# Highest power of a number that divides other number

Given two numbers N and M, the task is to find the highest power of M that divides N.

Note: M > 1

Examples:

Input: N = 48, M = 4
Output: 2
48 % (4^2) = 0

Input: N = 32, M = 20
Output: 0
32 % (20^0) = 0

Approach: Initially prime factorize both the numbers N and M and store the count of prime factors in freq1[] and freq2[] respectively for N and M. For every prime factor of M, check if its freq2[num] is greater than freq1[num] or not. If it is for any prime factor of M, then max power will be 0. Else the maximum power will be minimum of all freq1[num] / freq2[num] for every prime factor of M.

For a number N = 24, the prime factors will 2^3 * 3^1. Hence freq1 = 3 and freq1 = 1.

Below is the implementation of the above approach:

## C++

 `// C++ program to implement ` `// the above approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function to get the prime factors ` `// and its count of times it divides ` `void` `primeFactors(``int` `n, ``int` `freq[]) ` `{ ` ` `  `    ``int` `cnt = 0; ` ` `  `    ``// Count the number of 2s that divide n ` `    ``while` `(n % 2 == 0) { ` `        ``cnt++; ` `        ``n = n / 2; ` `    ``} ` ` `  `    ``freq = cnt; ` ` `  `    ``// n must be odd at this point. So we can skip ` `    ``// one element (Note i = i +2) ` `    ``for` `(``int` `i = 3; i <= ``sqrt``(n); i = i + 2) { ` `        ``cnt = 0; ` ` `  `        ``// While i divides n, count i and divide n ` `        ``while` `(n % i == 0) { ` `            ``cnt++; ` `            ``n = n / i; ` `        ``} ` ` `  `        ``freq[i] = cnt; ` `    ``} ` ` `  `    ``// This condition is to handle the case when n ` `    ``// is a prime number greater than 2 ` `    ``if` `(n > 2) ` `        ``freq[n] = 1; ` `} ` ` `  `// Function to return the highest power ` `int` `getMaximumPower(``int` `n, ``int` `m) ` `{ ` ` `  `    ``// Initialize two arrays ` `    ``int` `freq1[n + 1], freq2[m + 1]; ` ` `  `    ``memset``(freq1, 0, ``sizeof` `freq1); ` `    ``memset``(freq2, 0, ``sizeof` `freq2); ` ` `  `    ``// Get the prime factors of n and m ` `    ``primeFactors(n, freq1); ` `    ``primeFactors(m, freq2); ` ` `  `    ``int` `maxi = 0; ` ` `  `    ``// Iterate and find the maximum power ` `    ``for` `(``int` `i = 2; i <= m; i++) { ` ` `  `        ``// If i not a prime factor of n and m ` `        ``if` `(freq1[i] == 0 && freq2[i] == 0) ` `            ``continue``; ` ` `  `        ``// If i is a prime factor of n and m ` `        ``// If count of i dividing m is more ` `        ``// than i dividing n, then power will be 0 ` `        ``if` `(freq2[i] > freq1[i]) ` `            ``return` `0; ` ` `  `        ``// If i is a prime factor of M ` `        ``if` `(freq2[i]) { ` ` `  `            ``// get the maximum power ` `            ``maxi = max(maxi, freq1[i] / freq2[i]); ` `        ``} ` `    ``} ` ` `  `    ``return` `maxi; ` `} ` ` `  `// Drivers code ` `int` `main() ` `{ ` `    ``int` `n = 48, m = 4; ` `    ``cout << getMaximumPower(n, m); ` `    ``return` `0; ` `} `

