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Maximum count of unique index 10 or 01 substrings in given Binary string

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  • Last Updated : 09 Feb, 2022

Given a binary string str of length N, the task is to count the maximum number of adjacent pairs of form “01” or “10” that can be formed from the given binary string when one character can be considered for only one pair.

Note: Adjacent pair means pair formed using adjacent characters.

Examples:

Input: str = “0101110”
Output: 3
Explanation: The three pairs are “01” at the starting first,  
“01” starting at 2nd index (0 based indexing)
and “10” at the end of the string.
Notice 2 pairs can also be formed by using “10” from index 1 and “10” at last.
But that does not give the maximum number of adjacent pairs.

Input: str = “0011”
Output: 1

Input: str  = “11”
Output: 0

 

Approach: This is a implementation based problem. Follow the steps mentioned here to solve the problem:

  • Traverse string from left to right.
  • Initialize count of pairs with 0 and consider the previous character as free.
  • Run a loop from 1 to size of the string.
  • Check if the previous character is opposite to the current character or not and also if it is free or not
    • If yes then increment count of pairs and set the character as not free.
    • Else continue the traversal in the loop considering the character as free.
  • Print the count of pairs.

Below is the implementation of the above approach.

C++




// C++ code to implement the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Count pairs function
void check_pairs(string str)
{
    // Initialize pairs with 0
    int pairs = 0;
 
    // Previous char is free to pair
    bool prev_c = true;
 
    // Traverse string from second position
    for (int i = 1; i < str.size(); i++) {
        // Check both char are opposite or not
        // and also check previous char
        // is free or not
        if (str[i] != str[i - 1] && prev_c) {
 
            // Once previous char paired
            // with other make it false
            prev_c = false;
 
            // Increment pairs count
            pairs++;
        }
        else {
 
            // Previous char is free for pair
            prev_c = true;
        }
    }
 
    // Print count of pairs of two characters
    cout << pairs;
}
 
// Driver Code
int main()
{
    string str = "0101110";
 
    // Function call
    check_pairs(str);
    return 0;
}


Java




// Java code to implement the above approach
class GFG
{
 
  // Count pairs function
  static void check_pairs(String str)
  {
 
    // Initialize pairs with 0
    int pairs = 0;
 
    // Previous char is free to pair
    boolean prev_c = true;
 
    // Traverse String from second position
    for (int i = 1; i < str.length(); i++)
    {
 
      // Check both char are opposite or not
      // and also check previous char
      // is free or not
      if (str.charAt(i) != str.charAt(i - 1) && prev_c) {
 
        // Once previous char paired
        // with other make it false
        prev_c = false;
 
        // Increment pairs count
        pairs++;
      }
      else {
 
        // Previous char is free for pair
        prev_c = true;
      }
    }
 
    // Print count of pairs of two characters
    System.out.println(pairs);
  }
 
  // Driver Code
  public static void main(String args[])
  {
    String str = "0101110";
 
    // Function call
    check_pairs(str);
  }
}
 
// This code is contributed by gfgking


Python3




# python3 code to implement the above approach
 
# Count pairs function
def check_pairs(str):
 
    # Initialize pairs with 0
    pairs = 0
 
    # Previous char is free to pair
    prev_c = True
 
    # Traverse string from second position
    for i in range(1, len(str)):
       
        # Check both char are opposite or not
        # and also check previous char
        # is free or not
        if (str[i] != str[i - 1] and prev_c):
 
            # Once previous char paired
            # with other make it false
            prev_c = False
 
            # Increment pairs count
            pairs += 1
 
        else:
 
            # Previous char is free for pair
            prev_c = True
 
    # Print count of pairs of two characters
    print(pairs)
 
# Driver Code
if __name__ == "__main__":
 
    str = "0101110"
 
    # Function call
    check_pairs(str)
 
# This code is contributed by rakeshsahni


C#




// C# code to implement the above approach
using System;
class GFG
{
 
  // Count pairs function
  static void check_pairs(string str)
  {
 
    // Initialize pairs with 0
    int pairs = 0;
 
    // Previous char is free to pair
    bool prev_c = true;
 
    // Traverse string from second position
    for (int i = 1; i < str.Length; i++)
    {
 
      // Check both char are opposite or not
      // and also check previous char
      // is free or not
      if (str[i] != str[i - 1] && prev_c) {
 
        // Once previous char paired
        // with other make it false
        prev_c = false;
 
        // Increment pairs count
        pairs++;
      }
      else {
 
        // Previous char is free for pair
        prev_c = true;
      }
    }
 
    // Print count of pairs of two characters
    Console.Write(pairs);
  }
 
  // Driver Code
  public static int Main()
  {
    string str = "0101110";
 
    // Function call
    check_pairs(str);
    return 0;
  }
}
 
// This code is contributed by Taranpreet


Javascript




<script>
        // JavaScript code for the above approach
 
        // Count pairs function
        function check_pairs(str)
        {
         
            // Initialize pairs with 0
            let pairs = 0;
 
            // Previous char is free to pair
            let prev_c = true;
 
            // Traverse string from second position
            for (let i = 1; i < str.length; i++)
            {
             
                // Check both char are opposite or not
                // and also check previous char
                // is free or not
                if (str[i] != str[i - 1] && prev_c)
                {
 
                    // Once previous char paired
                    // with other make it false
                    prev_c = false;
 
                    // Increment pairs count
                    pairs++;
                }
                else {
 
                    // Previous char is free for pair
                    prev_c = true;
                }
            }
 
            // Print count of pairs of two characters
            document.write(pairs);
        }
 
        // Driver Code
        let str = "0101110";
 
        // Function call
        check_pairs(str);
 
       // This code is contributed by Potta Lokesh
    </script>


 
 

Output

3

 

Time Complexity: O(N)
Auxiliary Space: O(1)

 


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