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.
C++
// CPP 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 // required int 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 program int 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 characters import 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 characters MAX_CHAR = 26 # return the count of manipulations # required def 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 Code if __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 characters using 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. |
Output:
2
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