## Java

 `// Java program to implement ` `// the above approach ` ` `  `class` `GFG  ` `{ ` ` `  `// Function to get the prime factors ` `// and its count of times it divides ` `static` `void` `primeFactors(``int` `n, ``int` `freq[]) ` `{ ` ` `  `    ``int` `cnt = ``0``; ` ` `  `    ``// Count the number of 2s that divide n ` `    ``while` `(n % ``2` `== ``0``)  ` `    ``{ ` `        ``cnt++; ` `        ``n = n / ``2``; ` `    ``} ` ` `  `    ``freq[``2``] = cnt; ` ` `  `    ``// n must be odd at this point. So we can skip ` `    ``// one element (Note i = i +2) ` `    ``for` `(``int` `i = ``3``; i <= Math.sqrt(n); i = i + ``2``) ` `    ``{ ` `        ``cnt = ``0``; ` ` `  `        ``// While i divides n, count i and divide n ` `        ``while` `(n % i == ``0``) ` `        ``{ ` `            ``cnt++; ` `            ``n = n / i; ` `        ``} ` ` `  `        ``freq[i] = cnt; ` `    ``} ` ` `  `    ``// This condition is to handle the case when n ` `    ``// is a prime number greater than 2 ` `    ``if` `(n > ``2``) ` `        ``freq[n] = ``1``; ` `} ` ` `  `// Function to return the highest power ` `static` `int` `getMaximumPower(``int` `n, ``int` `m) ` `{ ` ` `  `    ``// Initialize two arrays ` `    ``int` `freq1[] = ``new` `int``[n + ``1``], freq2[] = ``new` `int``[m + ``1``]; ` ` `  `    ``// Get the prime factors of n and m ` `    ``primeFactors(n, freq1); ` `    ``primeFactors(m, freq2); ` ` `  `    ``int` `maxi = ``0``; ` ` `  `    ``// Iterate and find the maximum power ` `    ``for` `(``int` `i = ``2``; i <= m; i++) ` `    ``{ ` ` `  `        ``// If i not a prime factor of n and m ` `        ``if` `(freq1[i] == ``0` `&& freq2[i] == ``0``) ` `            ``continue``; ` ` `  `        ``// If i is a prime factor of n and m ` `        ``// If count of i dividing m is more ` `        ``// than i dividing n, then power will be 0 ` `        ``if` `(freq2[i] > freq1[i]) ` `            ``return` `0``; ` ` `  `        ``// If i is a prime factor of M ` `        ``if` `(freq2[i] != ``0``)  ` `        ``{ ` ` `  `            ``// get the maximum power ` `            ``maxi = Math.max(maxi, freq1[i] / freq2[i]); ` `        ``} ` `    ``} ` ` `  `    ``return` `maxi; ` `} ` ` `  `// Drivers code ` `public` `static` `void` `main(String[] args)  ` `{ ` `    ``int` `n = ``48``, m = ``4``; ` `    ``System.out.println(getMaximumPower(n, m)); ` ` `  `} ` `} ` ` `  `// This code contributed by Rajput-Ji `

## Python 3

 `import` `math ` ` `  `# Python program to implement ` `# the above approach ` ` `  `# Function to get the prime factors ` `# and its count of times it divides ` `def` `primeFactors(n, freq): ` `    ``cnt ``=` `0` ` `  `    ``# Count the number of 2s that divide n ` `    ``while` `n ``%` `2` `=``=` `0``: ` `        ``cnt ``=` `cnt ``+` `1` `        ``n ``=` `int``(n ``/``/` `2``) ` ` `  `    ``freq[``2``] ``=` `cnt ` ` `  `    ``# n must be odd at this point. So we can skip ` `    ``# one element (Note i = i+2) ` `    ``i``=``3` `    ``while` `i<``=``math.sqrt(n): ` `        ``cnt ``=` `0` ` `  `        ``# While i divides n, count i and divide n ` `        ``while` `(n ``%` `i ``=``=` `0``): ` `            ``cnt ``=` `cnt``+``1` `            ``n ``=` `int``(n ``/``/` `i) ` `             `  `        ``freq[``int``(i)] ``=` `cnt ` `        ``i``=``i ``+` `2` `         `  `    ``# This condition is to handle the case when n ` `    ``# is a prime number greater than 2 ` `    ``if` `(n > ``2``): ` `        ``freq[``int``(n)] ``=` `1` ` `  ` `  `# Function to return the highest power ` `def` `getMaximumPower(n, m): ` ` `  `    ``# Initialize two arrays ` `    ``freq1 ``=` `[``0``] ``*` `(n ``+` `1``) ` `    ``freq2 ``=` `[``0``] ``*` `(m ``+` `1``) ` ` `  ` `  `    ``# Get the prime factors of n and m ` `    ``primeFactors(n, freq1) ` `    ``primeFactors(m, freq2) ` ` `  `    ``maxi ``=` `0` ` `  `    ``# Iterate and find the maximum power ` `    ``i ``=` `2` `    ``while` `i <``=` `m: ` ` `  `        ``# If i not a prime factor of n and m ` `        ``if` `(freq1[i] ``=``=` `0` `and` `freq2[i] ``=``=` `0``): ` `            ``i ``=` `i ``+` `1` `            ``continue` ` `  `        ``# If i is a prime factor of n and m ` `        ``# If count of i dividing m is more ` `        ``# than i dividing n, then power will be 0 ` `        ``if` `(freq2[i] > freq1[i]): ` `            ``return` `0` ` `  `        ``# If i is a prime factor of M ` `        ``if` `(freq2[i]): ` ` `  `            ``# get the maximum power ` `            ``maxi ``=` `max``(maxi, ``int``(freq1[i] ``/``/` `freq2[i])) ` `         `  `        ``i ``=` `i ``+` `1` `     `  ` `  `    ``return` `maxi ` ` `  ` `  `# Drivers code ` `n ``=` `48` `m ``=` `4` `print``(getMaximumPower(n, m)) ` ` `  `# This code is contributed by Shashank_Sharma `

