Skip to content
Related Articles

Related Articles

Improve Article

Check whether str1 can be converted to str2 with the given operations

  • Last Updated : 13 May, 2021
Geek Week

Given two binary strings str1 and str2. The task is to check whether it is possible to convert str1 to str2 by applying the below operation an arbitrary number of times on str1. 
In each operation, one can combine two consecutive 0’s into a single 1
Examples: 
 

Input: str1 = “00100”, str2 = “111” 
Output: Yes 
Combine first two zeros to one and combine last two zeros to one.
Input: str1 = “00”, str2 = “000” 
Output: No 
It is not possible to convert str1 to str2. 
 

 

Approach: Let’s process str1 and str2 character by character from left to right in parallel. Let’s use two indices i and j: the index i is for str1 and the index j is for str2. Now, there are two cases: 
 

  1. If str1[i] = str2[j] then increment both i and j.
  2. If str1[i] != str2[j] then, 
    • If there are two consecutive 0’s in str1 i.e. str1[i] = 0 and str1[i + 1] = 0 and str2[j] = 1 which means both the zeroes can be combined to match with the 1 in str2. So, increment i by 2 and j by 1.
    • Else str1 cannot be converted to str2.
  3. If in the end both i and j are at the end of their respective strings i.e. str1 and str2 then the answer is Yes else the answer is No.

Below is the implementation of the above approach: 
 



C++




// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
// Function that returns true if str1 can be
// converted to str2 with the given operations
bool canConvert(string str1, string str2)
{
    int i = 0, j = 0;
 
    // Traverse from left to right
    while (i < str1.size() && j < str2.size()) {
 
        // If the two characters do not match
        if (str1[i] != str2[j]) {
 
            // If possible to combine
            if (str1[i] == '0' && str2[j] == '1'
                && i + 1 < str1.size()
                && str1[i + 1] == '0') {
                i += 2;
                j++;
            }
 
            // If not possible to combine
            else {
                return false;
            }
        }
 
        // If the two characters match
        else {
            i++;
            j++;
        }
    }
 
    // If possible to convert one string to another
    if (i == str1.size() && j == str2.size())
        return true;
    return false;
}
 
// Driver code
int main()
{
    string str1 = "00100", str2 = "111";
 
    if (canConvert(str1, str2))
        cout << "Yes";
    else
        cout << "No";
 
    return 0;
}

Java




// Java implementation of the approach
class GFG
{
 
    // Function that returns true if str1 can be
    // converted to str2 with the given operations
    static boolean canConvert(String str1, String str2)
    {
        int i = 0, j = 0;
 
        // Traverse from left to right
        while (i < str1.length() && j < str2.length())
        {
 
            // If the two characters do not match
            if (str1.charAt(i) != str2.charAt(j))
            {
 
                // If possible to combine
                if (str1.charAt(i) == '0' && str2.charAt(j) == '1'
                    && i + 1 < str1.length() && str1.charAt(i+1) == '0')
                {
                    i += 2;
                    j++;
                }
 
                // If not possible to combine
                else
                {
                    return false;
                }
            }
 
            // If the two characters match
            else
            {
                i++;
                j++;
            }
        }
 
        // If possible to convert one string to another
        if (i == str1.length() && j == str2.length())
            return true;
        return false;
    }
 
    // Driver code
    public static void main(String[] args)
    {
 
        String str1 = "00100", str2 = "111";
 
        if (canConvert(str1, str2))
            System.out.println("Yes");
        else
            System.out.println("No");
    }
}
 
// This code contributed by Rajput-Ji

Python3




# Python implementation of the approach
 
# Function that returns true if str1 can be
# converted to str2 with the given operations
def canConvert(str1, str2):
    i, j = 0, 0;
 
    # Traverse from left to right
    while (i < len(str1) and j < len(str2)):
 
        # If the two characters do not match
        if (str1[i] != str2[j]):
 
            # If possible to combine
            if (str1[i] == '0' and str2[j] == '1'
                and i + 1 < len(str1)
                and str1[i + 1] == '0'):
                i += 2;
                j+=1;
 
            # If not possible to combine
            else:
                return False;
        # If the two characters match
        else:
            i += 1;
            j += 1;
             
    # If possible to convert one string to another
    if (i == len(str1) and j == len(str2)):
        return True;
    return False;
 
# Driver code
str1 = "00100";
str2 = "111";
 
if (canConvert(str1, str2)):
    print("Yes");
else:
    print("No");
 
# This code is contributed by 29AjayKumar

C#




// C# implementation of the approach
using System;
 
class GFG
{
 
    // Function that returns true if str1 can be
    // converted to str2 with the given operations
    static bool canConvert(string str1, string str2)
    {
        int i = 0, j = 0;
 
        // Traverse from left to right
        while (i < str1.Length && j < str2.Length)
        {
 
            // If the two characters do not match
            if (str1[i] != str2[j])
            {
 
                // If possible to combine
                if (str1[i] == '0' && str2[j] == '1'
                    && i + 1 < str1.Length && str1[i+1] == '0')
                {
                    i += 2;
                    j++;
                }
 
                // If not possible to combine
                else
                {
                    return false;
                }
            }
 
            // If the two characters match
            else
            {
                i++;
                j++;
            }
        }
 
        // If possible to convert one string to another
        if (i == str1.Length && j == str2.Length)
            return true;
        return false;
    }
 
    // Driver code
    public static void Main()
    {
 
        string str1 = "00100", str2 = "111";
 
        if (canConvert(str1, str2))
            Console.WriteLine("Yes");
        else
            Console.WriteLine("No");
    }
}
 
// This code is contributed by AnkitRai01

Javascript




<script>
      // JavaScript implementation of the approach
      // Function that returns true if str1 can be
      // converted to str2 with the given operations
      function canConvert(str1, str2) {
        var i = 0,
          j = 0;
 
        // Traverse from left to right
        while (i < str1.length && j < str2.length) {
          // If the two characters do not match
          if (str1[i] !== str2[j]) {
            // If possible to combine
            if (
              str1[i] === "0" &&
              str2[j] === "1" &&
              i + 1 < str1.length &&
              str1[i + 1] === "0"
            ) {
              i += 2;
              j++;
            }
 
            // If not possible to combine
            else {
              return false;
            }
          }
 
          // If the two characters match
          else {
            i++;
            j++;
          }
        }
 
        // If possible to convert one string to another
        if (i === str1.length && j === str2.length) return true;
        return false;
      }
 
      // Driver code
      var str1 = "00100",
        str2 = "111";
 
      if (canConvert(str1, str2)) document.write("Yes");
      else document.write("No");
    </script>
Output: 
Yes

 

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.




My Personal Notes arrow_drop_up
Recommended Articles
Page :