# Check if two Integer are anagrams of each other

Given two integers A and B, the task is to check whether the given numbers are anagrams of each other or not. Just like strings, a number is said to be an anagram of some other number if it can be made equal to the other number by just shuffling the digits in it.

Examples:

Input: A = 204, B = 240
Output: Yes

Input: A = 23, B = 959
Output: No

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: Create two arrays freqA[] and freqB[] where freqA[i] and freqB[i] will store the frequency of digit i in a and b respectively. Now traverse the frequency arrays and for any digit i if freqA[i] != freqB[i] then the numbers are not anagrams of each other else they are.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` ` `  `const` `int` `TEN = 10; ` ` `  `// Function to update the frequency array ` `// such that freq[i] stores the ` `// frequency of digit i in n ` `void` `updateFreq(``int` `n, ``int` `freq[]) ` `{ ` ` `  `    ``// While there are digits ` `    ``// left to process ` `    ``while` `(n) { ` `        ``int` `digit = n % TEN; ` ` `  `        ``// Update the frequency of ` `        ``// the current digit ` `        ``freq[digit]++; ` ` `  `        ``// Remove the last digit ` `        ``n /= TEN; ` `    ``} ` `} ` ` `  `// Function that returns true if a and b ` `// are anagarams of each other ` `bool` `areAnagrams(``int` `a, ``int` `b) ` `{ ` ` `  `    ``// To store the frequencies of ` `    ``// the digits in a and b ` `    ``int` `freqA[TEN] = { 0 }; ` `    ``int` `freqB[TEN] = { 0 }; ` ` `  `    ``// Update the frequency of ` `    ``// the digits in a ` `    ``updateFreq(a, freqA); ` ` `  `    ``// Update the frequency of ` `    ``// the digits in b ` `    ``updateFreq(b, freqB); ` ` `  `    ``// Match the frequencies of ` `    ``// the common digits ` `    ``for` `(``int` `i = 0; i < TEN; i++) { ` ` `  `        ``// If frequency differs for any digit ` `        ``// then the numbers are not ` `        ``// anagrams of each other ` `        ``if` `(freqA[i] != freqB[i]) ` `            ``return` `false``; ` `    ``} ` ` `  `    ``return` `true``; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `a = 240, b = 204; ` ` `  `    ``if` `(areAnagrams(a, b)) ` `        ``cout << ``"Yes"``; ` `    ``else` `        ``cout << ``"No"``; ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation of the approach ` `class` `GFG ` `{ ` `    ``static` `final` `int` `TEN = ``10``; ` `     `  `    ``// Function to update the frequency array ` `    ``// such that freq[i] stores the ` `    ``// frequency of digit i in n ` `    ``static` `void` `updateFreq(``int` `n, ``int` `[] freq) ` `    ``{ ` `     `  `        ``// While there are digits ` `        ``// left to process ` `        ``while` `(n > ``0``) ` `        ``{ ` `            ``int` `digit = n % TEN; ` `     `  `            ``// Update the frequency of ` `            ``// the current digit ` `            ``freq[digit]++; ` `     `  `            ``// Remove the last digit ` `            ``n /= TEN; ` `        ``} ` `    ``} ` `     `  `    ``// Function that returns true if a and b ` `    ``// are anagarams of each other ` `    ``static` `boolean` `areAnagrams(``int` `a, ``int` `b) ` `    ``{ ` `     `  `        ``// To store the frequencies of ` `        ``// the digits in a and b ` `        ``int` `[] freqA = ``new` `int``[TEN]; ` `        ``int` `[] freqB = ``new` `int``[TEN]; ` `     `  `        ``// Update the frequency of ` `        ``// the digits in a ` `        ``updateFreq(a, freqA); ` `     `  `        ``// Update the frequency of ` `        ``// the digits in b ` `        ``updateFreq(b, freqB); ` `     `  `        ``// Match the frequencies of ` `        ``// the common digits ` `        ``for` `(``int` `i = ``0``; i < TEN; i++)  ` `        ``{ ` `     `  `            ``// If frequency differs for any digit ` `            ``// then the numbers are not ` `            ``// anagrams of each other ` `            ``if` `(freqA[i] != freqB[i]) ` `                ``return` `false``; ` `        ``} ` `        ``return` `true``; ` `    ``} ` `     `  `    ``// Driver code ` `    ``public` `static` `void` `main (String[] args) ` `    ``{ ` `        ``int` `a = ``204``, b = ``240``; ` `     `  `        ``if` `(areAnagrams(a, b)) ` `            ``System.out.println(``"Yes"``); ` `        ``else` `            ``System.out.println(``"No"``); ` `    ``} ` `} ` ` `  `// This code is contributed by ihirtik ` `        `

