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Check if a string can be emptied by removing all subsequences of the form “10”

  • Difficulty Level : Medium
  • Last Updated : 13 Apr, 2021

Given a binary string str, the task is to check if the string can be emptied by removing all subsequences of the form “10”
 

Examples:

Input: str = “11011000”
Output: Yes
Explanation:Following steps are required to be performed to empty the given string “11011000” → “111000” → “1100” → “10” → “”

Input: 101001
Output: No

Approach: Follow the steps below to solve the problem:



  1. Traverse the string and store all 1s in a Stack.
  2. If str[i] = 1, then store it in the stack.
  3. If str[i] = 0 and stack is not empty, then pop the character at the top of the stack.
  4. Otherwise, return false.
  5. If stack is empty, return true.
  6. Otherwise, return false

Below is the implementation of the above approach:
 

C++




// C++ program for
// the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to find if string
// is reducible to NULL
bool isReducible(string str)
{
    // Length of string
    int N = str.size();
 
    // Stack to store all 1s
    stack<char> s;
 
    // Iterate over the characters
    // of the string
    for (int i = 0; i < N; i++) {
 
        // If current character is 1
        if (str[i] == '1')
 
            // Push it into the stack
            s.push(str[i]);
 
        else if (!s.empty())
 
            // Pop from the stack
            s.pop();
 
        else
            // If the stack is empty
            return false;
    }
 
    return s.empty();
}
 
// Driver Code
int main()
{
    string str = "11011000";
 
    if (isReducible(str))
        cout << "Yes";
 
    else
        cout << "No";
    return 0;
}

Java




// Java program for
// the above approach
import java.io.*;
import java.util.*;
 
class GFG
{
 
  // Function to find if string
  // is reducible to NULL
  static boolean isReducible(String str)
  {
 
    // Length of string
    int N = str.length();
 
    // Stack to store all 1s
    ArrayList<Character> s = new ArrayList<Character>();
 
    // Iterate over the characters
    // of the string
    for (int i = 0; i < N; i++)
    {
 
      // If current character is 1
      if (str.charAt(i) == '1')
 
        // Push it into the stack
        s.add(str.charAt(i));
      else if (s.size() > 0)
 
        // Pop from the stack
        s.remove(s.size() - 1);
 
      else
 
        // If the stack is empty
        return false;
    }
 
    if(s.size() == 0)
    {
      return true;
    }
    else{
      return false;
    }
 
  }
 
  // Driver code
  public static void main (String[] args)
  {
    String str = "11011000";
    if (isReducible(str))
    {
      System.out.println("Yes");
    }
    else
    {
      System.out.println("No");
    }
  }
}
 
// This code is contributed by avanitrachhadiya2155

Python3




# Python3 program for the above approach
 
# Function to find if string
# is reducible to NULL
def isReducible(Str) :
 
    # Length of string
    N = len(Str)
     
    # Stack to store all 1s
    s = []
     
    # Iterate over the characters
    # of the string
    for i in range(N) :
     
      # If current character is 1
      if (Str[i] == '1') :
     
        # Push it into the stack
        s.append(Str[i])
     
      elif(len(s) > 0) :
     
        # Pop from the stack
        del s[len(s) - 1]
     
      else :
        # If the stack is empty
        return False
     
    if(len(s) == 0) :
      return True
       
    else :
      return False
       
      # Driver code
Str = "11011000"
  
if (isReducible(Str)) :
  print("Yes")
 
else :
  print("No")
   
  # This code is contributed by divyeshrabadiya07.

C#




// C# program for
// the above approach
using System;
using System.Collections.Generic;
class GFG {
 
  // Function to find if string
  // is reducible to NULL
  static bool isReducible(string str)
  {
    // Length of string
    int N = str.Length;
 
    // Stack to store all 1s
    List<char> s = new List<char>();
 
    // Iterate over the characters
    // of the string
    for (int i = 0; i < N; i++) {
 
      // If current character is 1
      if (str[i] == '1')
 
        // Push it into the stack
        s.Add(str[i]);
 
      else if (s.Count > 0)
 
        // Pop from the stack
        s.RemoveAt(s.Count - 1);
 
      else
        // If the stack is empty
        return false;
    }
 
    if(s.Count == 0)
    {
      return true;
    }
    else{
      return false;
    }
  }
 
  // Driver code
  static void Main()
  {
    string str = "11011000";
 
    if (isReducible(str))
      Console.WriteLine("Yes");
 
    else
      Console.WriteLine("No");
  }
}
 
// This code is contributed by divyesh072019.

Javascript




<script>
    // Javascript program for
    // the above approach
     
    // Function to find if string
    // is reducible to NULL
    function isReducible(str)
    {
      // Length of string
      let N = str.length;
 
      // Stack to store all 1s
      let s = [];
 
      // Iterate over the characters
      // of the string
      for (let i = 0; i < N; i++) {
 
        // If current character is 1
        if (str[i] == '1')
 
          // Push it into the stack
          s.push(str[i]);
 
        else if (s.length > 0)
 
          // Pop from the stack
          s.pop();
 
        else
          // If the stack is empty
          return false;
      }
 
      if(s.length == 0)
      {
        return true;
      }
      else{
        return false;
      }
    }
       
    let str = "11011000";
  
    if (isReducible(str))
      document.write("Yes");
  
    else
      document.write("No");
       
      // This code is contributed by suresh07.
</script>

 
 

Output: 
Yes

 

Time Complexity: O(N)
Auxiliary Space: O(N)

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