Minimum number of given operations required to make two strings equal

Last Updated : 11 Jul, 2025

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

If _ is at 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.

Examples:

Input: A = "aba_a", B = "_baaa"
Output: 2
Move 1 : A = "ab_aa" (Swapped A[2] with A[3]) Move 2 : A = "_baaa" (Swapped A[0] with A[2])

Input: A = "a_b", B = "ab_"
Output: 1

Source: Directi Interview Set 7

Approach:

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.
Also maintain a map vis which will store the string as key and the minimum moves to get to the string as value.
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.
Repeat the above steps until the queue is empty or the required string is generated i.e. tmp == B
If the required string is generated then return vis[str] + 1 which is the minimum number of operations required to change A to B.

Below is the implementation of the above approach:

C++

// C++ implementation of the above approach
#include <bits/stdc++.h>
using namespace std;

// Function to return the minimum number of
// operations to convert string A to B
int minOperations(string s, string f)
{

    // If both the strings are equal then
    // no operation needs to be performed
    if (s == f)
        return 0;

    unordered_map<string, int> vis;

    int n;

    n = s.length();
    int pos = 0;
    for (int i = 0; i < s.length(); i++) {
        if (s[i] == '_') {

            // store the position of '_'
            pos = i;
            break;
        }
    }

    // to store the generated string at every
    // move and the position of '_' within it
    queue<pair<string, int> > q;

    q.push({ s, pos });

    // vis will store the minimum operations
    // to reach that particular string
    vis[s] = 0;

    while (!q.empty()) {
        string ss = q.front().first;
        int pp = q.front().second;

        // minimum moves to reach the string ss
        int dist = vis[ss];
        q.pop();

        // try all 4 possible operations

        // if '_' can be swapped with
        // the character on it's left
        if (pp > 0) {

            // swap with the left character
            swap(ss[pp], ss[pp - 1]);

            // if the string is generated
            // for the first time
            if (!vis.count(ss)) {

                // if generated string is
                // the required string
                if (ss == f) {
                    return dist + 1;
                    break;
                }

                // update the distance for the
                // currently generated string
                vis[ss] = dist + 1;
                q.push({ ss, pp - 1 });
            }

            // restore the string before it was
            // swapped to check other cases
            swap(ss[pp], ss[pp - 1]);
        }

        // swap '_' with the character
        // on it's right this time
        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 '_' can be swapped
        // with the character 'i+2'
        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 '_' can be swapped
        // with the character at 'i+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]);
        }
    }
}

// Driver code
int main()
{

    string A = "aba_a";
    string B = "_baaa";

    cout << minOperations(A, B);

    return 0;
}

Java

// Java implementation of the above approach
import java.util.*;

class GFG {

  // Function to return the minimum number of
  // operations to convert string A to B
  static int minOperations(String s, String f)
  {

    // If both the strings are equal then
    // no operation needs to be performed
    if (s.equals(f))
      return 0;

    Map<String, Integer> vis = new HashMap<>();

    int n;

    n = s.length();
    int pos = 0;
    for (int i = 0; i < s.length(); i++) {
      if (s.charAt(i) == '_') {

        // store the position of '_'
        pos = i;
        break;
      }
    }

    // to store the generated string at every
    // move and the position of '_' within it
    Queue<Map.Entry<String, Integer>> q
      = new LinkedList<Map.Entry<String, Integer>>();

    q.add(new AbstractMap.SimpleEntry<String, Integer>(s, pos));

    // vis will store the minimum operations
    // to reach that particular string
    vis.put(s, 0);

    while (!q.isEmpty()) {
      String ss = q.peek().getKey();
      int pp = q.peek().getValue();

      // minimum moves to reach the string ss
      int dist = vis.get(ss);
      q.remove();

      // try all 4 possible operations

      // if '_' can be swapped with
      // the character on it's left
      if (pp > 0) {

        // swap with the left character
        char[] ch = ss.toCharArray();
        char temp = ch[pp];
        ch[pp] = ch[pp - 1];
        ch[pp - 1] = temp;
        ss = String.valueOf(ch);

        // if the string is generated
        // for the first time
        if (!vis.containsKey(ss)) {

          // if generated string is
          // the required string
          if (ss.equals(f)) {
            return dist + 1;
          }

          // update the distance for the
          // currently generated string
          vis.put(ss, dist + 1);
          q.add(new AbstractMap.SimpleEntry<String, Integer>(ss, pp - 1));
        }

        // restore the string before it was
        // swapped to check other cases
        ch = ss.toCharArray();
        temp = ch[pp];
        ch[pp] = ch[pp - 1];
        ch[pp - 1] = temp;
        ss = String.valueOf(ch);
      }

