Check if a String can be converted to another by inserting character same as both neighbours

Last Updated : 23 Jul, 2025

Given 2 strings A and B . The task is to check if A can be converted into B by performing the following operation any number of times :

Choose an index from 0 to N - 2 (suppose N is the length of A). If the character at ith index is equal to the character at (i + 1 )th index, then insert the same character between the ith and (i + 1)th index.

Examples:

Input: A = "abbaac", B = "abbbbaaac"
Output: YES
Explanation: A can be converted into B by performing the operations as follows:

Insert b between 2nd and 3rd characters of A. ( A becomes abbbaac )
Insert b between 2nd and 3rd characters of A.( A becomes abbbbaac )
Insert a between 6th and 7th characters of A. ( A becomes abbbbaaac )

Input: A = "xyzz", B = "xyyzz"
Output: NO

Approach: The given problem can be solved by applying the concept of Run Length Encoding on given strings. For example, strings in 1st example can be encoded as follows :

A = {(a, 1), (b, 2), (a, 2), (c, 1)}
B = {(a, 1), (b, 4), (a, 3), (c, 1)}

Let's generalize the encoding of 2 strings as :

A = {(a1, x1), (a2, x2), (a3, x3).....(aN, xN)}
B = {(b1, y1), (b2, y2), (b3, y3)....(bM, yM)}

Now, A can be made equal to B, if the following 3 conditions are satisfied:

N = M
a_i= b_i, for i = 1, 2, 3....N
Either (x_i= y_i) or (x_i< y_iand x_i ? 2) for i = 1, 2, 3....N

So, the problem can be solved easily by following the below steps:

Apply Run Length Encoding on both strings.
Check if the encodings of the strings satisfy above mentioned 3 conditions.

Below is the implementation for the above approach:

C++

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

// Function to perform Run Length Encoding
// on a string S and store that
// in a vector V
void runLengthEncoding(string S, vector<pair<char, int> >& V)
{
    int count = 1;
    for (int i = 1; i < S.length(); i++) {
        if (S[i] != S[i - 1]) {
            V.push_back({ S[i - 1], count });
            count = 0;
        }
        count++;
    }
    V.push_back({ S.back(), count });
}

// Function to Check if a string can be
// converted into another by performing
// the given operation any number of times
bool isPossible(string A, string B)
{

    // Declaring vectors V1 and V2 to store
    // Run Length Encoding of Strings A and B
    vector<pair<char, int> > V1, V2;
    runLengthEncoding(A, V1);
    runLengthEncoding(B, V2);

    // Checking for 1st condition
    if (V1.size() != V2.size()) {
        return false;
    }

    for (int i = 0; i < V1.size(); i++) {
        // Checking for second condition
        if (V1[i].first != V2[i].first) {
            return false;
        }

        // Checking for third condition
        if (!(V1[i].second == V2[i].second || (V1[i].second < V2[i].second && V1[i].second >= 2))) {
            return false;
        }
    }

    // If all three conditions are
    // satisfied, return true
    return true;
}

// Driver Code
int main()
{
    string A = "abbaac";
    string B = "abbbbaaac";

    // Function Call
    bool answer = isPossible(A, B);
    if (answer == 1) {
        cout << "YES" << endl;
    }
    else {
        cout << "NO" << endl;
    }
}

Java

// Java code for the above approach
import java.util.*;

public class GFG
{
  
    // Function to perform Run Length Encoding
    // on a string S and store that
    // in a vector V
    static void runLengthEncoding(
        String S, ArrayList<pair<Character, Integer> > V)
    {
        int count = 1;
        for (int i = 1; i < S.length(); i++) {
            if (S.charAt(i) != S.charAt(i - 1)) {
                V.add(new pair<Character, Integer>(
                    S.charAt(i - 1), count));
                count = 0;
            }
            count++;
        }
        V.add(new pair<Character, Integer>(
            S.charAt(S.length() - 1), count));
    }

    // Function to Check if a string can be
    // converted into another by performing
    // the given operation any number of times
    static boolean isPossible(String A, String B)
    {
      
        // Declaring vectors V1 and V2 to store
        // Run Length Encoding of Strings A and B
        ArrayList<pair<Character, Integer> > V1
            = new ArrayList<pair<Character, Integer> >();
        ArrayList<pair<Character, Integer> > V2
            = new ArrayList<pair<Character, Integer> >();
        runLengthEncoding(A, V1);
        runLengthEncoding(B, V2);

        // Checking for 1st condition
        if (V1.size() != V2.size()) {
            return false;
        }

        for (int i = 0; i < V1.size(); i++) {
            // Checking for second condition
            if (V1.get(i).first != V2.get(i).first) {
                return false;
            }

            // Checking for third condition
            if (!(V1.get(i).second == V2.get(i).second
                  || (V1.get(i).second < V2.get(i).second
                      && V1.get(i).second >= 2))) {
                return false;
            }
        }
      
        // If all three conditions are satisfied, return
        // true
        return true;
    }

    // Driver Code
    public static void main(String[] args)
    {
        String A = "abbaac";
        String B = "abbbbaaac";

        // Function Call
        boolean answer = isPossible(A, B);
        if (answer == true) {
            System.out.println("YES");
        }
        else {
            System.out.println("NO");
        }
    }
}

// Pair Class
class pair<T, U> {
    T first;
    U second;

    public pair(T first, U second)
    {
        this.first = first;
        this.second = second;
    }
}

