Check if characters of a string can be made non-decreasing by replacing '?'s

Given a string S of length N consisting of lowercase English alphabets and '?', the task is to check if it is possible to make the string non-decreasing by replacing all the '?' characters with English alphabets. If found to be true, print "Yes". Otherwise, print "No".

Examples:

Input: S = "abb?xy?" 
Output: Yes
Explanation:
Replacing the '?'s at index 3 and 6 with 'b' and 'z', modifies the string S to "abbbxyz", which is non-decreasing.

Input: S = "??z?a?"
Output : No

Naive Approach: The simplest approach to solve the problem is to generate all possible strings using English alphabets and check if the resulting string is non-decreasing or not. If found to be true, print "Yes", Otherwise, print "No".

Time Complexity: O(N*26^N)
Auxiliary Space: O(N)

Efficient Approach: The given problem can be solved based on the following observations: 

• It can be observed that the only position that matters is the position with S[i]! = '?', as one can replace the '?' with the character at the nearest index that is not '?'. 
• Therefore, the task is reduced to checking if characters of the strings that are not '?', are in non-decreasing order or not.

Follow the steps below to solve the problem:

• Initialize a variable, say last, to store the last character which is not a '?'.
• Traverse the string and for every i^th character, check if S[i]! = '?' and S[i]<last, then print "No". 
• Otherwise, update last as last = S[i] if S[i]!='?'.
• Print "Yes" if none of the above cases are satisfied.

Below is the implementation of the above approach:

// C++ program for the above approach

#include <bits/stdc++.h>
using namespace std;

// Function to check if it's possible
// to make the string s non decreasing
int nonDecreasing(string s)
{
    // Stores length of string
    int size = s.length();

    // Stores the last character
    // that is not '?'
    char c = 'a';

    // Traverse the string S
    for (int i = 0; i < size; i++) {

        // If current character
        // of the string is '?'
        if (s[i] == '?') {
            continue;
        }

        // If s[i] is not '?'
        // and is less than C
        else if ((s[i] != '?')
                 && (s[i] < c)) {

            return 0;
        }

        // Otherwise
        else {

            // Update C
            c = s[i];
        }
    }

    // Return 1
    return 1;
}

// Driver Code
int main()
{
    string S = "abb?xy?";

    if (nonDecreasing(S))
        cout << "Yes" << endl;
    else
        cout << "No" << endl;

    return 0;
}

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

class GFG
{

// Function to check if it's possible
// to make the String s non decreasing
static int nonDecreasing(char []s)
{
  
    // Stores length of String
    int size = s.length;

    // Stores the last character
    // that is not '?'
    char c = 'a';

    // Traverse the String S
    for (int i = 0; i < size; i++)
    {

        // If current character
        // of the String is '?'
        if (s[i] == '?') 
        {
            continue;
        }

        // If s[i] is not '?'
        // and is less than C
        else if ((s[i] != '?')
                 && (s[i] < c))
        {

            return 0;
        }

        // Otherwise
        else {

            // Update C
            c = s[i];
        }
    }

    // Return 1
    return 1;
}

// Driver Code
public static void main(String[] args)
{
    String S = "abb?xy?";

    if (nonDecreasing(S.toCharArray())==1)
        System.out.print("Yes" +"\n");
    else
        System.out.print("No" +"\n");

}
}

// This code is contributed by 29AjayKumar 

# python 3 program for the above approach

# Function to check if it's possible
# to make the string s non decreasing
def nonDecreasing(s):
  
    # Stores length of string
    size = len(s)

    # Stores the last character
    # that is not '?'
    c = 'a'

    # Traverse the string S
    for i in range(size):
      
        # If current character
        # of the string is '?'
        if (s[i] == '?'):
            continue

        # If s[i] is not '?'
        # and is less than C
        elif((s[i] != '?') and (s[i] < c)):
            return 0

        # Otherwise
        else:
          
            # Update C
            c = s[i]

    # Return 1
    return 1

# Driver Code
if __name__ == '__main__':
    S = "abb?xy?"

    if(nonDecreasing(S)):
        print("Yes")
    else:
        print("No")
        
        # This code is contributed by SURENDRA_GANGWAR.

// C# program for the above approach
using System;

public class GFG
{

  // Function to check if it's possible
  // to make the String s non decreasing
  static int nonDecreasing(char []s)
  {

    // Stores length of String
    int size = s.Length;

    // Stores the last character
    // that is not '?'
    char c = 'a';

    // Traverse the String S
    for (int i = 0; i < size; i++)
    {

      // If current character
      // of the String is '?'
      if (s[i] == '?') 
      {
        continue;
      }

      // If s[i] is not '?'
      // and is less than C
      else if ((s[i] != '?')
               && (s[i] < c))
      {

        return 0;
      }

      // Otherwise
      else {

        // Update C
        c = s[i];
      }
    }

    // Return 1
    return 1;
  }

  // Driver Code
  public static void Main(String[] args)
  {
    String S = "abb?xy?";

    if (nonDecreasing(S.ToCharArray())==1)
      Console.Write("Yes" +"\n");
    else
      Console.Write("No" +"\n");

  }
}

// This code is contributed by 29AjayKumar 

<script>

// Javascript program for the above approach

// Function to check if it's possible
// to make the String s non decreasing
function nonDecreasing(s)
{
    
    // Stores length of String
    let size = s.length;

    // Stores the last character
    // that is not '?'
    var c = 'a';

    // Traverse the String S
    for(let i = 0; i < size; i++)
    {
        
        // If current character
        // of the String is '?'
        if (s[i] == '?') 
        {
            continue;
        }

        // If s[i] is not '?'
        // and is less than C
        else if ((s[i] != '?') &&
                 (s[i] < c))
        {
            return 0;
        }

        // Otherwise
        else 
        {
            
            // Update C
            c = s[i];
        }
    }

    // Return 1
    return 1;
}

// Driver Code
let S = "abb?xy?";

if (nonDecreasing(S.split('')))
    document.write("Yes" + "\n");
else
    document.write("No" + "\n");

// This code is contributed by shivanisinghss2110 

</script>

Yes

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