      // swap '_' with the character
      // on it's right this time
      if (pp < n - 1) {
        char[] ch = ss.toCharArray();
        char temp = ch[pp];
        ch[pp] = ch[pp + 1];
        ch[pp + 1] = temp;
        ss = String.valueOf(ch);
        if (!vis.containsKey(ss)) {
          if (ss.equals(f)) {
            return dist + 1;
          }
          vis.put(ss, dist + 1);
          q.add(new AbstractMap.SimpleEntry<String, Integer>(ss, pp + 1));
        }
        ch = ss.toCharArray();
        temp = ch[pp];
        ch[pp] = ch[pp + 1];
        ch[pp + 1] = temp;
        ss = String.valueOf(ch);
      }

      // if '_' can be swapped
      // with the character 'i+2'
      if (pp > 1 && ss.charAt(pp - 1) != ss.charAt(pp - 2)) {
        char[] ch = ss.toCharArray();
        char temp = ch[pp];
        ch[pp] = ch[pp - 2];
        ch[pp - 2] = temp;
        ss = String.valueOf(ch);
        if (!vis.containsKey(ss)) {
          if (ss.equals(f)) {
            return dist + 1;
          }
          vis.put(ss, dist + 1);
          q.add(new AbstractMap.SimpleEntry<String, Integer>(ss, pp - 2));
        }
        ch = ss.toCharArray();
        temp = ch[pp];
        ch[pp] = ch[pp - 2];
        ch[pp - 2] = temp;
        ss = String.valueOf(ch);
      }

      // if '_' can be swapped
      // with the character at 'i+2'
      if (pp < n - 2 && ss.charAt(pp + 1) != ss.charAt(pp + 2)) {
        char[] ch = ss.toCharArray();
        char temp = ch[pp];
        ch[pp] = ch[pp + 2];
        ch[pp + 2] = temp;
        ss = String.valueOf(ch);
        if (!vis.containsKey(ss)) {
          if (ss.equals(f)) {
            return dist + 1;
          }
          vis.put(ss, dist + 1);
          q.add(new AbstractMap.SimpleEntry<String, Integer>(ss, pp + 2));
        }
        ch = ss.toCharArray();
        temp = ch[pp];
        ch[pp] = ch[pp + 2];
        ch[pp + 2] = temp;
        ss = String.valueOf(ch);
      }
    }
    return -1;
  }

  // Driver code
  public static void main(String[] args)
  {

    String A = "aba_a";
    String B = "_baaa";

    System.out.println(minOperations(A, B));
  }
}

Python3

# Python3 implementation of the approach
from collections import deque

# Function to return the minimum number of
# operations to convert string A to B
def minOperations(s: str, f: str) -> int:

    # If both the strings are equal then
    # no operation needs to be performed
    if s == f:
        return 0

    vis = dict()
    n = len(s)
    pos = 0
    for i in range(n):
        if s[i] == '_':

            # store the position of '_'
            pos = i
            break

    # to store the generated string at every
    # move and the position of '_' within it
    q = deque()
    q.append((s, pos))

    # vis will store the minimum operations
    # to reach that particular string
    vis[s] = 0

    while q:
        ss = q[0][0]
        pp = q[0][1]

        # minimum moves to reach the string ss
        dist = vis[ss]
        q.popleft()

        ss = list(ss)

        # try all 4 possible operations

        # if '_' can be swapped with
        # the character on it's left
        if pp > 0:

            # swap with the left character
            ss[pp], ss[pp - 1] = ss[pp - 1], ss[pp]
            ss = ''.join(ss)

            # if the string is generated
            # for the first time
            if ss not in vis:

                # if generated string is
                # the required string
                if ss == f:
                    return dist + 1

                # update the distance for the
                # currently generated string
                vis[ss] = dist + 1
                q.append((ss, pp - 1))

            ss = list(ss)

            # restore the string before it was
            # swapped to check other cases
            ss[pp], ss[pp - 1] = ss[pp - 1], ss[pp]
            ss = ''.join(ss)

        # swap '_' with the character
        # on it's right this time
        if pp < n - 1:
            ss = list(ss)
            ss[pp], ss[pp + 1] = ss[pp + 1], ss[pp]
            ss = ''.join(ss)

            if ss not in vis:
                if ss == f:
                    return dist + 1

                vis[ss] = dist + 1
                q.append((ss, pp + 1))

            ss = list(ss)
            ss[pp], ss[pp + 1] = ss[pp + 1], ss[pp]
            ss = ''.join(ss)

