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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 −
#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"));
}Input
"yxy" "xyx"
Output
1
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