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

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.

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Method 1 (Use Sorting):

1. Sort both strings
2. 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 code
int 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)

Auxiliary space: O(1). 

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. 

1. Create count arrays of size 256 for both strings. Initialize all values in count arrays as 0.
2. Iterate through every character of both strings and increment the count of character in the corresponding count arrays.
3. 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 code
int 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

Time Complexity: O(n)

Auxiliary space: O(n). 

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 code
int 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)

Auxiliary space: O(n). 

Method 4 (Using unordered_map):

We can optimize the space complexity of the above method by using unordered_map instead of initializing 256 characters array. So in this approach, we will first count the occurrences of each unique character with the help of unordered_map for the first string. Then we will reduce the count of each character while we encounter them in the second string. Finally, if the count of each character in the unordered_map is 0 then it means both strings are anagrams else not.

Below is the code for the above approach.

C++

// C++ program to check if two
// strings are anagrams of each other
#include <bits/stdc++.h>
using namespace std;

bool isAnagram(string a,string b)
{
// Check if both strings has same length or not
if (a.length() != b.length()) {
    return false;
}

// Initialising unordered_map
unordered_map<char,int> m;

// Storing the count of each character
// present in first String
for (int i = 0; i < a.length(); i++) {
    m[a[i]]++;
}

// Now iterating over second String
for (int i = 0; i < b.length(); i++) {
    // Check if unordered_map already contain the current
    // character or not
    if (m[b[i]]) {
    // If contains reduce count of that
    // character by 1 to indicate that current
    // character has been already counted as
    // idea here is to check if in last count of
    // all characters in last is zero which
    // means all characters in String a are
    // present in String b.
    m[b[i]] -= 1;
    }
}

// Loop over all keys and check if all keys are 0
// as it means that all the characters are present
// in equal count in both strings.
for (auto items : m) {
    if (items.second != 0) {
    return false;
    }
}
// Returning True as all keys are zero
return true;
}

// Driver code
int main()
{
string str1 = "geeksforgeeks";
string str2 = "forgeeksgeeks";

// Function call
if (isAnagram(str1, str2))
    cout<<"The two strings are anagram of each other"<<endl;
else
    cout<<"The two strings are not anagram of each other"<<endl;
}

// This code is contributed by Pushpesh Raj

Output

The two strings are anagram of each other

Time Complexity: O(n)

Auxiliary space: O(m) where m is the number of unique characters in the first string. 

Please suggest if someone has a better solution which is more efficient in terms of space and time.
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