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Checking Anagrams (check whether two string is anagrams or not)

Checking Anagrams: In the following we are going to learn how to check whether two string is anagrams or not?
Submitted by Radib Kar, on November 19, 2018 [Last updated : March 20, 2023]

Problem Description

Given two strings, check whether two given strings are anagram of each other or not. An anagram of a string is another string that contains same characters, only the order of characters can be different.

For example, "act" and "cat" are anagram of each other.

Solution of Checking Anagrams

Algorithm

Anagram means both the string contains the same set of character, only their orders are different. Thus, in both the string the frequency of each letter must be the same if they are an anagram of each other. Thus our algorithm checks and compare frequency of each letter in both the strings.

1. The strings to be anagram of each other, their length must be same.

Let n1 be the length of first string & n2 be the length of second string.
If (n1!=n2)
	Strings are not anagram. Return.
Else
Proceed to step2.

2. Initialize two arrays of length 26 to store frequency of characters in each string.

array1[26]={0}, array2[26]={0};
//for the first string
For i=1:n1 //n1 be the length of first string
	// for each letter of the string their corresponding
	array1[string1[i]-'a']++; 
//frequencies are being stored

End for loop

//for the second string
For i=1:n2 //n2 be the length of second string
	// for each letter of the string their corresponding
	array2[string2[i]-'a']++; 
//frequencies are being stored

End for loop

3. Comparison step

Compare the frequency of each letter in both the strings
If 
all the letter in both of the string have same frequency (number of occurrence)
Then they are anagrams of each other
Else
They are not anagrams of each other

4. Print result and return.

C++ program to check whether two string is anagrams or not

#include <bits/stdc++.h>
using namespace std;

int anagram(string s1, string s2)
{
    int array1[26] = { 0 }, array2[26] = { 0 };
    //if string lengths are different
    if (s1.length() != s2.length())
        return 0; //they are not anagrams
    //for string1
    for (int i = 0; s1[i] != '\0'; i++) {
        //storing frequency for each letter in the string
        array1[s1[i] - 'a']++;
    }

    //for string2
    for (int i = 0; s2[i] != '\0'; i++) {
        //storing frequency for each letter in the string
        array2[s2[i] - 'a']++;
    }
    //comparison step
    for (int i = 0; i < 26; i++) {
        // if any letter has different no of occurence,
        // then strings are not anagrams
        if (array1[i] != array2[i])
            return 0;
    }

    return 1; // else they are anagrams
}
int main()
{
    int n;
    string s1, s2;

    //input the strings
    cout << "enter string1\n";
    cin >> s1;
    cout << "enter string2\n";
    cin >> s2;

    if (anagram(s1, s2))
        printf("strings are anagrams of each other\n");
    else
        printf("strings are not anagrams of each other\n");

    return 0;
}

Output

First run:
enter string1
includehelp
enter string2
cnldeehpiul
strings are anagrams of each other 

Second run:
enter string1
includehelp
enter string2
helpincludr
strings are not anagrams of each other 

Explanation

In the first output,

String1 ='includehelp'
String2='cnldeehpiul'
So, in the first string there is:
one 'I', one 'n', one 'c', two 'l', one 'u', one 'd', two 'e', one 'h' & one 'p'
in the second string there is:
one 'c', one 'n', two 'l', one 'd', two 'e', one 'h' & one 'p', one 'i' &one 'u'
Thus they are anagrams.

But in the second output,

The second string contains 'r' which is not in the first string, thus they are not anagrams.

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