All Factors or Divisors (original) (raw)

Last Updated : 2 Apr, 2026

Given a positive integer **n, find all the **distinct divisors of **n.

**Examples:

**Input: n = 10
**Output: [1, 2, 5, 10]
**Explanation: 1, 2, 5 and 10 are the divisors of 10.

**Input: n = 100
**Output: [1, 2, 4, 5, 10, 20, 25, 50, 100]
**Explanation: 1, 2, 4, 5, 10, 20, 25, 50 and 100 are divisors of 100.

Try It Yourself

Table of Content

[Naive Approach] Iterating till n - O(n) Time and O(1) Space
[Expected Approach] Finding all factors in pairs - O(sqrt(n)) Time and O(1) Space

[Naive Approach] Iterating till n - O(n) Time and O(1) Space

The idea is to iterate over all the numbers from 1 to n and for each number check if the number divides n. If the number divides n, print it.

C++ `

#include #include using namespace std;

vector printDivisors(int n) { vectordivisors;

for (int i = 1; i <= n; i++) {
    
    // i is a divisor of n
    if (n % i == 0) {
        divisors.push_back(i);
    }
}

return divisors;

}

int main() { vectordivisors = printDivisors(10); for(auto &div: divisors) { cout << div << " " ; }

return 0;

}

C

#include <stdio.h> #include <stdlib.h>

int* printDivisors(int n, int* size) { int* divisors = (int*)malloc(n * sizeof(int)); *size = 0;

for (int i = 1; i <= n; i++) {
   
    // i is a divisor of n
    if (n % i == 0) {
        divisors[(*size)++] = i;
    }
}

return divisors;

}

int main() { int size; int* divisors = printDivisors(10, &size); for (int i = 0; i < size; i++) { printf("%d ", divisors[i]); }

free(divisors);
return 0;

}

Java

import java.util.*;

public class Main {

public static ArrayList<Integer> printDivisors(int n) {
    
    ArrayList<Integer> divisors = new ArrayList<>();

    // Iterate from 1 to n and check divisibility
    for (int i = 1; i <= n; i++) {
        if (n % i == 0) {
            
            // If 'i' divides 'n' evenly, it's a divisor
            divisors.add(i);
        }
    }

    // Return the list of divisors
    return divisors;
}   
 
public static void main(String[] args) {
    int number = 10;

    ArrayList<Integer> divisors = printDivisors(number);

    for (int div : divisors) {
        System.out.print(div + " ");
    }
}

}

Python

def printDivisors(n):

# Create a list to store divisors
divisors = []

# Iterate from 1 to n and check divisibility
for i in range(1, n + 1):
    if n % i == 0:
        
        # If 'i' divides 'n' evenly, it's a divisor
        divisors.append(i)

return divisors

if name == "main": number = 10 divisors = printDivisors(number) for div in divisors: print(div, end=" ")

C#

using System; using System.Collections.Generic;

class GfG { static List printDivisors(int n) { List divisors = new List();

    for (int i = 1; i <= n; i++) {
        // i is a divisor of n
        if (n % i == 0) {
            divisors.Add(i);
        }
    }

    return divisors;
}

static void Main() {
    List<int> divisors = printDivisors(10);
    foreach (int div in divisors) {
        Console.Write(div + " ");
    }
}

}

JavaScript

function printDivisors(n) {

let divisors = [];

for (let i = 1; i <= n; i++) {
    
    // i is a divisor of n
    if (n % i === 0) {
        divisors.push(i);
    }
}

return divisors;

}

// Driver Code let divisors = printDivisors(10); for (let div of divisors) { process.stdout.write(div + " "); }

[Expected Approach] Finding all factors in pairs - O(sqrt(n)) Time and O(1) Space

If we look carefully, all the divisors of a number appear in pairs.
For example, if n = 100, then the divisor pairs are:
(1, 100), (2, 50), (4, 25), (5, 20), (10, 10).

We need to be careful in cases like (10, 10)—i.e., when both divisors in a pair are equal (which happens when n is a perfect square). In such cases, we should include that divisor only once.
Using this fact, we can optimize our program significantly.

