Given two integers **K** and **N**, the task is to find the count of integers from the range **[2, N – 1]** whose sum of prime divisors is **K****Example:**

Input:N = 20, K = 7Output:2

7 and 10 are the only valid numbers.

sumPFactors(7) = 7

sumPFactors(10) = 2 + 5 = 7Input:N = 25, K = 5Output:5

**Approach:** Create an array **sumPF[]** where **sumPF[i]** stores the sum of prime divisors of **i** which can be easily calculated using the approach used in this article. Now, initialise a variable **count = 0** and run a loop from **2** to **N – 1** and for every element **i** if **sumPF[i] = K** then increment the **count**.

Below is the implementation of the above approach:

## C++

`// C++ implementation of the approach` `#include <iostream>` `using` `namespace` `std;` `#define MAX 1000001` `// Function to return the count of numbers` `// below N whose sum of prime factors is K` `int` `countNum(` `int` `N, ` `int` `K)` `{` ` ` `// To store the sum of prime factors` ` ` `// for all the numbers` ` ` `int` `sumPF[MAX] = { 0 };` ` ` `for` `(` `int` `i = 2; i < N; i++) {` ` ` `// If i is prime` ` ` `if` `(sumPF[i] == 0) {` ` ` `// Add i to all the numbers` ` ` `// which are divisible by i` ` ` `for` `(` `int` `j = i; j < N; j += i) {` ` ` `sumPF[j] += i;` ` ` `}` ` ` `}` ` ` `}` ` ` `// To store the count of required numbers` ` ` `int` `count = 0;` ` ` `for` `(` `int` `i = 2; i < N; i++) {` ` ` `if` `(sumPF[i] == K)` ` ` `count++;` ` ` `}` ` ` `// Return the required count` ` ` `return` `count;` `}` `// Driver code` `int` `main()` `{` ` ` `int` `N = 20, K = 7;` ` ` `cout << countNum(N, K);` ` ` `return` `0;` `}` |

## Java

`// Java implementation of the approach` `import` `java.util.*;` `class` `GFG` `{` `static` `int` `MAX = ` `1000001` `;` `// Function to return the count of numbers` `// below N whose sum of prime factors is K` `static` `int` `countNum(` `int` `N, ` `int` `K)` `{` ` ` `// To store the sum of prime factors` ` ` `// for all the numbers` ` ` `int` `[]sumPF = ` `new` `int` `[MAX];` ` ` `for` `(` `int` `i = ` `2` `; i < N; i++)` ` ` `{` ` ` `// If i is prime` ` ` `if` `(sumPF[i] == ` `0` `)` ` ` `{` ` ` `// Add i to all the numbers` ` ` `// which are divisible by i` ` ` `for` `(` `int` `j = i; j < N; j += i)` ` ` `{` ` ` `sumPF[j] += i;` ` ` `}` ` ` `}` ` ` `}` ` ` `// To store the count of required numbers` ` ` `int` `count = ` `0` `;` ` ` `for` `(` `int` `i = ` `2` `; i < N; i++)` ` ` `{` ` ` `if` `(sumPF[i] == K)` ` ` `count++;` ` ` `}` ` ` `// Return the required count` ` ` `return` `count;` `}` `// Driver code` `public` `static` `void` `main(String[] args)` `{` ` ` `int` `N = ` `20` `, K = ` `7` `;` ` ` `System.out.println(countNum(N, K));` `}` `}` `// This code is contributed by Rajput-Ji` |

## Python3

`# Python3 implementation of the approach` `MAX` `=` `1000001` `# Function to return the count of numbers` `# below N whose sum of prime factors is K` `def` `countNum(N, K) :` ` ` `# To store the sum of prime factors` ` ` `# for all the numbers` ` ` `sumPF ` `=` `[` `0` `] ` `*` `MAX` `;` ` ` `for` `i ` `in` `range` `(` `2` `, N) :` ` ` ` ` `# If i is prime` ` ` `if` `(sumPF[i] ` `=` `=` `0` `) :` ` ` `# Add i to all the numbers` ` ` `# which are divisible by i` ` ` `for` `j ` `in` `range` `(i, N, i) :` ` ` `sumPF[j] ` `+` `=` `i;` ` ` `# To store the count of required numbers` ` ` `count ` `=` `0` `;` ` ` `for` `i ` `in` `range` `(` `2` `, N) :` ` ` `if` `(sumPF[i] ` `=` `=` `K) :` ` ` `count ` `+` `=` `1` `;` ` ` `# Return the required count` ` ` `return` `count;` `# Driver code` `if` `__name__ ` `=` `=` `"__main__"` `:` ` ` `N ` `=` `20` `; K ` `=` `7` `;` ` ` ` ` `print` `(countNum(N, K));` `# This code is contributed by AnkitRai01` |

## C#

`// C# implementation of the approach` `using` `System;` ` ` `class` `GFG` `{` `static` `int` `MAX = 1000001;` `// Function to return the count of numbers` `// below N whose sum of prime factors is K` `static` `int` `countNum(` `int` `N, ` `int` `K)` `{` ` ` `// To store the sum of prime factors` ` ` `// for all the numbers` ` ` `int` `[]sumPF = ` `new` `int` `[MAX];` ` ` `for` `(` `int` `i = 2; i < N; i++)` ` ` `{` ` ` `// If i is prime` ` ` `if` `(sumPF[i] == 0)` ` ` `{` ` ` `// Add i to all the numbers` ` ` `// which are divisible by i` ` ` `for` `(` `int` `j = i; j < N; j += i)` ` ` `{` ` ` `sumPF[j] += i;` ` ` `}` ` ` `}` ` ` `}` ` ` `// To store the count of required numbers` ` ` `int` `count = 0;` ` ` `for` `(` `int` `i = 2; i < N; i++)` ` ` `{` ` ` `if` `(sumPF[i] == K)` ` ` `count++;` ` ` `}` ` ` `// Return the required count` ` ` `return` `count;` `}` `// Driver code` `public` `static` `void` `Main(String[] args)` `{` ` ` `int` `N = 20, K = 7;` ` ` `Console.WriteLine(countNum(N, K));` `}` `}` `// This code is contributed by 29AjayKumar` |

## Javascript

`<script>` `// Javascript implementation of the approach` `const MAX = 1000001;` `// Function to return the count of numbers` `// below N whose sum of prime factors is K` `function` `countNum(N, K)` `{` ` ` `// To store the sum of prime factors` ` ` `// for all the numbers` ` ` `let sumPF = ` `new` `Array(MAX).fill(0);` ` ` `for` `(let i = 2; i < N; i++) {` ` ` `// If i is prime` ` ` `if` `(sumPF[i] == 0) {` ` ` `// Add i to all the numbers` ` ` `// which are divisible by i` ` ` `for` `(let j = i; j < N; j += i) {` ` ` `sumPF[j] += i;` ` ` `}` ` ` `}` ` ` `}` ` ` `// To store the count of required numbers` ` ` `let count = 0;` ` ` `for` `(let i = 2; i < N; i++) {` ` ` `if` `(sumPF[i] == K)` ` ` `count++;` ` ` `}` ` ` `// Return the required count` ` ` `return` `count;` `}` `// Driver code` ` ` `let N = 20, K = 7;` ` ` `document.write(countNum(N, K));` `</script>` |

**Output:**

2

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