# Program to find value of K for K-Similar Strings in C++

PythonServer Side ProgrammingProgramming

Suppose we have two strings s and t. These two strings are K-similar when we can swap the positions of two letters in s exactly K times so that the resulting string is t. We have two anagrams s and t, and we have to find the smallest K for which s and t are K-similar.

So, if the input is like s = "abc", t = "bac", then the output will be 1.

To solve this, we will follow these steps −

• Define a function swapp(), this will take string s, i, j,

• x := s[i], y := s[j]

• s[i] := y, s[j] := x

• From the main method do the following −

• if A is same as B, then:, return 0

• Define one set visited

• insert A into visited

• Define one queue q, insert A into q

• for initialize lvl := 1, when not q is empty, update (increase lvl by 1), do −

• sz := size of q

• while sz is non-zero, decrease sz by 1 in each iteration, do:

• curr := first element of q

• delete element from q

• i := 0

• while (i < size of curr and curr[i] is same as B[i]), do

• (increase i by 1)

• for initialize j := i + 1, when j < size of curr, update (increase j by 1), do

• if curr[i] is same as curr[j], then:

• if curr[j] is not equal to B[i], then:

• if curr[j] is same as B[j], then:

• swapp(curr, i, j)

• if curr is same as B, then:

• return lvl

• if not call count(curr) of visited, then:

• insert curr into visited

• insert curr into q

• swapp(curr, i, j)

• return -1

## Example

Let us see the following implementation to get better understanding

#include <bits/stdc++.h>
using namespace std;

class Solution {
public:
int kSimilarity(string A, string B) {
if (A == B)
return 0;
unordered_set<string> visited;
visited.insert(A);
queue<string> q;
q.push(A);
for (int lvl = 1; !q.empty(); lvl++) {
int sz = q.size();
while (sz--) {
string curr = q.front();
q.pop();
int i = 0;
while (i < curr.size() && curr[i] == B[i])
i++;
for (int j = i + 1; j < curr.size(); j++) {
if (curr[i] == curr[j])
continue;
if (curr[j] != B[i])
continue;
if (curr[j] == B[j])
continue;
swapp(curr, i, j);
if (curr == B)
return lvl;
if (!visited.count(curr)) {
visited.insert(curr);
q.push(curr);
}
swapp(curr, i, j);
}
}
}
return -1;
}
void swapp(string &s, int i, int j) {
char x = s[i];
char y = s[j];
s[i] = y;
s[j] = x;
}
};

main(){
Solution ob;
cout << (ob.kSimilarity("abc", "bac"));
}

## Input

"abc", "bac"

## Output

1
Published on 07-Oct-2021 08:22:56