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: YesInput: A = 23, B = 959
Output: No
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 <bits/stdc++.h> 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; } |
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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 |
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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 = 240b = 204 if (areAnagrams(a, b)): print("Yes") else: print("No") # This code is contributed by # divyamohan123 |
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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 |
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Output:
Yes
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