// This code is contributed by Tapesh(tapeshdua420)

Python3

# Python code for the above approach

# Function to perform Run Length Encoding
# on a string S and store that
# in a vector V
def runLengthEncoding(S, V):
    count = 1
    for i in range(1, len(S)):
        if S[i] != S[i - 1]:
            V.append((S[i - 1], count))
            count = 0
        count += 1
    V.append((S[-1], count))

# Function to Check if a string can be
# converted into another by performing
# the given operation any number of times


def isPossible(A, B):
    # Declaring vectors V1 and V2 to store
    # Run Length Encoding of Strings A and B
    V1 = []
    V2 = []
    runLengthEncoding(A, V1)
    runLengthEncoding(B, V2)

    # Checking for 1st condition
    if len(V1) != len(V2):
        return False

    for i in range(len(V1)):
        # Checking for second condition
        if V1[i][0] != V2[i][0]:
            return False

        # Checking for third condition
        if not (V1[i][1] == V2[i][1] or (V1[i][1] < V2[i][1] and V1[i][1] >= 2)):
            return False

        # If all three conditions are satisfied, return true
        return True

# Driver Code
if __name__ == '__main__':
    A = "abbaac"
    B = "abbbbaaac"
    answer = isPossible(A, B)
    print("YES") if answer else print("NO")

# This code is contributed by Tapesh(tapeshdua420)

// C# implementation 
using System;
using System.Collections.Generic;

// represent pair
public class Pair<T, U> {
  public Pair() {
  }

  public Pair(T first, U second) {
    this.First = first;
    this.Second = second;
  }

  public T First { get; set; }
  public U Second { get; set; }
};

public class GFG{

  // Function to perform Run Length Encoding
  // on a string S and store that
  // in a vector V
  public static void runLengthEncoding(string S, List<Pair<char, int>> V)
  {
    // Pair<String, int> pair = new Pair<String, int>("test", 2);
    int count = 1;
    for (int i = 1; i < S.Length; i++) {
      if (S[i] != S[i - 1]) {
        Pair<char, int> pair = new Pair<char, int>(S[i - 1], count);
        V.Add(pair);
        count = 0;
      }
      count++;
    }
    Pair<char, int> pair2 = new Pair<char, int>(S[S.Length -1], count);

    V.Add(pair2);
  }

  // Function to Check if a string can be
  // converted into another by performing
  // the given operation any number of times
  public static bool isPossible(string A, string B)
  {

    // Declaring vectors V1 and V2 to store
    // Run Length Encoding of Strings A and B
    //vector<pair<char, int> > V1, V2;
    var V1 = new List<Pair<char, int>>();
    var V2 = new List<Pair<char, int>>();
    runLengthEncoding(A, V1);
    runLengthEncoding(B, V2);

    // Checking for 1st condition
    if (V1.Count != V2.Count) {
      return false;
    }

    for (int i = 0; i < V1.Count; i++) {
      // Checking for second condition
      if (V1[i].First != V2[i].First) {
        return false;
      }

      // Checking for third condition
      if (!(V1[i].Second == V2[i].Second || (V1[i].Second < V2[i].Second && V1[i].Second >= 2))) {
        return false;
      }
    }

    // If all three conditions are
    // satisfied, return true
    return true;
  }


  static public void Main (){

    // Code
    string A = "abbaac";
    string B = "abbbbaaac";

    // Function Call
    bool answer = isPossible(A, B);
    if (answer == true) {
      Console.WriteLine("YES");
    }
    else {
      Console.WriteLine("NO");
    }
  }
}

// this code is contributed by ksam24000

JavaScript

// Javascript code for the above approach

// Function to perform Run Length Encoding
// on a string S and store that
// in a vector V
function runLengthEncoding(S, V)
{
    let count = 1;
    for (let i = 1; i < S.length; i++) {
        if (S[i] != S[i - 1]) {
            V.push({ "first":S[i - 1],
                     "second":count });
            count = 0;
        }
        count++;
    }
    V.push({ "first":S[S.length-1], 
             "second":count });
}

// Function to Check if a string can be
// converted into another by performing
// the given operation any number of times
function isPossible( A, B)
{

    // Declaring vectors V1 and V2 to store
    // Run Length Encoding of Strings A and B
    let V1 = [];
    let V2 = [];
    runLengthEncoding(A, V1);
    runLengthEncoding(B, V2);

    // Checking for 1st condition
    if (V1.length != V2.length) {
        return false;
    }

    for (let i = 0; i < V1.length; i++) {
        // Checking for second condition
        if (V1[i].first != V2[i].first) {
            return false;
        }

        // Checking for third condition
        if (!(V1[i].second == V2[i].second || (V1[i].second < V2[i].second && V1[i].second >= 2))) {
            return false;
        }
    }

    // If all three conditions are
    // satisfied, return true
    return true;
}

// Driver Code

let A = "abbaac";
let B = "abbbbaaac";

// Function Call
let answer = isPossible(A, B);
if (answer == 1) {
    console.log("YES");
}
else {
    console.log("NO");
}

// This code is contributed by akashish__

Output

YES

Time Complexity: O(M + N), where M and N are lengths of string A and B respectively
Auxiliary Space: O(N)

Analysis of Algorithms

kamabokogonpachiro

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Check if a String can be converted to another by inserting character same as both neighbours

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