Remove minimum number of characters so that two strings become anagram

Given two strings in lowercase, the task is to make them anagram. The only allowed operation is to remove a character from any string. Find minimum number of characters to be deleted to make both the strings anagram?
If two strings contains same data set in any order then strings are called Anagrams.

Examples :

Input : str1 = "bcadeh" str2 = "hea"
Output: 3
We need to remove b, c and d from str1.

Input : str1 = "cddgk" str2 = "gcd"
Output: 2

Input : str1 = "bca" str2 = "acb"
Output: 0

Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

The idea is to make character count arrays for both the strings and store frequency of each character. Now iterate the count arrays of both strings and difference in frequency of any character abs(count1[str1[i]-‘a’] – count2[str2[i]-‘a’]) in both the strings is the number of character to be removed in either string.

C++

// C++ program to find minimum number of characters 
// to be removed to make two strings anagram. 
#include<bits/stdc++.h> 
using namespace std; 
const int CHARS = 26; 
  
// function to calculate minimum numbers of characters 
// to be removed to make two strings anagram 
int remAnagram(string str1, string str2) 
{ 
    // make hash array for both string and calculate 
    // frequency of each character 
    int count1[CHARS] = {0}, count2[CHARS] = {0}; 
  
    // count frequency of each character in first string 
    for (int i=0; str1[i]!='\0'; i++) 
        count1[str1[i]-'a']++; 
  
    // count frequency of each character in second string 
    for (int i=0; str2[i]!='\0'; i++) 
        count2[str2[i]-'a']++; 
  
    // traverse count arrays to find number of characters 
    // to be removed 
    int result = 0; 
    for (int i=0; i<26; i++) 
        result += abs(count1[i] - count2[i]); 
    return result; 
} 
  
// Driver program to run the case 
int main() 
{ 
    string str1 = "bcadeh", str2 = "hea"; 
    cout << remAnagram(str1, str2); 
    return 0; 
} 

Java

// Java program to find minimum number of 
// characters to be removed to make two 
// strings anagram. 
import java.util.*; 
  
class GFG { 
  
    // function to calculate minimum numbers 
    // of characters to be removed to make 
    // two strings anagram 
    static int remAnagram(String str1, String str2) 
    { 
        // make hash array for both string  
        // and calculate frequency of each 
        // character 
        int count1[] = new int[26];  
        int count2[] = new int[26]; 
  
        // count frequency of each character  
        // in first string 
        for (int i = 0; i < str1.length() ; i++) 
            count1[str1.charAt(i) -'a']++; 
      
        // count frequency of each character  
        // in second string 
        for (int i = 0; i < str2.length() ; i++) 
            count2[str2.charAt(i) -'a']++; 
  
        // traverse count arrays to find  
        // number of characters to be removed 
        int result = 0; 
        for (int i = 0; i < 26; i++) 
            result += Math.abs(count1[i] - 
                               count2[i]); 
        return result; 
    } 
  
    // Driver program to run the case 
    public static void main(String[] args) 
    { 
        String str1 = "bcadeh", str2 = "hea"; 
        System.out.println(remAnagram(str1, str2)); 
    } 
} 
// This code is contributed by Prerna Saini 

Python3

# Python 3 program to find minimum  
# number of characters 
# to be removed to make two  
# strings anagram. 
  
CHARS = 26
  
# function to calculate minimum  
# numbers of characters 
# to be removed to make two  
# strings anagram 
def remAnagram(str1, str2): 
  
    # make hash array for both string  
    # and calculate 
    # frequency of each character 
    count1 = [0]*CHARS 
    count2 = [0]*CHARS 
  
    # count frequency of each character  
    # in first string 
    i = 0
    while i < len(str1): 
        count1[ord(str1[i])-ord('a')] += 1
        i += 1
  
    # count frequency of each character  
    # in second string 
    i =0
    while i < len(str2): 
        count2[ord(str2[i])-ord('a')] += 1
        i += 1
  
    # traverse count arrays to find  
    # number of characters 
    # to be removed 
    result = 0
    for i in range(26): 
        result += abs(count1[i] - count2[i]) 
    return result 
  
# Driver program to run the case 
if __name__ == "__main__": 
    str1 = "bcadeh"
    str2 = "hea"
    print(remAnagram(str1, str2)) 
      
# This code is contributed by  
# ChitraNayal 

C#

// C# program to find minimum  
// number of characters to be  
// removed to make two strings  
// anagram. 
using System; 
  
class GFG  
{ 
  
    // function to calculate  
    // minimum numbers of  
    // characters to be removed  
    // to make two strings anagram 
    static int remAnagram(string str1,  
                          string str2) 
    { 
        // make hash array for both  
        // string and calculate frequency  
        // of each character 
        int []count1 = new int[26];  
        int []count2 = new int[26]; 
  
        // count frequency of each  
        // character in first string 
        for (int i = 0; i < str1.Length ; i++) 
            count1[str1[i] -'a']++; 
      
        // count frequency of each   
        // character in second string 
        for (int i = 0; i < str2.Length ; i++) 
            count2[str2[i] -'a']++; 
  
        // traverse count arrays to  
        // find number of characters  
        // to be removed 
        int result = 0; 
        for (int i = 0; i < 26; i++) 
            result += Math.Abs(count1[i] - 
                               count2[i]); 
        return result; 
    } 
  
    // Driver Code 
    public static void Main() 
    { 
        string str1 = "bcadeh",  
               str2 = "hea"; 
        Console.Write(remAnagram(str1, str2)); 
    } 
} 
  
// This code is contributed 
// by nitin mittal. 

