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

• Last Updated : 26 Nov, 2022

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 using namespace std; // Function to check it possible to get// "10" as substring after all possible// replacementsstring 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 codeint 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 approachimport 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// replacementsstatic 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 Codepublic 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# replacementsdef 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 Codeif __name__ == '__main__':     S = "1?0"    ans = check(S, len(S))    print(ans)

## C#

 // C# program for the approachusing 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// replacementsstatic 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



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.

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