prime = []
 
# Sieve array of prime
isprime = [True] * 1000
 
# DP array
dp = [[['0' for col in range(200)] 
            for col in range(20)]
            for row in range(1000)]
 
def sieve():
 
    #  Sieve of Eratosthenes.
    for i in range(2, 1000):
         
        if (isprime[i]):
            for j in range(i * i, 1000, i):
                isprime[j] = False
 
    # Push all the primes into
    # prime vector
    for i in range(2, 1000):
        if (isprime[i]):
            prime.append(i)
 
# Function to get the number of
# distinct ways to get sum
# as K different primes
def CountWays(i, j, sums, n, k):
 
    # If index went out of prime
    # array size or the sum became
    # larger than n return 0
    if (i >= len(prime) or sums > n):
        return 0
 
    # If sum becomes equal to n and
    # j becomes exactly equal to k.
    # Return 1, else if j is still
    # not equal to k, return 0
    if (sums == n):
        if (j == k):
            return 1
             
        return 0
 
    # If sum!=n and still j as
    # exceeded, return 0
    if j == k:
        return 0
 
    # If that state is already
    # calculated, return directly
    # the ans
    if dp[i][j][sums] == 0:
        return dp[i][j][sums]
 
    inc = 0
    exc = 0
     
    # Include the current prime
    inc = CountWays(i + 1, j + 1, 
                 sums + prime[i], 
                 n, k)
 
    # Exclude the current prime
    exc = CountWays(i + 1, j, sums, n, k)
 
    # Return by memoizing the ans
    dp[i][j][sums] = inc + exc
    return dp[i][j][sums]
 
# Driver code
if __name__ == "__main__":
     
    # Precompute primes by sieve
    sieve()
 
    N = 100
    K = 5
 
    print(CountWays(0, 0, 0, N, K))
 
# This code is contributed by akhilsaini

C#

// C# program to count the number
// of distinct ways to represent
// a number as K different primes
using System;
using System.Collections.Generic;
 
class GFG{
 
// Prime vector
static List<int> prime = new List<int>();
 
// Sieve array of prime
static bool[] isprime = new bool[1000];
 
// DP array
static int[, , ] dp = new int[200, 20, 1000];
 
static void sieve()
{
     
    // Initialise all numbers
    // as prime
    for(int i = 0; i < 1000; i++)
        isprime[i] = true;
 
    //  Sieve of Eratosthenes.
    for(int i = 2; i * i < 1000; i++)
    {
        if (isprime[i])
        {
            for(int j = i * i; 
                    j < 1000; j += i)
            {
                isprime[j] = false;
            }
        }
    }
     
    // Push all the primes into
    // prime vector
    for(int i = 2; i < 1000; i++)
    {
        if (isprime[i]) 
        {
            prime.Add(i);
        }
    }
}
 
// Function to get the number of
// distinct ways to get sum
// as K different primes
static int CountWays(int i, int j, int sum,
                     int n, int k)
{
     
    // If index went out of prime
    // array size or the sum became
    // larger than n return 0
    if (i >= prime.Count - 1 || sum > n)
    {
        return 0;
    }
 
    // If sum becomes equal to n and
    // j becomes exactly equal to k.
    // Return 1, else if j is still
    // not equal to k, return 0
    if (sum == n)
    {
        if (j == k)
        {
            return 1;
        }
        return 0;
    }
 
    // If sum!=n and still j as
    // exceeded, return 0
    if (j == k)
        return 0;
 
    // If that state is already
    // calculated, return directly
    // the ans
    if (dp[i, j, sum] != 0)
        return dp[i, j, sum];
 
    int inc = 0, exc = 0;
     
    // Include the current prime
    inc = CountWays(i + 1, j + 1, 
                  sum + prime[i], n, k);
 
    // Exclude the current prime
    exc = CountWays(i + 1, j, sum, n, k);
 