PHP

<?php 
// PHP program to find minimum number of 
// characters to be removed to make two 
// strings anagram. 
  
// function to calculate minimum numbers 
// of characters to be removed to make 
// two strings anagram 
function remAnagram($str1, $str2) 
{ 
    // make hash array for both string  
    // and calculate frequency of each 
    // character 
    $count1 = array(26);  
    $count2 = array(26); 
  
    // count frequency of each character  
    // in first string 
    for ($i = 0; $i < strlen($str1) ; $i++) 
        $count1[$str1[$i] - 'a']++; 
  
    // count frequency of each character  
    // in second string 
    for ($i = 0; $i < strlen($str2) ; $i++) 
        $count2[$str2[$i] -'a']++; 
  
    // traverse count arrays to find  
    // number of characters to be removed 
    $result = 0; 
    for ($i = 0; $i < 26; $i++) 
        $result += abs($count1[$i] - 
                       $count2[$i]); 
    return $result; 
} 
  
// Driver Code 
{ 
    $str1 = "bcadeh"; $str2 = "hea"; 
    echo(remAnagram($str1, $str2)); 
} 
  
// This code is contributed by Code_Mech 

Output :

Time Complexity : O(n)
Auxiliary space : O(ALPHABET-SIZE)

The above solution can be optimized to work with single count array.

C++

// C++ implementation to count number of deletions 
// required from two strings to create an anagram  
#include <bits/stdc++.h> 
using namespace std; 
const int CHARS = 26; 
  
int countDeletions(string str1, string str2) 
{  
    int arr[CHARS] = {0}; 
    for (int i = 0; i < str1.length(); i++)  
        arr[str1[i] - 'a']++;  
      
    for (int i = 0; i < str2.length(); i++)  
        arr[str2[i] - 'a']--; 
      
    long long int ans = 0; 
    for(int i = 0; i < CHARS; i++) 
        ans +=abs(arr[i]); 
    return ans; 
} 
  
int main() { 
    string str1 = "bcadeh", str2 = "hea"; 
    cout << countDeletions(str1, str2); 
    return 0; 
}  

Java

// Java implementation to count number of deletions 
// required from two strings to create an anagram  
  
class GFG { 
  
    final static int CHARS = 26; 
  
    static int countDeletions(String str1, String str2) { 
        int arr[] = new int[CHARS]; 
        for (int i = 0; i < str1.length(); i++) { 
            arr[str1.charAt(i) - 'a']++; 
        } 
  
        for (int i = 0; i < str2.length(); i++) { 
            arr[str2.charAt(i) - 'a']--; 
        } 
  
        int ans = 0; 
        for (int i = 0; i < CHARS; i++) { 
            ans += Math.abs(arr[i]); 
        } 
        return ans; 
    } 
  
    static public void main(String[] args) { 
        String str1 = "bcadeh", str2 = "hea"; 
        System.out.println(countDeletions(str1, str2)); 
    } 
} 
  
// This code is contributed by 29AjayKumar 

Python3

# Python3 program to find minimum  
# number of characters to be  
# removed to make two strings  
# anagram. 
  
# function to calculate minimum  
# numbers of characters to be  
# removed to make two strings anagram  
def makeAnagram(a, b): 
    buffer = [0] * 26
    for char in a: 
        buffer[ord(char) - ord('a')] += 1
    for char in b: 
        buffer[ord(char) - ord('a')] -= 1
    return sum(map(abs, buffer)) 
  
# Driver Code 
if __name__ == "__main__" :  
  
    str1 = "bcadeh"
    str2 = "hea"
    print(makeAnagram(str1, str2)) 
      
# This code is contributed  
# by Raghib Ahsan 

C#

      
// C# implementation to count number of deletions 
// required from two strings to create an anagram  
using System; 
public class GFG { 
   
    readonly static int CHARS = 26; 
   
    static int countDeletions(String str1, String str2) { 
        int []arr = new int[CHARS]; 
        for (int i = 0; i < str1.Length; i++) { 
            arr[str1[i]- 'a']++; 
        } 
   
        for (int i = 0; i < str2.Length; i++) { 
            arr[str2[i] - 'a']--; 
        } 
   
        int ans = 0; 
        for (int i = 0; i < CHARS; i++) { 
            ans += Math.Abs(arr[i]); 
        } 
        return ans; 
    } 
   
    static public void Main() { 
        String str1 = "bcadeh", str2 = "hea"; 
        Console.WriteLine(countDeletions(str1, str2)); 
    } 
} 
  //This code is contributed by PrinciRaj1992 

PHP

<?php 
// PHP implementation to count number of deletions 
// required from two strings to create an anagram  
  
function countDeletions($str1, $str2)  
{ 
    $CHARS = 26; 
    $arr = array(); 
    for ($i = 0; $i < strlen($str1); $i++)  
    { 
        $arr[ord($str1[$i]) - ord('a')]++; 
    } 
  
    for ($i = 0; $i < strlen($str2); $i++)  
    { 
        $arr[ord($str2[$i]) - ord('a')]--; 
    } 
  
    $ans = 0; 
    for ($i = 0; $i < $CHARS; $i++)  
    { 
        $ans += abs($arr[$i]); 
    } 
    return $ans; 
} 
  
// Driver Code 
$str1 = "bcadeh"; $str2 = "hea"; 
echo(countDeletions($str1, $str2)); 
  
// This code is contributed by Code_Mech 
?> 

Output :

Thanks to vishal9619 for suggesting this optimized solution.

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Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

C++

Java

Python3

C#

PHP

C++

Java

Python3

C#

PHP

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