Minimum Number of Operations to Make Two Strings Equal using C++

Problem statement

Given two strings str1 and str2, both strings contain characters ‘a’ and ‘b’. Both strings are of equal lengths. There is one _ (empty space) in both the strings. The task is to convert the first string into the second string by doing the minimum number of the following operations −

• If _ is at a position I then _ can be swapped with a character at position i+1 or i-1

• If characters at positions i+1 and i+2 are different then _ can be swapped with a character at position i+1 or i+2

• Similarly, if characters at positions i-1 and i-2 are different then _ can be swapped with a character at position i-1 or i-2

If str1 = “aba_a” and str2 = “_baaa” then we require 2 moves to transform str1 to str2 −

1. str1 = “ab_aa” (Swapped str1[2] with str1[3])
2. str2 = “_baaa” (Swapped str1[0] with str1[2])

Algorithm

1. Apply a simple Breadth First Search over the string and an element of the queue used for BFS will contain the pair str, pos where pos is the position of _ in the string str.
2. Also maintain a map namely ‘vis’ which will store the string as key and the minimum moves to get to the string as value.
3. For every string str from the queue, generate a new string tmp based on the four conditions given and update the vis map as vis[tmp] = vis[str] + 1.
4. Repeat the above steps until the queue is empty or the required string is generated i.e. tmp == B
5. If the required string is generated, then return vis[str] + 1 which is the minimum number of operations required to change A to B.

Example

#include <iostream>
#include <string>
#include <unordered_map>
#include <queue>
using namespace std;
int transformString(string str, string f){
   unordered_map<string, int> vis;
   int n;
   n = str.length();
   int pos = 0;
   for (int i = 0; i < str.length(); i++) {
      if (str[i] == '_') {
         pos = i;
         break;
      }
   }
   queue<pair<string, int> > q;
   q.push({ str, pos });
   vis[str] = 0;
   while (!q.empty()) {
      string ss = q.front().first;
      int pp = q.front().second;
      int dist = vis[ss];
      q.pop();
      if (pp > 0) {
         swap(ss[pp], ss[pp - 1]);
         if (!vis.count(ss)) {
            if (ss == f) {
               return dist + 1;
               break;
            }
            vis[ss] = dist + 1;
            q.push({ ss, pp - 1 });
         }
         swap(ss[pp], ss[pp - 1]);
      }
      if (pp < n - 1) {
         swap(ss[pp], ss[pp + 1]);
         if (!vis.count(ss)) {
         if (ss == f) {
            return dist + 1;
            break;
         }
         vis[ss] = dist + 1;
         q.push({ ss, pp + 1 });
      }
      swap(ss[pp], ss[pp + 1]);
   }
   if (pp > 1 && ss[pp - 1] != ss[pp - 2]) {
      swap(ss[pp], ss[pp - 2]);
      if (!vis.count(ss)) {
         if (ss == f) {
            return dist + 1;
            break;
         }
         vis[ss] = dist + 1;
         q.push({ ss, pp - 2 });
      }
      swap(ss[pp], ss[pp - 2]);
   }
   if (pp < n - 2 && ss[pp + 1] != ss[pp + 2]) {
      swap(ss[pp], ss[pp + 2]);
      if (!vis.count(ss)) {
         if (ss == f) {
            return dist + 1;
            break;
         }
         vis[ss] = dist + 1;
         q.push({ ss, pp + 2 });
      }
      swap(ss[pp], ss[pp + 2]);
      }
   }
   return 0;
}
int main(){
   string str1 = "aba_a";
   string str2 = "_baaa";
   cout << "Minimum required moves: " << transformString(str1, str2) << endl;
   return 0;
}

Output

When you compile and execute the above program. It generates the following output −

Minimum required moves: 2