## Python3

 `# Python3 implementation of the approach  ` `TEN ``=` `10` ` `  `# Function to update the frequency array  ` `# such that freq[i] stores the  ` `# frequency of digit i in n  ` `def` `updateFreq(n, freq) : ` ` `  `    ``# While there are digits  ` `    ``# left to process  ` `    ``while` `(n) :  ` `        ``digit ``=` `n ``%` `TEN  ` ` `  `        ``# Update the frequency of  ` `        ``# the current digit  ` `        ``freq[digit] ``+``=` `1` ` `  `        ``# Remove the last digit  ` `        ``n ``/``/``=` `TEN  ` ` `  `# Function that returns true if a and b  ` `# are anagarams of each other  ` `def` `areAnagrams(a, b):  ` ` `  `    ``# To store the frequencies of  ` `    ``# the digits in a and b  ` `    ``freqA ``=` `[ ``0` `] ``*` `TEN  ` `    ``freqB ``=` `[ ``0` `] ``*` `TEN ` ` `  `    ``# Update the frequency of  ` `    ``# the digits in a  ` `    ``updateFreq(a, freqA)  ` ` `  `    ``# Update the frequency of  ` `    ``# the digits in b  ` `    ``updateFreq(b, freqB)  ` ` `  `    ``# Match the frequencies of  ` `    ``# the common digits  ` `    ``for` `i ``in` `range``(TEN):  ` ` `  `        ``# If frequency differs for any digit  ` `        ``# then the numbers are not  ` `        ``# anagrams of each other  ` `        ``if` `(freqA[i] !``=` `freqB[i]):  ` `            ``return` `False` `             `  `    ``return` `True` ` `  `# Driver code  ` `a ``=` `240` `b ``=` `204` ` `  `if` `(areAnagrams(a, b)):  ` `    ``print``(``"Yes"``)  ` `else``: ` `    ``print``(``"No"``)  ` ` `  `# This code is contributed by ` `# divyamohan123 `

## C#

 `// C# implementation of the approach ` `using` `System; ` ` `  `class` `GFG ` `{ ` `    ``static` `int` `TEN = 10; ` `     `  `    ``// Function to update the frequency array ` `    ``// such that freq[i] stores the ` `    ``// frequency of digit i in n ` `    ``static` `void` `updateFreq(``int` `n, ``int` `[] freq) ` `    ``{ ` `     `  `        ``// While there are digits ` `        ``// left to process ` `        ``while` `(n > 0) ` `        ``{ ` `            ``int` `digit = n % TEN; ` `     `  `            ``// Update the frequency of ` `            ``// the current digit ` `            ``freq[digit]++; ` `     `  `            ``// Remove the last digit ` `            ``n /= TEN; ` `        ``} ` `    ``} ` `     `  `    ``// Function that returns true if a and b ` `    ``// are anagarams of each other ` `    ``static` `bool` `areAnagrams(``int` `a, ``int` `b) ` `    ``{ ` `     `  `        ``// To store the frequencies of ` `        ``// the digits in a and b ` `        ``int` `[] freqA = ``new` `int``[TEN]; ` `        ``int` `[] freqB = ``new` `int``[TEN];; ` `     `  `        ``// Update the frequency of ` `        ``// the digits in a ` `        ``updateFreq(a, freqA); ` `     `  `        ``// Update the frequency of ` `        ``// the digits in b ` `        ``updateFreq(b, freqB); ` `     `  `        ``// Match the frequencies of ` `        ``// the common digits ` `        ``for` `(``int` `i = 0; i < TEN; i++)  ` `        ``{ ` `     `  `            ``// If frequency differs for any digit ` `            ``// then the numbers are not ` `            ``// anagrams of each other ` `            ``if` `(freqA[i] != freqB[i]) ` `                ``return` `false``; ` `        ``} ` `        ``return` `true``; ` `    ``} ` `     `  `    ``// Driver code ` `    ``public` `static` `void` `Main () ` `    ``{ ` `        ``int` `a = 204, b = 240; ` `     `  `        ``if` `(areAnagrams(a, b)) ` `            ``Console.WriteLine(``"Yes"``); ` `        ``else` `            ``Console.WriteLine(``"No"``); ` `    ``} ` `} ` ` `  `// This code is contributed by ihirtik `

Output:

```Yes
```

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