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# Find all divisors of first N natural numbers

• Last Updated : 29 Apr, 2021

Given an integer N, the task is to find all the divisors of numbers from 1 to N.
Note: 1 ? N ? 100000

Examples:

Input: N = 2
Output:
1 –>1
2 –>1, 2

Input: N = 5
Output:
1 –>1
2 –>1, 2
3 –>1, 3
4 –>1, 2, 4
5 –>1, 5

Naive Approach:

• Iterate over first N natural numbers using a loop variable (say i)
• Iterate over the natural numbers from 1 to i with a loop variable (say j) and check that i % j == 0. Then j is a divisor of the natural number i.

Below is the implementation of the above approach:

## C++

 `// C++ implementation to find all``// the divisors of the first N``// natural numbers` `#include ` `using` `namespace` `std;` `// Function to find the divisors``// of the first N natural numbers``void` `factors(``int` `n)``{``    ``int` `i, j;``    ``cout << ``"1 -->1\n"``;``    ` `    ``// Loop to find the divisors``    ``for` `(i = 2; i <= n; i++) {``        ``cout << i << ``" -->"``;``        ``for` `(j = 1; j <= i / 2; j++) {``            ``if` `(i % j == 0)``                ``cout << j << ``", "``;``        ``}``        ``cout << i << ``"\n"``;``    ``}``}` `// Driver Code``int` `main()``{``    ``int` `n = 5;``    ``factors(n);``}`

## Java

 `// Java implementation to find all``// the divisors of the first N``// natural numbers``class` `GFG{` `// Function to find the divisors``// of the first N natural numbers``static` `void` `factors(``int` `n)``{``    ``int` `i, j;``    ``System.out.print(``"1 -->1\n"``);``    ` `    ``// Loop to find the divisors``    ``for``(i = ``2``; i <= n; i++)``    ``{``       ``System.out.print(i + ``" -->"``);``       ``for``(j = ``1``; j <= i / ``2``; j++)``       ``{``          ``if` `(i % j == ``0``)``              ``System.out.print(j + ``", "``);``       ``}``       ``System.out.print(i + ``"\n"``);``    ``}``}` `// Driver Code``public` `static` `void` `main(String[] args)``{``    ``int` `n = ``5``;``    ``factors(n);``}``}` `// This code is contributed by Rohit_ranjan`

## Python3

 `# Python3 implementation to find all``# the divisors of the first N``# natural numbers` `# Function to find the divisors``# of the first N natural numbers``def` `factors(n):` `    ``i ``=` `0``; j ``=` `0``;``    ``print``(``"1 -->1"``);``    ` `    ``# Loop to find the divisors``    ``for` `i ``in` `range``(``2``, n ``+` `1``):``        ``print``(i, ``"-->"``, end ``=` `"");``        ``for` `j ``in` `range``(``1``, (i ``/``/` `2``) ``+` `1``):``            ``if` `(i ``%` `j ``=``=` `0``):``                ``print``(j, ``","``, end ``=` `"");``        ` `        ``print``(i, end ``=` `"\n"``);``    ` `# Driver Code``n ``=` `5``;``factors(n);` `# This code is contributed by Code_Mech`

## C#

 `// C# implementation to find all``// the divisors of the first N``// natural numbers``using` `System;``class` `GFG{` `// Function to find the divisors``// of the first N natural numbers``static` `void` `factors(``int` `n)``{``    ``int` `i, j;``    ``Console.Write(``"1 -->1\n"``);``    ` `    ``// Loop to find the divisors``    ``for``(i = 2; i <= n; i++)``    ``{``        ``Console.Write(i + ``" -->"``);``        ``for``(j = 1; j <= i / 2; j++)``        ``{``            ``if` `(i % j == 0)``                ``Console.Write(j + ``", "``);``        ``}``        ``Console.Write(i + ``"\n"``);``    ``}``}` `// Driver Code``public` `static` `void` `Main()``{``    ``int` `n = 5;``    ``factors(n);``}``}` `// This code is contributed by Nidhi_biet`

