Minimum Number of Steps to Make Two Strings Anagram in C++

Suppose we have two equal-size strings s and t. In one step we can choose any character of t and replace it with another character. We have to find the minimum number of steps required to make t an anagram of s. Note: An Anagram of a string is a string that contains the same characters with a different (or the same) ordering.

So if the input is like - “yxy” and “xyx”, then the output will be 1, as only one character is needed to be replaced.

To solve this, we will follow these steps −

n := size of characters in s
make a map m, and fill this with the frequency of each character present in s, make another map m2, and fill this with the frequency of each character present in t
ret := n
for each key-value pair it in m
- x := minimum of value of it, and value of m2[value of it]
- decrease ret by 1
return ret

Example (C++)

Let us see the following implementation to get a better understanding −

Live Demo

#include <bits/stdc++.h>
using namespace std;
class Solution {
public:
   int minSteps(string s, string t) {
      int n = s.size();
      map <char, int> m1;
      for(int i = 0; i < s.size(); i++){
         m1[s[i]]++;
      }
      int ret = n;
      map <char, int> m2;
      for(int i = 0; i < t.size(); i++){
         m2[t[i]]++;
      }
      map <char, int> :: iterator it = m1.begin();
      while(it != m1.end()){
         //cout << it->first << " " << it->second << " " << m2[it->first] << endl;
         int x = min(it->second, m2[it->first]);
         ret -= x;
         it++;
      }
      return ret;
   }
};
main(){
   Solution ob;
   cout << (ob.minSteps("yxy", "xyx"));
}