    // Return by memoizing the ans
    return dp[i, j, sum] = inc + exc;
}
 
// Driver code
static public void Main()
{
     
    // Precompute primes by sieve
    sieve();
 
    int N = 100, K = 5;
 
    Console.WriteLine(CountWays(0, 0, 0, N, K));
}
}
 
// This code is contributed by akhilsaini

Javascript

<script>
// Javascript program to count the Number 
// of distinct ways to represent 
// a number as K different primes 
 
// Prime vector 
var prime = [];  
 
// Sieve array of prime 
var isprime = Array(1000).fill(true);
 
// DP array 
var dp = Array.from(Array(200), ()=>Array(20));
for(var i =0; i<200; i++)
        for(var j =0; j<20; j++)
            dp[i][j] = new Array(1000).fill(0);
 
function sieve() 
{ 
    // Initialise all numbers 
    // as prime 
 
    // Sieve of Eratosthenes. 
    for (var i = 2; i * i <= 1000; 
        i++) 
    { 
        if (isprime[i]) 
        { 
            for (var j = i * i; 
                j <= 1000; j += i) 
            { 
                isprime[j] = false; 
            } 
        } 
    } 
    // Push all the primes into 
    // prime vector 
    for (var i = 2; i <= 1000; i++) 
    { 
        if (isprime[i]) 
        { 
            prime.push(i); 
        } 
    } 
} 
 
// Function to get the number of 
// distinct ways to get sum 
// as K different primes 
function CountWays(i,  j, sum, n, k) 
{ 
 
    // If index went out of prime 
    // array size or the sum became 
    // larger than n return 0 
    if (i > prime.length || sum > n) 
    { 
        return 0; 
    } 
 
    // If sum becomes equal to n and 
    // j becomes exactly equal to k. 
    // Return 1, else if j is still 
    // not equal to k, return 0 
    if (sum == n) { 
        if (j == k) { 
            return 1; 
        } 
        return 0; 
    } 
 
    // If sum!=n and still j as 
    // exceeded, return 0 
    if (j == k) 
        return 0; 
 
    // If that state is already 
    // calculated, return directly 
    // the ans 
    if (dp[i][j][sum]) 
        return dp[i][j][sum]; 
 
    var inc = 0, exc = 0; 
    // Include the current prime 
    inc = CountWays(i + 1, j + 1, 
                sum + prime[i], 
                n, k); 
 
    // Exclude the current prime 
    exc = CountWays(i + 1, j, sum, 
                n, k); 
 
    // Return by memoizing the ans 
    return dp[i][j][sum] = inc + exc; 
} 
 
// Driver code 
// Precompute primes by sieve 
sieve(); 
var N = 100, K = 5; 
document.write( CountWays(0, 0, 0, N, K)); 
 
 
</script>

Output:

Time Complexity: O(N*K).

Auxiliary Space: O(20*200*1000).

Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!

Number of distinct ways to represent a number as sum of K unique primes

C++

Java

Python3

C#

Javascript

From Word2Vec to LLM2Vec: How to Choose the Right Embedding Model for RAG

How to Debug Slow Search Requests in Milvus

When Context Engineering Is Done Right, Hallucinations Can Be the Spark of AI Creativity

LEAVE A REPLY Cancel reply

Most Popular

The Google Pixel Buds 2a look great, but I wanted more out of them

Samsung’s Galaxy S26 Pro is gearing up for the big Exynos comeback

The Chromecast is still king, here’s why I won’t give it up

I double-wristed all weekend, and don’t regret a minute of it

EDITOR PICKS

The Google Pixel Buds 2a look great, but I wanted more out of them

Samsung’s Galaxy S26 Pro is gearing up for the big Exynos comeback

The Chromecast is still king, here’s why I won’t give it up

POPULAR POSTS

The Google Pixel Buds 2a look great, but I wanted more out of them

Samsung’s Galaxy S26 Pro is gearing up for the big Exynos comeback

The Chromecast is still king, here’s why I won’t give it up

POPULAR CATEGORY

ABOUT US

FOLLOW US