## Javascript

 ``
Output:
```1 -->1
2 -->1, 2
3 -->1, 3
4 -->1, 2, 4
5 -->1, 5```

Time Complexity: O(N2)

Better Approach:

• Iterate over the first N natural numbers using a loop variable.
• For the number to find its divisors iterate from 2 to that number and check any one of them is a divisor of the given number.

Below is the implementation of the above approach:

## C++

 `// C++ implementation to find all``// the divisors of the first``// N natural numbers` `#include ` `using` `namespace` `std;` `// Function to find the factors of``// the numbers from 1 to N``void` `factors(``int` `n)``{``    ``int` `i, j;``    ``cout << ``"1 -->1\n"``;``    ` `    ``// Loop to find the factors``    ``// of the first N natural``    ``// numbers of the integer``    ``for` `(i = 2; i <= n; i++) {``        ``cout << i << ``" -->"``;``        ``for` `(j = 1; j * j <= i; j++) {``            ``if` `(i % j == 0){``                ``cout << j << ``", "``;``                ``if` `(i / j != j)``                ``cout << i/j << ``", "``;``            ``}``        ``}``        ``cout << ``"\n"``;``    ``}``}` `// Driver Code``int` `main()``{``    ``int` `n = 5;``    ``factors(n);``}`

## Java

 `// Java implementation to find all``// the divisors of the first``// N natural numbers``import` `java.util.*;``class` `GFG{` `// Function to find the factors of``// the numbers from 1 to N``static` `void` `factors(``int` `n)``{``    ``int` `i, j;``    ``System.out.print(``"1 -->1\n"``);``    ` `    ``// Loop to find the factors``    ``// of the first N natural``    ``// numbers of the integer``    ``for` `(i = ``2``; i <= n; i++)``    ``{``        ``System.out.print(i + ``" -->"``);``        ``for` `(j = ``1``; j * j <= i; j++)``        ``{``            ``if` `(i % j == ``0``)``            ``{``                ``System.out.print(j + ``", "``);``                ``if` `(i / j != j)``                    ``System.out.print(i / j + ``", "``);``            ``}``        ``}``        ``System.out.print(``"\n"``);``    ``}``}` `// Driver Code``public` `static` `void` `main(String args[])``{``    ``int` `n = ``5``;``    ``factors(n);``}``}` `// This code is contributed by Code_Mech`

## Python3

 `# Python3 implementation to find all``# the divisors of the first``# N natural numbers` `# Function to find the factors of``# the numbers from 1 to N``def` `factors(n):``    ` `    ``print``(``"1 -->1"``);` `    ``# Loop to find the factors``    ``# of the first N natural``    ``# numbers of the integer``    ``for` `i ``in` `range``(``2``, n ``+` `1``):``        ``print``(i, ``" -->"``, end ``=` `"");``        ` `        ``for` `j ``in` `range``(``1``, ``int``(``pow``(i, ``1``))):``            ``if` `(i ``%` `j ``=``=` `0``):``                ``print``(j, ``", "``, end ``=` `"");``                ` `                ``if` `(i ``/``/` `j !``=` `j):``                    ``print``(i ``/``/` `j, ``", "``, end ``=` `"");``            ` `        ``print``(end ``=` `"\n"``);``    ` `# Driver Code``if` `__name__ ``=``=` `'__main__'``:``    ` `    ``n ``=` `5``;``    ``factors(n);` `# This code is contributed by gauravrajput1`

## C#

 `// C# implementation to find all``// the divisors of the first``// N natural numbers``using` `System;``class` `GFG{` `// Function to find the factors of``// the numbers from 1 to N``static` `void` `factors(``int` `n)``{``    ``int` `i, j;``    ``Console.Write(``"1 -->1\n"``);``    ` `    ``// Loop to find the factors``    ``// of the first N natural``    ``// numbers of the integer``    ``for` `(i = 2; i <= n; i++)``    ``{``        ``Console.Write(i + ``" -->"``);``        ``for` `(j = 1; j * j <= i; j++)``        ``{``            ``if` `(i % j == 0)``            ``{``                ``Console.Write(j + ``", "``);``                ``if` `(i / j != j)``                    ``Console.Write(i / j + ``", "``);``            ``}``        ``}``        ``Console.Write(``"\n"``);``    ``}``}` `// Driver Code``public` `static` `void` `Main()``{``    ``int` `n = 5;``    ``factors(n);``}``}` `// This code is contributed by Code_Mech`

## Javascript

 ``
Output:
```1 -->1
2 -->1, 2,
3 -->1, 3,
4 -->1, 4, 2,
5 -->1, 5, ```

Time Complexity: O(N*sqrt(N))

Efficient Approach: The idea is to precompute the factors of the numbers with the help of the Sieve of Eratosthenes. Then finally iterate over the first N natural numbers to find the factors.

