Check if K ‘0’s can be flipped such that Binary String contains no pair of adjacent ‘1’s

Given a binary string S of length N and an integer K, the task is to check if it is possible to flip K 0s such that the resulting string does not contain any pair of adjacent 1s. If it is possible to do so, then print “Yes”. Otherwise, print “No”.

Input: S = “01001001000”, K = 1
Output: Yes
Explanation:
Modifying S = “01001001000” to “01001001001” or modifying S = “01001001000″ to “01001001010″ satisfies the given condition.

Input: S = “000001000”, K = 4 
Output: No

Approach: The idea to traverse the string and replace those ‘0′s with ‘1′s whose both adjacent characters are ‘0′ and flip one of the ‘0′s to count all the possible positions of flips such that it doesn’t affect the original string. Follow the steps below to solve the problem:

• Initialize s variable, say cnt as 0, to store the count of positions that can be flipped.
• Traverse the string using a variable i and perform the following steps:
  • If the current character is ‘1′, then increment i by 2.
  • Otherwise:
    • If i = 0, and s[i + 1] = ‘0’: Increment cnt by 1 and i by 2. Otherwise, increment i by 1.
    • If i = (N – 1), and s[i – 1] = ‘0’: Increment cnt by 1 and i by 2. Otherwise, increment i by 1.
    • Otherwise, if s[i + 1] = ‘0’ and s[i – 1] = ‘0’: Increment cnt by 1 and i by 2. Otherwise, increment i by 1.
• After completing the above steps, if the value of cnt is at least K, then print “Yes”, Otherwise, print “No”.

Below is the implementation of the above approach:

C++

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

// Function to check if k '0's can
// be flipped such that the string
// does not contain any pair of adjacent '1's
void canPlace(string s, int n, int k)
{
    // Store the count of flips
    int cnt = 0;

    // Variable to iterate the string
    int i = 0;

    // Iterate over characters
    // of the string
    while (i < n) {

        // If the current character
        // is '1', increment i by 2
        if (s[i] == '1') {
            i += 2;
        }

        // Otherwise, 3 cases arises
        else {

            // If the current index
            // is the starting index
            if (i == 0) {

                // If next character is '0'
                if (s[i + 1] == '0') {
                    cnt++;
                    i += 2;
                }

                // Increment i by 1
                else
                    i++;
            }

            // If the current index
            // is the last index
            else if (i == n - 1) {

                // If previous character is '0'
                if (s[i - 1] == '0') {

                    cnt++;
                    i += 2;
                }
                else
                    i++;
            }

            // For remaining characters
            else {

                // If both the adjacent
                // characters are '0'
                if (s[i + 1] == '0' && s[i - 1] == '0') {

                    cnt++;
                    i += 2;
                }
                else
                    i++;
            }
        }
    }

    // If cnt is at least K, print "Yes"
    if (cnt >= k) {
        cout << "Yes";
    }

    // Otherwise, print "No"
    else {
        cout << "No";
    }
}

// Driver Code
int main()
{
    string S = "10001";
    int K = 1;
    int N = S.size();

    canPlace(S, N, K);

    return 0;
}

Java

/*package whatever //do not write package name here */
import java.io.*;
class GFG
{

  // Function to check if k '0's can
  // be flipped such that the string
  // does not contain any pair of adjacent '1's
  public static void canPlace(String s, int n, int k)
  {

    // Store the count of flips
    int cnt = 0;

    // Variable to iterate the string
    int i = 0;

    // Iterate over characters
    // of the string
    while (i < n)
    {

      // If the current character
      // is '1', increment i by 2
      if (s.charAt(i) == '1')
      {
        i += 2;
      }

      // Otherwise, 3 cases arises
      else
      {

        // If the current index
        // is the starting index
        if (i == 0)
        {

          // If next character is '0'
          if (s.charAt(i + 1) == '0')
          {
            cnt++;
            i += 2;
          }

          // Increment i by 1
          else
            i++;
        }

        // If the current index
        // is the last index
        else if (i == n - 1)
        {

          // If previous character is '0'
          if (s.charAt(i - 1) == '0')
          {

            cnt++;
            i += 2;
          }
          else
            i++;
        }

        // For remaining characters
        else
        {

          // If both the adjacent
          // characters are '0'
          if (s.charAt(i + 1) == '0' && s.charAt(i - 1) == '0')
          {

            cnt++;
            i += 2;
          }
          else
            i++;
        }
      }
    }

    // If cnt is at least K, print "Yes"
    if (cnt >= k)
    {
      System.out.println("Yes");
    }

    // Otherwise, print "No"
    else
    {
      System.out.println("No");
    }
  }

  // Driver code
  public static void main (String[] args)
  {
    String S = "10001";
    int K = 1;
    int N = 5;
    canPlace(S, N, K);
  }
}

// This code is contributed by manupatharia.