        # if '_' can be swapped
        # with the character 'i+2'
        if pp > 1 and ss[pp - 1] != ss[pp - 2]:
            ss = list(ss)
            ss[pp], ss[pp - 2] = ss[pp - 2], ss[pp]
            ss = ''.join(ss)

            if ss not in vis:
                if ss == f:
                    return dist + 1

                vis[ss] = dist + 1
                q.append((ss, pp - 2))

            ss = list(ss)
            ss[pp], ss[pp - 2] = ss[pp - 2], ss[pp]
            ss = ''.join(ss)

        # if '_' can be swapped
        # with the character at 'i+2'
        if pp < n - 2 and ss[pp + 1] != ss[pp + 2]:
            ss = list(ss)
            ss[pp], ss[pp + 2] = ss[pp + 2], ss[pp]
            ss = ''.join(ss)

            if ss not in vis:
                if ss == f:
                    return dist + 1

                vis[ss] = dist + 1
                q.append((ss, pp + 2))

            ss = list(ss)
            ss[pp], ss[pp + 2] = ss[pp + 2], ss[pp]
            ss = ''.join(ss)

# Driver Code
if __name__ == "__main__":

    A = "aba_a"
    B = "_baaa"

    print(minOperations(A, B))

# This code is contributed by
# sanjeev2552

JavaScript

// JavaScript implementation of the approach

// Function to return the minimum number of
// operations to convert string A to B
function minOperations(s, f) {
    
    // If both the strings are equal then
    // no operation needs to be performed
    if (s === f) {
        return 0;
    }

    let vis = {};
    let n = s.length;
    let pos = 0;
    for (let i = 0; i < n; i++) {
        if (s[i] === '_') {
            // store the position of '_'
            pos = i;
            break;
        }
    }
    
    
    // to store the generated string at every
    // move and the position of '_' within it
    let q = [];
    q.push([s, pos]);
    
    
    // vis will store the minimum operations
    // to reach that particular string
    vis[s] = 0;

    while (q.length) {
        let [ss, pp] = q[0];
        
        // minimum moves to reach the string ss
        let dist = vis[ss];
        q.shift();
        ss = ss.split('');
        
        
        // try all 4 possible operations

        // if '_' can be swapped with
        // the character on it's left
        if (pp > 0) {
            
            // swap with the left character
            [ss[pp], ss[pp - 1]] = [ss[pp - 1], ss[pp]];
            ss = ss.join('');
            
            
            // if the string is generated
            // for the first time
            if (!(ss in vis)) {
                
                
                // if generated string is
                // the required string
                if (ss === f) {
                    return dist + 1;
                }
                
                
                // update the distance for the
                // currently generated string
                vis[ss] = dist + 1;
                q.push([ss, pp - 1]);
            }

            ss = ss.split('');
            
            // restore the string before it was
            // swapped to check other cases
            [ss[pp], ss[pp - 1]] = [ss[pp - 1], ss[pp]];
            ss = ss.join('');
        }

        
        // swap '_' with the character
        // on it's right this time
        if (pp < n - 1) {
            ss = ss.split('');
            [ss[pp], ss[pp + 1]] = [ss[pp + 1], ss[pp]];
            ss = ss.join('');

            if (!(ss in vis)) {
                if (ss === f) {
                    return dist + 1;
                }

                vis[ss] = dist + 1;
                q.push([ss, pp + 1]);
            }

            ss = ss.split('');
            [ss[pp], ss[pp + 1]] = [ss[pp + 1], ss[pp]];
            ss = ss.join('');
        }

        
        
        // if '_' can be swapped
        // with the character 'i+2'
        if (pp > 1 && ss[pp - 1] !== ss[pp - 2]) {
            ss = ss.split('');
            [ss[pp], ss[pp - 2]] = [ss[pp - 2], ss[pp]];
            ss = ss.join('');

            if (!(ss in vis)) {
                if (ss === f) {
                    return dist + 1;
                }

                vis[ss] = dist + 1;
                q.push([ss, pp - 2]);
            }

            ss = ss.split('');
            [ss[pp], ss[pp - 2]] = [ss[pp - 2], ss[pp]];
            ss = ss.join('');
        }
        
    
        // if '_' can be swapped
        // with the character at 'i+2'
        if (pp < n - 2 && ss[pp + 1] !== ss[pp + 2]) {
            ss = ss.split('');
            [ss[pp], ss[pp + 2]] = [ss[pp + 2], ss[pp]];
            ss = ss.join('');

            if (!(ss in vis)) {
                if (ss === f) {
                    return dist + 1;
                }

                vis[ss] = dist + 1;
                q.push([ss, pp + 2]);
            }

            ss = ss.split('');
            [ss[pp], ss[pp + 2]] = [ss[pp + 2], ss[pp]];
            ss = ss.join('');
        }
    }
}