Below is the implementation of the above approach:

## C++

 `// C++ implementation to find the``// factors of first N natural``// numbers` `#include ` `using` `namespace` `std;` `const` `int` `MAX = 1e5;``  ` `// Initialize global divisor vector``// array of sequence container``vector<``int``> divisor[MAX + 1];` `// Calculate all``// divisors of number``void` `sieve()``{``    ``for` `(``int` `i = 1; i <= MAX; ++i) {``        ``for` `(``int` `j = i; j <= MAX; j += i)``            ``divisor[j].push_back(i);``    ``}``}` `// Function to find the``// factors of first n``// natural numbers``void` `findNFactors(``int` `n){``    ``for``(``int` `i = 1; i <= n; i++){``        ``cout << i << ``"-->"``;``        ``for` `(``auto` `&divi: divisor[i]){``            ``cout << divi << ``", "``;``        ``}``        ``cout << ``"\n"``;``    ``}``}` `// Driver Code``int` `main()``{``    ``int` `n = 5;``    ``sieve();``    ` `    ``// Function Call``    ``findNFactors(n);``}`

## Java

 `// Java implementation to find the``// factors of first N natural``// numbers``import` `java.util.*;``class` `GFG{` `static` `int` `MAX = (``int``) 1e5;``  ` `// Initialize global divisor vector``// array of sequence container``static` `Vector []divisor = ``new` `Vector[MAX + ``1``];` `// Calculate all``// divisors of number``static` `void` `sieve()``{``    ``for` `(``int` `i = ``1``; i <= MAX; ++i)``    ``{``        ``for` `(``int` `j = i; j <= MAX; j += i)``            ``divisor[j].add(i);``    ``}``}` `// Function to find the``// factors of first n``// natural numbers``static` `void` `findNFactors(``int` `n)``{``    ``for``(``int` `i = ``1``; i <= n; i++)``    ``{``        ``System.out.print(i+ ``"-->"``);``        ``for` `(``int` `divi: divisor[i])``        ``{``            ``System.out.print(divi+ ``", "``);``        ``}``        ``System.out.print(``"\n"``);``    ``}``}` `// Driver Code``public` `static` `void` `main(String[] args)``{``    ``int` `n = ``5``;``    ``for` `(``int` `i = ``0``; i < divisor.length; i++)``        ``divisor[i] = ``new` `Vector();``    ``sieve();``    ` `    ``// Function Call``    ``findNFactors(n);``}``}` `// This code is contributed by Princi Singh`

## Python3

 `# Python3 program to find the factors``# of first N natural numbers``MAX` `=` `100001` `# Initialize divisor list(array)``# of sequence container``divisor ``=` `[[] ``for` `x ``in` `range``(``MAX``)]` `# Calculate all divisors of a number``def` `sieve():` `    ``for` `i ``in` `range``(``1``, ``MAX``):``        ``for` `j ``in` `range``(i, ``MAX``, i):``            ``divisor[j].append(i)` `# Function to find the factors of``# first n natural numbers``def` `findNFactors (n):` `    ``for` `i ``in` `range``(``1``, n ``+` `1``):``        ``print``(i, ``" --> "``, end ``=` `'')``        ` `        ``for` `divi ``in` `divisor[i]:``            ``print``(divi, ``", "``, end ``=` `'')``        ``print``()` `# Driver code``if` `__name__ ``=``=` `'__main__'``:` `    ``n ``=` `5``    ``sieve()` `    ``# Function call``    ``findNFactors(n)` `# This code is contributed by himanshu77`

## C#

 `// C# implementation to find the``// factors of first N natural``// numbers``using` `System;``using` `System.Collections.Generic;`` ` `public` `class` `GFG{`` ` `static` `int` `MAX = (``int``) 1e5;``   ` `// Initialize global divisor vector``// array of sequence container``static` `List<``int``> []divisor = ``new` `List<``int``>[MAX + 1];`` ` `// Calculate all``// divisors of number``static` `void` `sieve()``{``    ``for` `(``int` `i = 1; i <= MAX; ++i)``    ``{``        ``for` `(``int` `j = i; j <= MAX; j += i)``            ``divisor[j].Add(i);``    ``}``}`` ` `// Function to find the``// factors of first n``// natural numbers``static` `void` `findNFactors(``int` `n)``{``    ``for``(``int` `i = 1; i <= n; i++)``    ``{``        ``Console.Write(i+ ``"-->"``);``        ``foreach` `(``int` `divi ``in` `divisor[i])``        ``{``            ``Console.Write(divi+ ``", "``);``        ``}``        ``Console.Write(``"\n"``);``    ``}``}`` ` `// Driver Code``public` `static` `void` `Main(String[] args)``{``    ``int` `n = 5;``    ``for` `(``int` `i = 0; i < divisor.Length; i++)``        ``divisor[i] = ``new` `List<``int``>();``    ``sieve();``     ` `    ``// Function Call``    ``findNFactors(n);``}``}`` ` `// This code is contributed by shikhasingrajput`

## Javascript

 ``
Output:
```1-->1,
2-->1, 2,
3-->1, 3,
4-->1, 2, 4,
5-->1, 5,```

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