Minimum Number of Manipulations required to make two Strings Anagram Without Deletion of Character
Given two strings s1 and s2, we need to find the minimum number of manipulations required to make two strings anagram without deleting any character.
Note:- The anagram strings have same set of characters, sequence of characters can be different.
If deletion of character is allowed and cost is given, refer to Minimum Cost To Make Two Strings Identical
Question Source: Yatra.com Interview Experience | Set 7
Examples:
Input :
s1 = "aba"
s2 = "baa"
Output : 0
Explanation: Both String contains identical characters
Input :
s1 = "ddcf"
s2 = "cedk"
Output : 2
Explanation : Here, we need to change two characters
in either of the strings to make them identical. We
can change 'd' and 'f' in s1 or 'e' and 'k' in s2.
Assumption: Length of both the Strings is considered similar
C++
// C++ Program to find minimum number // of manipulations required to make // two strings identical #include <bits/stdc++.h> using namespace std; // Counts the no of manipulations // required int countManipulations(string s1, string s2) { int count = 0; // store the count of character int char_count[26]; for (int i = 0; i < 26; i++) { char_count[i] = 0; } // iterate though the first String // and update count for (int i = 0; i < s1.length(); i++) char_count[s1[i] - 'a']++; // iterate through the second string // update char_count. // if character is not found in // char_count then increase count for (int i = 0; i < s2.length(); i++) { char_count[s2[i] - 'a']--; } for(int i = 0; i < 26; ++i) { if(char_count[i] != 0) { count+=abs(char_count[i]); } } return count; } // Driver code int main() { string s1 = "ddcf"; string s2 = "cedk"; cout<<countManipulations(s1, s2); } // This code is contributed by vt_m. |
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Java
// Java Program to find minimum number of manipulations// required to make two strings identicalpublic class Similar_strings { // Counts the no of manipulations required static int countManipulations(String s1, String s2) { int count = 0; // store the count of character int char_count[] = new int[26]; // iterate though the first String and update // count for (int i = 0; i < s1.length(); i++) char_count[s1.charAt(i) - 'a']++; // iterate through the second string // update char_count. // if character is not found in char_count // then increase count for (int i = 0; i < s2.length(); i++) { char_count[s2.charAt(i) - 'a']--; } for(int i = 0; i < 26; ++i) { if(char_count[i] != 0) { count+=abs(char_count[i]); } } return count; } // Driver code public static void main(String[] args) { String s1 = "ddcf"; String s2 = "cedk"; System.out.println(countManipulations(s1, s2)); }} |
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Python 3
# Python3 Program to find minimum number # of manipulations required to make # two strings identical # Counts the no of manipulations # required def countManipulations(s1, s2): count = 0 # store the count of character char_count = [0] * 26 for i in range(26): char_count[i] = 0 # iterate though the first String # and update count for i in range(len( s1)): char_count[ord(s1[i]) - ord('a')] += 1 # iterate through the second string # update char_count. # if character is not found in # char_count then increase count for i in range(len(s2)): char_count[ord(s2[i]) - ord('a')] -= 1 for i in range(26): if char_count[i] != 0: count += abs(char_count[i]) return count# Driver code if __name__ == "__main__": s1 = "ddcf" s2 = "cedk" print(countManipulations(s1, s2))# This code is contributed by ita_c |
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C#
// C# Program to find minimum number// of manipulations required to make// two strings identicalusing System;public class GFG { // Counts the no of manipulations // required static int countManipulations(string s1, string s2) { int count = 0; // store the count of character int []char_count = new int[26]; // iterate though the first String // and update count for (int i = 0; i < s1.Length; i++) char_count[s1[i] - 'a']++; // iterate through the second string // update char_count. // if character is not found in // char_count then increase count for (int i = 0; i < s2.Length; i++) char_count[s2[i] - 'a']--; for(int i = 0; i < 26; ++i) { if(char_count[i] != 0) { count+=abs(char_count[i]); } } return count; } // Driver code public static void Main() { string s1 = "ddcf"; string s2 = "cedk"; Console.WriteLine( countManipulations(s1, s2)); }}// This code is contributed by vt_m. |
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PHP
<?php// PHP Program to find minimum number // of manipulations required to make // two strings identical // Counts the no of manipulations // required function countManipulations($s1, $s2) { $count = 0; // store the count of character $char_count = array_fill(0, 26, 0); // iterate though the first String // and update count for ($i = 0; $i < strlen($s1); $i++) $char_count[ord($s1[$i]) - ord('a')] += 1; // iterate through the second string // update char_count. // if character is not found in // char_count then increase count for ($i = 0; $i < strlen($s2); $i++) { $char_count[ord($s2[$i]) - ord('a')] -= 1; } for ($i = 0; $i < 26; $i++) { if($char_count[i]!=0) { $count+=abs($char_count[i]); } } return $count; } // Driver code$s1 = "ddcf"; $s2 = "cedk"; echo countManipulations($s1, $s2); // This code is contributed by Ryuga?> |
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Output:
2
Time Complexity: O(n), where n is the length of the string.
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