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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