## C#

 `// C# program to implement ` `// the above approach ` `using` `System; ` ` `  `class` `GFG  ` `{ ` ` `  `// Function to get the prime factors ` `// and its count of times it divides ` `static` `void` `primeFactors(``int` `n, ``int` `[]freq) ` `{ ` ` `  `    ``int` `cnt = 0; ` ` `  `    ``// Count the number of 2s that divide n ` `    ``while` `(n % 2 == 0)  ` `    ``{ ` `        ``cnt++; ` `        ``n = n / 2; ` `    ``} ` ` `  `    ``freq = cnt; ` ` `  `    ``// n must be odd at this point. So we can skip ` `    ``// one element (Note i = i +2) ` `    ``for` `(``int` `i = 3; i <= Math.Sqrt(n); i = i + 2) ` `    ``{ ` `        ``cnt = 0; ` ` `  `        ``// While i divides n, count i and divide n ` `        ``while` `(n % i == 0) ` `        ``{ ` `            ``cnt++; ` `            ``n = n / i; ` `        ``} ` ` `  `        ``freq[i] = cnt; ` `    ``} ` ` `  `    ``// This condition is to handle the case when n ` `    ``// is a prime number greater than 2 ` `    ``if` `(n > 2) ` `        ``freq[n] = 1; ` `} ` ` `  `// Function to return the highest power ` `static` `int` `getMaximumPower(``int` `n, ``int` `m) ` `{ ` ` `  `    ``// Initialize two arrays ` `    ``int` `[]freq1 = ``new` `int``[n + 1];``int` `[]freq2 = ``new` `int``[m + 1]; ` ` `  `    ``// Get the prime factors of n and m ` `    ``primeFactors(n, freq1); ` `    ``primeFactors(m, freq2); ` ` `  `    ``int` `maxi = 0; ` ` `  `    ``// Iterate and find the maximum power ` `    ``for` `(``int` `i = 2; i <= m; i++) ` `    ``{ ` ` `  `        ``// If i not a prime factor of n and m ` `        ``if` `(freq1[i] == 0 && freq2[i] == 0) ` `            ``continue``; ` ` `  `        ``// If i is a prime factor of n and m ` `        ``// If count of i dividing m is more ` `        ``// than i dividing n, then power will be 0 ` `        ``if` `(freq2[i] > freq1[i]) ` `            ``return` `0; ` ` `  `        ``// If i is a prime factor of M ` `        ``if` `(freq2[i] != 0)  ` `        ``{ ` ` `  `            ``// get the maximum power ` `            ``maxi = Math.Max(maxi, freq1[i] / freq2[i]); ` `        ``} ` `    ``} ` ` `  `    ``return` `maxi; ` `} ` ` `  `// Drivers code ` `public` `static` `void` `Main(String[] args)  ` `{ ` `    ``int` `n = 48, m = 4; ` `    ``Console.WriteLine(getMaximumPower(n, m)); ` ` `  `} ` `} ` ` `  `// This code has been contributed by 29AjayKumar `

## PHP

 ` 2)  ` `        ``\$freq``[``\$n``] = 1;  ` `     `  `    ``return` `\$freq` `; ` `}  ` ` `  `// Function to return the highest power  ` `function` `getMaximumPower(``\$n``, ``\$m``)  ` `{  ` ` `  `    ``\$freq1` `= ``array_fill``(0,``\$n` `+ 1,0);  ` `    ``\$freq2` `= ``array_fill``(0,``\$m` `+ 1,0);  ` ` `  `    ``// Get the prime factors of n and m  ` `    ``\$freq1` `= primeFactors(``\$n``, ``\$freq1``);  ` `    ``\$freq2` `= primeFactors(``\$m``, ``\$freq2``);  ` ` `  `    ``\$maxi` `= 0;  ` ` `  `    ``// Iterate and find the maximum power  ` `    ``for` `(``\$i` `= 2; ``\$i` `<= ``\$m``; ``\$i``++)  ` `    ``{  ` ` `  `        ``// If i not a prime factor of n and m  ` `        ``if` `(``\$freq1``[``\$i``] == 0 && ``\$freq2``[``\$i``] == 0)  ` `            ``continue``;  ` ` `  `        ``// If i is a prime factor of n and m  ` `        ``// If count of i dividing m is more  ` `        ``// than i dividing n, then power will be 0  ` `        ``if` `(``\$freq2``[``\$i``] > ``\$freq1``[``\$i``])  ` `            ``return` `0;  ` ` `  `        ``// If i is a prime factor of M  ` `        ``if` `(``\$freq2``[``\$i``])  ` `        ``{  ` ` `  `            ``// get the maximum power  ` `            ``\$maxi` `= max(``\$maxi``, ``floor``(``\$freq1``[``\$i``] / ``\$freq2``[``\$i``]));  ` `        ``}  ` `    ``}  ` ` `  `    ``return` `\$maxi``;  ` `}  ` ` `  `    ``// Drivers code  ` `    ``\$n` `= 48; ``\$m` `= 4;  ` `    ``echo` `getMaximumPower(``\$n``, ``\$m``);  ` ` `  `    ``// This code is contributed by Ryuga ` `?> `

Output:

```2
```

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.

My Personal Notes arrow_drop_up Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.

Article Tags :
Practice Tags :

Be the First to upvote.

Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.