C++ Program To Check Whether Two Strings Are Anagram Of Each Other - GeeksforGeeks

20 Dec, 2021
Write a function to check whether two given strings are anagram of each other or not. An anagram of a string is another string that contains the same characters, only the order of characters can be different. For example, “abcd” and “dabc” are an anagram of each other.
Method 1 (Use Sorting):  
Sort both stringsCompare the sorted strings
Sort both strings
Compare the sorted strings
Below is the implementation of the above idea:
C++
// C++ program to check whether two // strings are anagrams of each other#include <bits/stdc++.h>using namespace std;  /* Function to check whether two strings    are anagram of each other */bool areAnagram(string str1, string str2){    // Get lengths of both strings    int n1 = str1.length();    int n2 = str2.length();      // If length of both strings is not     // same, then they cannot be anagram    if (n1 != n2)        return false;      // Sort both the strings    sort(str1.begin(), str1.end());    sort(str2.begin(), str2.end());      // Compare sorted strings    for (int i = 0; i < n1; i++)        if (str1[i] != str2[i])            return false;      return true;}  // Driver codeint main(){    string str1 = "test";    string str2 = "ttew";      // Function Call    if (areAnagram(str1, str2))        cout <<         "The two strings are anagram of each other";    else        cout << "The two strings are not anagram of each "                "other";    return 0;}
Output: 
The two strings are not anagram of each other
Time Complexity: O(nLogn)
Method 2 (Count characters): This method assumes that the set of possible characters in both strings is small. In the following implementation, it is assumed that the characters are stored using 8 bit and there can be 256 possible characters. 
Create count arrays of size 256 for both strings. Initialize all values in count arrays as 0.Iterate through every character of both strings and increment the count of character in the corresponding count arrays.Compare count arrays. If both count arrays are same, then return true.
Create count arrays of size 256 for both strings. Initialize all values in count arrays as 0.
Iterate through every character of both strings and increment the count of character in the corresponding count arrays.
Compare count arrays. If both count arrays are same, then return true.
Below is the implementation of the above idea:
C++
// C++ program to check if two strings// are anagrams of each other#include <bits/stdc++.h>using namespace std;#define NO_OF_CHARS 256  /* Function to check whether two    strings are anagram of each other */bool areAnagram(char* str1, char* str2){    // Create 2 count arrays and initialize     // all values as 0    int count1[NO_OF_CHARS] = {0};    int count2[NO_OF_CHARS] = {0};    int i;      // For each character in input strings,     // increment count in the corresponding     // count array    for (i = 0; str1[i] && str2[i]; i++)     {        count1[str1[i]]++;        count2[str2[i]]++;    }      // If both strings are of different length.     // Removing this condition will make the     // program fail for strings like "aaca"     // and "aca"    if (str1[i] || str2[i])        return false;      // Compare count arrays    for (i = 0; i < NO_OF_CHARS; i++)        if (count1[i] != count2[i])            return false;      return true;}  // Driver codeint main(){    char str1[] = "geeksforgeeks";    char str2[] = "forgeeksgeeks";        // Function Call    if (areAnagram(str1, str2))        cout <<         "The two strings are anagram of each other";    else        cout << "The two strings are not anagram of each "                "other";      return 0;}// This is code is contributed by rathbhupendra
Output:
The two strings are anagram of each other
Method 3 (count characters using one array): The above implementation can be further to use only one count array instead of two. We can increment the value in count array for characters in str1 and decrement for characters in str2. Finally, if all count values are 0, then the two strings are anagram of each other. Thanks to Ace for suggesting this optimization.
C++
// C++ program to check if two strings// are anagrams of each other#include <bits/stdc++.h>using namespace std;#define NO_OF_CHARS 256  bool areAnagram(char* str1, char* str2){    // Create a count array and initialize     // all values as 0    int count[NO_OF_CHARS] = { 0 };    int i;      // For each character in input strings,     // increment count in the corresponding     // count array    for (i = 0; str1[i] && str2[i]; i++)     {        count[str1[i]]++;        count[str2[i]]--;    }      // If both strings are of different length.     // Removing this condition will make the     // program fail for strings like "aaca"     // and "aca"    if (str1[i] || str2[i])        return false;      // See if there is any non-zero value     // in count array    for (i = 0; i < NO_OF_CHARS; i++)        if (count[i])            return false;    return true;}  // Driver codeint main(){    char str1[] = "geeksforgeeks";    char str2[] = "forgeeksgeeks";        // Function call    if (areAnagram(str1, str2))        cout <<         "The two strings are anagram of each other";    else        cout << "The two strings are not anagram of each "                "other";    return 0;}
Output:
The two strings are anagram of each other
Time Complexity: O(n)
Please refer complete article on Check whether two strings are anagram of each other for more details!
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