Instead of iterating from 1 to n, we only need to iterate from 1 to √n.
Why? Because for any factor a of n, the corresponding factor b = n / a forms a pair (a, b).
At least one of the two values in any such pair must lie within the range [1, √n].

So, we can:

Iterate from 1 to √n to find all divisors less than or equal to √n.
For each such divisor d, also add n / d as the paired divisor.
Take care not to duplicate the square root divisor if n is a perfect square.

Factors-occurring-in-pairs

C++ `

#include #include #include #include using namespace std;

vector printDivisors(int n) { vectordivisors;

// Note that this loop runs till square root
for (int i = 1; i <= sqrt(n); i++) {
    if (n % i == 0) {
        
        // If divisors are equal, print only one
        if (n / i == i) {
            divisors.push_back(i) ;
        }
        // Otherwise print both 
        else {
            divisors.push_back(i) ;
            divisors.push_back(n/i) ;
        }
    }
}

return divisors;

}

int main() { vectordivisors = printDivisors(10); for(auto &divs: divisors) { cout << divs << " " ; } return 0; }

C

#include <stdio.h> #include <stdlib.h> #include <math.h>

int* printDivisors(int n, int *size) { *size = 0;

// Rough upper bound for number of divisors
int capacity = 2 * (int)sqrt(n); 
int* divisors = (int*)malloc(capacity * sizeof(int));

for (int i = 1; i <= sqrt(n); i++) {
    if (n % i == 0) {
        if (n / i == i) {
            
            // is both i and n/i is same
            divisors[(*size)++] = i;
        } else {
            
            // if i and n/i is different 
            divisors[(*size)++] = i;
            divisors[(*size)++] = n / i;
        }
    }
}

return divisors;

}

int main() { int size; int* divisors = printDivisors(10, &size);

for (int i = 0; i < size; i++) {
    printf("%d ", divisors[i]);
}

free(divisors);
return 0;

}

Java

import java.util.ArrayList;

public class Main {

static ArrayList<Integer> printDivisors(int n) {
    ArrayList<Integer> divisors = new ArrayList<>();
    
    // Note that this loop runs till square root
    for (int i = 1; i <= Math.sqrt(n); i++) {
        if (n % i == 0) {

            // If divisors are equal, add only once
            if (n / i == i) {
                divisors.add(i);
            }
            // Otherwise add both
            else {
                divisors.add(i);
                divisors.add(n / i);
            }
        }
    }
    return divisors;
}

public static void main(String[] args) {
    ArrayList<Integer> divisors = printDivisors(10);
    for (int divs : divisors) {
        System.out.print(divs + " ");
    }
}

}

Python

import math

def printDivisors(n): divisors = []

# Loop runs up to square root of n
for i in range(1, int(math.sqrt(n)) + 1):
    if n % i == 0:
        
        # If both divisors are same (perfect square), add only once
        if n // i == i:
            divisors.append(i)
        else:
            
            # Add both divisors
            divisors.append(i)
            divisors.append(n // i)
return divisors

if name == "main": number = 10 divisors = printDivisors(number)

for div in divisors:
    print(div, end=" ")

C#

using System; using System.Collections.Generic;

class GfG { static List printDivisors(int n) { List divisors = new List();

    // Note that this loop runs till square root
    for (int i = 1; i <= Math.Sqrt(n); i++) {
        if (n % i == 0) {

            // If divisors are equal, print only one
            if (n / i == i) {
                divisors.Add(i);
            }
            
            // Otherwise print both 
            else {
                divisors.Add(i);
                divisors.Add(n / i);
            }
        }
    }
    return divisors;
}

static void Main() {
    List<int> divisors = printDivisors(10);
    foreach (int divs in divisors) {
        Console.Write(divs + " ");
    }
}

}

JavaScript

// Function to print the divisors function printDivisors(n) { let divisors = []; // Note that this loop runs till square root for (let i = 1; i <= Math.sqrt(n); i++) { if (n % i === 0) {

        // If divisors are equal, print only one
        if (n / i === i) {
            divisors.push(i);
        }
        // Otherwise print both 
        else {
            divisors.push(i);
            divisors.push(n / i);
        }
    }
}
return divisors;

}

// Driver Code let n = 10; let divisors = printDivisors(n); for (let divs of divisors) { process.stdout.write(divs + " "); }