# Minimum number of given operations required 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

Updated on: 31-Oct-2019

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