Check if substring “10” occurs in the given binary string in all possible replacements of ‘?’ with 1 or 0

Given a string S consisting of only ‘0’, ‘1’ and ‘?’, the task is to check if there exists a substring “10” in every possible replacement of the character ‘?’ with either 1 or 0.

Examples:

Input: S = “1?0”
Output: Yes
Explanation:
Following are all the possible replacements of ‘?’:

1. Replacing the ‘?’ with 0 modifies the string to “100”. In the modifies string, the substring “10” occurs.
2. Replacing the ‘?’ with 1 modifies the string to “110”. In the modifies string, the substring “10” occurs.

From the above all possible replacements, the substring “10” occurs in all the replacements, therefore, print Yes.

Input: S= “??”
Output: No

Approach: The given problem can be solved by using a Greedy Approach which is based on the observation that if the string S contains many consecutive ‘?’, it can be replaced with a single ‘?’ as in the worst case we can replace it with all 1s or 0s.

Therefore, the idea is to create a new string from the given string S by replacing continuous ‘?’ with single ‘?’ and then check if there exists “10” or “1?0” as a substring, then it is possible to get “10” as substring after all possible replacements, therefore, print Yes. Otherwise, print No.

Below is the implementation of the above approach: 

C++

// C++ Program to implement
// the above approach
#include <iostream>
using namespace std;

// Function to check it possible to get
// "10" as substring after all possible
// replacements
string check(string S, int n)
{

  // Initialize empty string ans
  string ans = "";
  int c = 0;

  // Run loop n times
  for (int i = 0; i < n; i++) {

    // If char is "?", then increment
    // c by 1
    if (S[i] == '?') {
      c++;
    }
    else {

      // Continuous '?' characters
      if (c) {
        ans += "?";
      }
      c = 0;
      ans += S[i];
    }
  }

  // Their is no consecutive "?"
  if (c) {
    ans += "?";
  }

  // Check if "10" or "1?0" exists
  // in the string ans or not
  if (ans.find("10") != -1 || ans.find("1?0") != -1) {
    return "Yes";
  }
  else {
    return "No";
  }
}

// Driver code
int main()
{
  string S = "1?0";
  int n = S.size();
  string ans = check(S, n);
  cout << ans;
  return 0;
}
// This code is contributed by parthmanchanda81

Java

// Java program for the above approach
import java.io.*;

class GFG {


 // Returns true if s1 is substring of s2
    static int isSubstring(String s1, String s2)
    {
        int M = s1.length();
        int N = s2.length();

        /* A loop to slide pat[] one by one */
        for (int i = 0; i <= N - M; i++) {
            int j;

            /* For current index i, check for
            pattern match */
            for (j = 0; j < M; j++)
                if (s2.charAt(i + j) != s1.charAt(j))
                    break;

            if (j == M)
                return i;
        }

        return -1;
    }


// Function to check it possible to get
// "10" as substring after all possible
// replacements
static String check(String S, int n)
{

  // Initialize empty string ans
  String ans = "";
  int c = 0;

  // Run loop n times
  for (int i = 0; i < n; i++) {

    // If char is "?", then increment
    // c by 1
    if (S.charAt(i) == '?') {
      c++;
    }
    else {

      // Continuous '?' characters
      if (c != 0) {
        ans += "?";
      }
      c = 0;
      ans += S.charAt(i);
    }
  }

  // Their is no consecutive "?"
  if (c != 0) {
    ans += "?";
  }

  // Check if "10" or "1?0" exists
  // in the string ans or not
  if (isSubstring("10", S) != -1 || isSubstring("1?0", S) != -1) {
    return "Yes";
  }
  else {
    return "No";
  }
}

// Driver Code
public static void main (String[] args)
{
    String S = "1?0";
        int n = S.length();
        String ans = check(S, n);
        System.out.println(ans);
}
}

// This code is contributed by avijitmondal1998.

Python3

# Python program for the above approach

# Function to check it possible to get
# "10" as substring after all possible
# replacements
def check(S, n):

    # Initialize empty string ans
    ans = ""
    c = 0

    # Run loop n times
    for _ in range(n):

        # If char is "?", then increment
        # c by 1
        if S[_] == "?":
            c += 1
        else:

            # Continuous '?' characters
            if c:
                ans += "?"

            # Their is no consecutive "?"
            c = 0
            ans += S[_]

    # "?" still left
    if c:
        ans += "?"

    # Check if "10" or "1?0" exists
    # in the string ans or not
    if "10" in ans or "1?0" in ans:
        return "Yes"
    else:
        return "No"


# Driver Code
if __name__ == '__main__':

    S = "1?0"
    ans = check(S, len(S))
    print(ans)

C#

// C# program for the approach
using System;
using System.Collections.Generic;

class GFG {

 // Returns true if s1 is substring of s2
    static int isSubstring(string s1, string s2)
    {
        int M = s1.Length;
        int N = s2.Length;

        /* A loop to slide pat[] one by one */
        for (int i = 0; i <= N - M; i++) {
            int j;

            /* For current index i, check for
            pattern match */
            for (j = 0; j < M; j++)
                if (s2[i + j] != s1[j])
                    break;

            if (j == M)
                return i;
        }

        return -1;
    }


// Function to check it possible to get
// "10" as substring after all possible
// replacements
static string check(string S, int n)
{

  // Initialize empty string ans
  string ans = "";
  int c = 0;

  // Run loop n times
  for (int i = 0; i < n; i++) {

    // If char is "?", then increment
    // c by 1
    if (S[i] == '?') {
      c++;
    }
    else {

      // Continuous '?' characters
      if (c != 0) {
        ans += "?";
      }
      c = 0;
      ans += S[i];
    }
  }

  // Their is no consecutive "?"
  if (c != 0) {
    ans += "?";
  }

  // Check if "10" or "1?0" exists
  // in the string ans or not
  if (isSubstring("10", S) != -1 || isSubstring("1?0", S) != -1) {
    return "Yes";
  }
  else {
    return "No";
  }
}

    // Driver Code
    public static void Main()
    {
        string S = "1?0";
        int n = S.Length;
        string ans = check(S, n);
        Console.Write(ans);
    }
}

// This code is contributed by sanjoy_62.

Javascript

  <script>
        // JavaScript Program to implement
        // the above approach

        // Function to check it possible to get
        // "10" as substring after all possible
        // replacements
        function check(S, n) {

            // Initialize empty string ans
            ans = ""
            c = 0

            // Run loop n times
            for (let i = 0; i < n; i++) {

                // If char is "?", then increment
                // c by 1
                if (S[i] == "?")
                    c += 1
                else

                    // Continuous '?' characters
                    if (c != 0)
                        ans += "?"

                // Their is no consecutive "?"
                c = 0
                ans += S[i]

                // "?" still left
                if (c != 0)
                    ans += "?"
            }

            // Check if "10" or "1?0" exists
            // in the string ans or not

            if (ans.includes('10') || ans.includes('1?0'))
                return "Yes"
            else
                return "No"

        }

        // Driver Code
        S = "1?0"
        ans = check(S, S.length)
        document.write(ans)

// This code is contributed by Potta Lokesh
    </script>

Output:

Yes

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

Note: The same approach can be used for substrings “00”/”01″/”11″ as well with minor changes.