// Driver Code
let A = "aba_a";
let B = "_baaa";
console.log(minOperations(A, B))

// This code is contributed by codebraxnzt

using System;
using System.Collections.Generic;

class GFG
{
    // Function to return the minimum number of
    // operations to convert string A to B
    static int minOperations(string s, string f)
    {
        // If both the strings are equal then
        // no operation needs to be performed
        if (s == f)
            return 0;

        Dictionary<string, int> vis = new Dictionary<string, int>();

        int n = s.Length;
        int pos = 0;
        for (int i = 0; i < s.Length; i++)
        {
            if (s[i] == '_')
            {
                // store the position of '_'
                pos = i;
                break;
            }
        }

        // to store the generated string at every
        // move and the position of '_' within it
        Queue<Tuple<string, int>> q = new Queue<Tuple<string, int>>();

        q.Enqueue(new Tuple<string, int>(s, pos));

        // vis will store the minimum operations
        // to reach that particular string
        vis[s] = 0;

        while (q.Count > 0)
        {
            string ss = q.Peek().Item1;
            int pp = q.Peek().Item2;

            // minimum moves to reach the string ss
            int dist = vis[ss];
            q.Dequeue();

            // try all 4 possible operations

            // if '_' can be swapped with
            // the character on it's left
            if (pp > 0)
            {
                // swap with the left character
                char[] arr = ss.ToCharArray();
                char temp = arr[pp];
                arr[pp] = arr[pp - 1];
                arr[pp - 1] = temp;
                ss = new string(arr);

                // if the string is generated
                // for the first time
                if (!vis.ContainsKey(ss))
                {
                    // if generated string is
                    // the required string
                    if (ss == f)
                        return dist + 1;

                    // update the distance for the
                    // currently generated string
                    vis[ss] = dist + 1;
                    q.Enqueue(new Tuple<string, int>(ss, pp - 1));
                }

                // restore the string before it was
                // swapped to check other cases
                arr = ss.ToCharArray();
                temp = arr[pp];
                arr[pp] = arr[pp - 1];
                arr[pp - 1] = temp;
                ss = new string(arr);
            }

            // swap '_' with the character
            // on it's right this time
            if (pp < n - 1)
            {
                char[] arr = ss.ToCharArray();
                char temp = arr[pp];
                arr[pp] = arr[pp + 1];
                arr[pp + 1] = temp;
                ss = new string(arr);
                if (!vis.ContainsKey(ss))
                {
                    if (ss == f)
                        return dist + 1;

                    vis[ss] = dist + 1;
                    q.Enqueue(new Tuple<string, int>(ss, pp + 1));
                }
                arr = ss.ToCharArray();
                temp = arr[pp];
                arr[pp] = arr[pp + 1];
                arr[pp + 1] = temp;
                ss = new string(arr);
            }
            
            // if '_' can be swapped
            // with the character 'i+2'
            if (pp > 1 && ss[pp - 1] != ss[pp - 2]) {
                char[] arr = ss.ToCharArray();
                char temp = arr[pp];
                arr[pp] = arr[pp - 2];
                arr[pp - 2] = temp;
                ss = new string(arr);
                if (!vis.ContainsKey(ss)) {
                    if (ss == f) {
                        return dist + 1;
                    }
                    vis[ss] = dist + 1;
                    q.Enqueue(new Tuple<string, int>(ss, pp - 2));

                }
                arr = ss.ToCharArray();
                temp = arr[pp];
                arr[pp] = arr[pp - 2];
                arr[pp - 2] = temp;
                ss = new string(arr);
            }
    
            // if '_' can be swapped
            // with the character at 'i+2'
            if (pp < n - 2 && ss[pp + 1] != ss[pp + 2]) {
                char[] arr = ss.ToCharArray();
                char temp = arr[pp];
                arr[pp] = arr[pp + 2];
                arr[pp + 2] = temp;
                ss = new string(arr);
                
                if (!vis.ContainsKey(ss)) {
                    if (ss == f) {
                        return dist + 1;
                    }
                    vis[ss] = dist + 1;
                    q.Enqueue(new Tuple<string, int>(ss, pp + 2));
                }
                arr = ss.ToCharArray();
                temp = arr[pp];
                arr[pp] = arr[pp + 2];
                arr[pp + 2] = temp;
                ss = new string(arr);
                
            }
        }
        return -1;
    }
    
    // Driver code
    public static void Main()
    {
    
        string A = "aba_a";
        string B = "_baaa";
    
        Console.Write(minOperations(A, B));
    
    }
}

Output

Analysis of Algorithms

Kushdeep_Mittal

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Article Tags :

Minimum number of given operations required to make two strings equal

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