kth prime factor of a given number (original) (raw)

Last Updated : 9 Jun, 2026

Given two numbers **n and **k, find the kth prime factor of **n.
Prime factors are considered with repetition and are taken in non-decreasing order. If **k is greater than the total number of prime factors of **n, return -1.

**Examples:

**Input: n = 225, k = 2
**Output: 3
**Explanation: The prime factorization of 225 is 3 × 3 × 5 × 5. The prime factors in non-decreasing order are [3, 3, 5, 5]. The 2nd prime factor is 3.

**Input: n = 81, k = 5
**Output: -1
**Explanation: The prime factorization of 81 is 3 × 3 × 3 × 3. The prime factors in non-decreasing order are [3, 3, 3, 3]. There is no 5th prime factor, so return -1.

Try It Yourself

Using Trial Division - O(√n) Time and O(1) Space

The idea is to divide n by every number starting from 2 up to √n. For each divisor that divides n, we keep dividing and decrementing k for every occurrence. Since we start from the smallest divisor and move upward, prime factors are naturally extracted in non-decreasing order. If k reaches 0 during this process we return that factor. After the loop, if n is still greater than 1 then n itself is a prime factor - we check if it is the kth one. If k is still not 0 after full factorization, we return -1.

C++ `

#include using namespace std;

int kthPrimeFactor(int n, int k) {

// Traverse from 2 to sqrt(n) to find prime factors
for (int div = 2; div * div <= n; div++) {

    // Divide out all occurrences of current factor
    while (n % div == 0) {

        // Decrement k for each prime factor found
        k--;

        // If k reaches 0 we found our answer
        if (k == 0)
            return div;

        n /= div;
    }
}

// If n is still greater than 1 it is a prime factor itself
if (n > 1) {
    k--;

    // Check if this remaining prime is the kth factor
    if (k == 0)
        return n;
}

// k is greater than total number of prime factors
return -1;

}

int main() { int n = 225, k = 2; cout << kthPrimeFactor(n, k) << endl; return 0; }

Java

class GFG {

static int kthPrimeFactor(int n, int k) {

    // Traverse from 2 to sqrt(n) to find prime factors
    for (int div = 2; div * div <= n; div++) {

        // Divide out all occurrences of current factor
        while (n % div == 0) {

            // Decrement k for each prime factor found
            k--;

            // If k reaches 0 we found our answer
            if (k == 0)
                return div;

            n /= div;
        }
    }

    // If n is still greater than 1 it is a prime factor itself
    if (n > 1) {
        k--;

        // Check if this remaining prime is the kth factor
        if (k == 0)
            return n;
    }

    // k is greater than total number of prime factors
    return -1;
}

public static void main(String[] args) {
    int n = 225, k = 2;
    System.out.println(kthPrimeFactor(n, k));
}

}

Python

def kthPrimeFactor(n, k):

# Traverse from 2 to sqrt(n) to find prime factors
div = 2
while div * div <= n:

    # Divide out all occurrences of current factor
    while n % div == 0:

        # Decrement k for each prime factor found
        k -= 1

        # If k reaches 0 we found our answer
        if k == 0:
            return div

        n //= div

    div += 1

# If n is still greater than 1 it is a prime factor itself
if n > 1:
    k -= 1

    # Check if this remaining prime is the kth factor
    if k == 0:
        return n

# k is greater than total number of prime factors
return -1

if name == "main": n = 225 k = 2 print(kthPrimeFactor(n, k))

C#

using System;

class GFG {

static int kthPrimeFactor(int n, int k) {

    // Traverse from 2 to sqrt(n) to find prime factors
    for (int div = 2; div * div <= n; div++) {

        // Divide out all occurrences of current factor
        while (n % div == 0) {

            // Decrement k for each prime factor found
            k--;

            // If k reaches 0 we found our answer
            if (k == 0)
                return div;

            n /= div;
        }
    }

    // If n is still greater than 1 it is a prime factor itself
    if (n > 1) {
        k--;

        // Check if this remaining prime is the kth factor
        if (k == 0)
            return n;
    }

    // k is greater than total number of prime factors
    return -1;
}

static void Main() {
    int n = 225, k = 2;
    Console.WriteLine(kthPrimeFactor(n, k));
}

}

JavaScript

function kthPrimeFactor(n, k) {

// Traverse from 2 to sqrt(n) to find prime factors
for (let div = 2; div * div <= n; div++) {

    // Divide out all occurrences of current factor
    while (n % div === 0) {

        // Decrement k for each prime factor found
        k--;

        // If k reaches 0 we found our answer
        if (k === 0)
            return div;

        n /= div;
    }
}

// If n is still greater than 1 it is a prime factor itself
if (n > 1) {
    k--;

    // Check if this remaining prime is the kth factor
    if (k === 0)
        return n;
}

// k is greater than total number of prime factors
return -1;

}

// Driver code let n = 225; let k = 2; console.log(kthPrimeFactor(n, k));

`