Given two strings return the value of least number of manipulations needed to ensure both strings have identical characters, i.e., both string become anagram of each other.
Examples:
Input : s1 = "aab"
s2 = "aba"
Output : 2
Explanation : string 1 contains 2 a's and 1 b,
also string 2 contains same characters
Input : s1 = "abc"
s2 = "cdd"
Output : 2
Explanation : string 1 contains 1 a, 1 b, 1 c
while string 2 contains 1 c and 2 d's
so there are 2 different characters
Question Source : Yatra.com Interview Experience | Set 7
The idea is to create a extra count array for both the strings separately and then count the difference in characters.
Implementation:
C++
// C++ program to count least number// of manipulations to have two strings// set of same characters#include <bits/stdc++.h>using namespace std;const int MAX_CHAR = 26;// return the count of manipulations// requiredint leastCount(string s1, string s2, int n){ int count1[MAX_CHAR] = { 0 }; int count2[MAX_CHAR] = { 0 }; // count the number of different // characters in both strings for (int i = 0; i < n; i++) { count1[s1[i] - 'a'] += 1; count2[s2[i] - 'a'] += 1; } // check the difference in characters // by comparing count arrays int res = 0; for (int i = 0; i < MAX_CHAR; i++) { if (count1[i] != 0) { res += abs(count1[i] - count2[i]); } } return res;}// driver programint main(){ string s1 = "abc"; string s2 = "cdd"; int len = s1.length(); int res = leastCount(s1, s2, len); cout << res << endl; return 0;} |
Java
// Java program to count least number// of manipulations to have two// strings set of same charactersimport java.io.*;public class GFG { static int MAX_CHAR = 26; // return the count of manipulations // required static int leastCount(String s1, String s2, int n) { int[] count1 = new int[MAX_CHAR]; int[] count2 = new int[MAX_CHAR]; // count the number of different // characters in both strings for (int i = 0; i < n; i++) { count1[s1.charAt(i) - 'a'] += 1; count2[s2.charAt(i) - 'a'] += 1; } // check the difference in characters // by comparing count arrays int res = 0; for (int i = 0; i < MAX_CHAR; i++) { if (count1[i] != 0) { res += Math.abs(count1[i] - count2[i]); } } return res; } // driver program static public void main(String[] args) { String s1 = "abc"; String s2 = "cdd"; int len = s1.length(); int res = leastCount(s1, s2, len); System.out.println(res); }}// This code is contributed by vt_m. |
Python3
# Python3 program to count least number# of manipulations to have two strings# set of same charactersMAX_CHAR = 26# return the count of manipulations# requireddef leastCount(s1, s2, n): count1 = [0] * MAX_CHAR count2 = [0] * MAX_CHAR # count the number of different # characters in both strings for i in range ( n): count1[ord(s1[i]) - ord('a')] += 1 count2[ord(s2[i]) - ord('a')] += 1 # check the difference in characters # by comparing count arrays res = 0 for i in range (MAX_CHAR): if (count1[i] != 0): res += abs(count1[i] - count2[i]) return res# Driver Codeif __name__ == "__main__": s1 = "abc" s2 = "cdd" l = len(s1) res = leastCount(s1, s2, l) print (res)# This code is contributed by ita_c |
C#
// C# program to count least number// of manipulations to have two strings// set of same charactersusing System;public class GFG { static int MAX_CHAR = 26; // return the count of manipulations // required static int leastCount(string s1, string s2, int n) { int[] count1 = new int[MAX_CHAR]; int[] count2 = new int[MAX_CHAR]; // count the number of different // characters in both strings for (int i = 0; i < n; i++) { count1[s1[i] - 'a'] += 1; count2[s2[i] - 'a'] += 1; } // check the difference in characters // by comparing count arrays int res = 0; for (int i = 0; i < MAX_CHAR; i++) { if (count1[i] != 0) { res += Math.Abs(count1[i] - count2[i]); } } return res; } // driver program static public void Main() { string s1 = "abc"; string s2 = "cdd"; int len = s1.Length; int res = leastCount(s1, s2, len); Console.WriteLine(res); }}// This code is contributed by vt_m. |
Javascript
<script>// Javascript program to count least number// of manipulations to have two// strings set of same characterslet MAX_CHAR = 26;// Return the count of manipulations// requiredfunction leastCount(s1, s2, n){ let count1 = new Array(MAX_CHAR); let count2 = new Array(MAX_CHAR); for(let i = 0; i < MAX_CHAR; i++) { count1[i] = 0; count2[i] = 0; } // Count the number of different // characters in both strings for(let i = 0; i < n; i++) { count1[s1[i].charCodeAt(0) - 'a'.charCodeAt(0)] += 1; count2[s2[i].charCodeAt(0) - 'a'.charCodeAt(0)] += 1; } // Check the difference in characters // by comparing count arrays let res = 0; for(let i = 0; i < MAX_CHAR; i++) { if (count1[i] != 0) { res += Math.abs(count1[i] - count2[i]); } } return res;}// Driver Codelet s1 = "abc";let s2 = "cdd";let len = s1.length;let res = leastCount(s1, s2, len);document.write(res);// This code is contributed by avanitrachhadiya2155 </script> |
Output
2
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