Python3

# Python3 program for the above approach

# Function to check if k '0's can
# be flipped such that the string
# does not contain any pair of adjacent '1's
def canPlace(s, n, k) :

    # Store the count of flips
    cnt = 0

    # Variable to iterate the string
    i = 0

    # Iterate over characters
    # of the string
    while (i < n) :

      # If the current character
      # is '1', increment i by 2
      if (s[i] == '1') :
        i += 2

      # Otherwise, 3 cases arises
      else :

        # If the current index
        # is the starting index
        if (i == 0) :

          # If next character is '0'
          if (s[i + 1] == '0') :
            cnt += 1
            i += 2

          # Increment i by 1
          else :
            i += 1

        # If the current index
        # is the last index
        elif (i == n - 1) :

          # If previous character is '0'
          if (s[i - 1] == '0') :

            cnt += 1
            i += 2

          else :
            i += 1

        # For remaining characters
        else :

          # If both the adjacent
          # characters are '0'
          if (s[i + 1] == '0' and s[i - 1] == '0') :

            cnt += 1
            i += 2

          else :
            i += 1

    # If cnt is at least K, print "Yes"
    if (cnt >= k) :

      print("Yes")

    # Otherwise, print "No"
    else :

      print("No")

      # Driver code
S = "10001"
K = 1
N = len(S)

canPlace(S, N, K)

# This code is contributed by divyeshrabadiya07.

C#

// C# program for the above approach
using System;
class GFG
{

  // Function to check if k '0's can
  // be flipped such that the string
  // does not contain any pair of adjacent '1's
  static void canPlace(string s, int n, int k)
  {

    // Store the count of flips
    int cnt = 0;

    // Variable to iterate the string
    int i = 0;

    // Iterate over characters
    // of the string
    while (i < n) {

      // If the current character
      // is '1', increment i by 2
      if (s[i] == '1') {
        i += 2;
      }

      // Otherwise, 3 cases arises
      else {

        // If the current index
        // is the starting index
        if (i == 0) {

          // If next character is '0'
          if (s[i + 1] == '0') {
            cnt++;
            i += 2;
          }

          // Increment i by 1
          else
            i++;
        }

        // If the current index
        // is the last index
        else if (i == n - 1) {

          // If previous character is '0'
          if (s[i - 1] == '0') {

            cnt++;
            i += 2;
          }
          else
            i++;
        }

        // For remaining characters
        else {

          // If both the adjacent
          // characters are '0'
          if (s[i + 1] == '0' && s[i - 1] == '0') {

            cnt++;
            i += 2;
          }
          else
            i++;
        }
      }
    }

    // If cnt is at least K, print "Yes"
    if (cnt >= k)
    {
      Console.WriteLine("Yes");
    }

    // Otherwise, print "No"
    else
    {
      Console.WriteLine("No");
    }
  }

  // Driver Code
  public static void Main(String[] args)
  {
    string S = "10001";
    int K = 1;
    int N = S.Length;

    canPlace(S, N, K);
  }
}

// This code is contributed by susmitakundugoaldanga.

Javascript

<script>
/*package whatever //do not write package name here */
  // Function to check if k '0's can
  // be flipped such that the string
  // does not contain any pair of adjacent '1's
  function canPlace(s,  n,  k)
  {

    // Store the count of flips
    var cnt = 0;

    // Variable to iterate the string
    var i = 0;

    // Iterate over characters
    // of the string
    while (i < n)
    {

      // If the current character
      // is '1', increment i by 2
      if (s.charAt(i) == '1')
      {
        i += 2;
      }

      // Otherwise, 3 cases arises
      else
      {

        // If the current index
        // is the starting index
        if (i == 0)
        {

          // If next character is '0'
          if (s.charAt(i + 1) == '0')
          {
            cnt++;
            i += 2;
          }

          // Increment i by 1
          else
            i++;
        }

        // If the current index
        // is the last index
        else if (i == n - 1)
        {

          // If previous character is '0'
          if (s.charAt(i - 1) == '0')
          {

            cnt++;
            i += 2;
          }
          else
            i++;
        }

        // For remaining characters
        else
        {

          // If both the adjacent
          // characters are '0'
          if (s.charAt(i + 1) == '0' && s.charAt(i - 1) == '0')
          {

            cnt++;
            i += 2;
          }
          else
            i++;
        }
      }
    }

    // If cnt is at least K, print "Yes"
    if (cnt >= k)
    {
      document.write("Yes");
    }

    // Otherwise, print "No"
    else
    {
      document.write("No");
    }
  }

  // Driver code
    var S = "10001";
    var K = 1;
    var N = 5;
    canPlace(S, N, K);

// This code is contributed by shivanisinghss2110
</script>

Output:

Yes

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