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:

          • Ignore following part, skip to the next iteration

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

          • Ignore following part, skip to the next iteration

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

          • Ignore following part, skip to the next iteration

        • 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


Let us see the following implementation to get better understanding

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

class Solution {
   int kSimilarity(string A, string B) {
      if (A == B)
         return 0;
      unordered_set<string> visited;
      queue<string> q;
      for (int lvl = 1; !q.empty(); lvl++) {
         int sz = q.size();
         while (sz--) {
            string curr = q.front();
            int i = 0;
            while (i < curr.size() && curr[i] == B[i])
            for (int j = i + 1; j < curr.size(); j++) {
               if (curr[i] == curr[j])
               if (curr[j] != B[i])
               if (curr[j] == B[j])
               swapp(curr, i, j);
               if (curr == B)
                  return lvl;
               if (!visited.count(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;

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


"abc", "bac"


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