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Remove minimum number of characters so that two strings become anagram

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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 Practice

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.  

Implementation:

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

                    

Javascript

<script>
// javascript 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
    var count1 = Array.from({length: 26}, (_, i) => 0); 
    var count2 = Array.from({length: 26}, (_, i) => 0);
  
    // count frequency of each character 
    // in first string
    for (i = 0; i < str1.length ; i++)
        count1[str1.charAt(i).charCodeAt(0) -'a'.charCodeAt(0)]++;
  
    // count frequency of each character 
    // in second string
    for (i = 0; i < str2.length ; i++)
        count2[str2.charAt(i).charCodeAt(0) -'a'.charCodeAt(0)]++;
  
    // traverse count arrays to find 
    // number of characters to be removed
    var result = 0;
    for (i = 0; i < 26; i++)
        result += Math.abs(count1[i] -
                           count2[i]);
    return result;
}
  
// Driver program to run the case
var str1 = "bcadeh", str2 = "hea";
document.write(remAnagram(str1, str2));
  
// This code is contributed by Amit Katiyar 
</script>

                    

Output
3

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
?>

                    

Javascript

<script>
  
// Javascript implementation to count
// number of deletions required from 
// two strings to create an anagram 
CHARS = 26;
  
function countDeletions(str1, str2)
{
    var arr = Array.from({length: CHARS}, (_, i) => 0);
      
    for(i = 0; i < str1.length; i++)
    {
        arr[str1.charAt(i).charCodeAt(0) -
                       'a'.charCodeAt(0)]++;
    }
  
    for(i = 0; i < str2.length; i++)
    {
        arr[str2.charAt(i).charCodeAt(0) - 
                       'a'.charCodeAt(0)]--;
    }
  
    var ans = 0;
    for(i = 0; i < CHARS; i++) 
    {
        ans += Math.abs(arr[i]);
    }
    return ans;
}
  
// Driver code
str1 = "bcadeh", str2 = "hea";
  
document.write(countDeletions(str1, str2));
  
// This code is contributed by Rajput-Ji 
  
</script>

                    

Output
3

Time complexity: O(n) where n is the total number of characters in both strings. 
Auxiliary space: O(1) as we only use a constant size array.

Thanks to vishal9619 for suggesting this optimized solution.

Another Approach: Using Counter from collections module

Explanation:

  • The collections module provides a Counter class that works like a dictionary to count the frequency of elements in a list or string.
  • The makeAnagram function takes two strings a and b as input.
  • freq = Counter(a) initializes a counter object freq with the characters in string a.
  • freq.subtract(Counter(b)) subtracts the characters in string b from the counter object. This operation will update the counts of the characters in freq.
  • sum(abs(count) for count in freq.values()) calculates the sum of absolute values of counts in the counter object freq.

C++

#include <iostream>
#include <map>
#include <string>
using namespace std;
  
int makeAnagram(string a, string b)
{
    // Initialize a map object with the characters in string
    // a
    map<char, int> freq;
    for (char c : a) {
        freq++;
    }
  
    // Subtract the characters in string b from the map
    // object
    for (char c : b) {
        freq--;
    }
  
    // Calculate the sum of absolute values of counts in the
    // map object
    int count = 0;
    for (auto p : freq) {
        count += abs(p.second);
    }
    return count;
}
  
int main()
{
    string str1 = "bcadeh";
    string str2 = "hea";
    cout << makeAnagram(str1, str2) << endl;
  
    return 0;
}
// This code is contributed by user_dtewbxkn77n

                    

Java

import java.util.HashMap;
import java.util.Map;
  
public class Main {
  
    public static int makeAnagram(String a, String b) {
        // Initialize a map object with the characters in string a
        Map<Character, Integer> freq = new HashMap<>();
        for (char c : a.toCharArray()) {
            freq.put(c, freq.getOrDefault(c, 0) + 1);
        }
  
        // Subtract the characters in string b from the map object
        for (char c : b.toCharArray()) {
            freq.put(c, freq.getOrDefault(c, 0) - 1);
        }
  
        // Calculate the sum of absolute values of counts in the map object
        int count = 0;
        for (int val : freq.values()) {
            count += Math.abs(val);
        }
        return count;
    }
  
    public static void main(String[] args) {
        String str1 = "bcadeh";
        String str2 = "hea";
        System.out.println(makeAnagram(str1, str2));
    }
}

                    

Python3

from collections import Counter
  
def makeAnagram(a, b):
    # Initialize a counter object with the characters in string a
    freq = Counter(a)
      
    # Subtract the characters in string b from the counter object
    freq.subtract(Counter(b))
      
    # Calculate the sum of absolute values of counts in the counter object
    return sum(abs(count) for count in freq.values())
  
if __name__ == "__main__":
    str1 = "bcadeh"
    str2 = "hea"
    print(makeAnagram(str1, str2))

                    

C#

using System;
using System.Collections.Generic;
  
namespace ConsoleApp {
  class Program {
    static int MakeAnagram(string a, string b)
    {
  
      // Initialize a dictionary object with the
      // characters in string a
      Dictionary<char, int> freq
        = new Dictionary<char, int>();
      foreach(char c in a)
      {
        if (freq.ContainsKey(c)) {
          freq++;
        }
        else {
          freq.Add(c, 1);
        }
      }
  
      // Subtract the characters in string b from the
      // dictionary object
      foreach(char c in b)
      {
        if (freq.ContainsKey(c)) {
          freq--;
        }
        else {
          freq.Add(c, -1);
        }
      }
  
      // Calculate the sum of absolute values of counts in
      // the dictionary object
      int count = 0;
      foreach(KeyValuePair<char, int> p in freq)
      {
        count += Math.Abs(p.Value);
      }
      return count;
    }
  
    static void Main(string[] args)
    {
      string str1 = "bcadeh";
      string str2 = "hea";
      Console.WriteLine(MakeAnagram(str1, str2));
    }
  }
}

                    

Javascript

function makeAnagram(string1, string2) {
    // Initialize a map object with the characters in string1
    let freq = new Map();
    for (let c of string1) {
        freq.set(c, (freq.get(c) || 0) + 1);
    }
  
    // Subtract the characters in string2 from the map object
    for (let c of string2) {
        freq.set(c, (freq.get(c) || 0) - 1);
    }
  
    // Calculate the sum of absolute values of counts in the map object
    let count = 0;
    for (let [key, value] of freq) {
        count += Math.abs(value);
    }
    return count;
}
  
let str1 = "bcadeh";
let str2 = "hea";
console.log(makeAnagram(str1, str2));

                    

Output
3

Time complexity: O(len(a) + len(b)) 

The time complexity of the function is O(len(a) + len(b)), where len(a) and len(b) are the lengths of the input strings.

Auxiliary Space: O(len(a) + len(b)) 

The space complexity of the function is O(len(a) + len(b)). This is because the function creates a frequency buffer of length 26 to store the count of each character in the input strings

 



Last Updated : 11 Sep, 2023
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