Check if given string is a substring of string formed by repeated concatenation of z to a

Given a string str, the task is to check if string str is a substring of an infinite length string S in which lowercase alphabets are concatenated in reverse order as: 
 

S = “zyxwvutsrqponmlkjihgfedcbazyxwvutsrqponmlkjihgfedcba….”

Examples:

Input: str = “cbaz”
Output: YES 
Explanation:
Given string “cbaz” is a valid sub string of S.

Input: str = “ywxtuv”
Output: NO
Explanation:
Given string “ywxtuv” is a valid sub string of S. 



 

Approach: It can be observed that every next character has a lower ASCII value than the previous character except when ‘a’ is followed by ‘z’. The best way to solve this problem is to simply check for every character if the character following it has a lower ASCII value. Ignore this comparison when the current character is ‘a’. If the current character ‘a’ occurred then check if it is followed by character is ‘z’.

Below are the steps:

  1. Create a flag variable to mark if a given string is a valid substring or not. Initially set it to true.
  2. Traverse the given string str, and for every character, str[i] do the following:
    • If str[i+1] + 1 < str[i], continue with the loop.
    • If str[i] = ‘a’ and str[i+1] = ‘z’, again continue with the loop.
    • Else mark flag variable as false and break from the loop.
  3. Finally, if the flag is true print YES else print NO.

Below is the implementation of the above approach : 

C++

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// C++ program for the above approach
#include <iostream> 
using namespace std;
  
// Function checks if a given string is
// valid or not and prints the output
void checkInfinite(string s)
{
    // Boolean flag variable to mark
    // if given string is valid
    bool flag = 1;
  
    int N = s.length();
  
    // Traverse the given string
    for (int i = 0; i < N - 1; i++) {
  
        // If adjacent character
        // differ by 1
        if (s[i] == char(int(s[i + 1]) + 1)) {
            continue;
        }
  
        // If character 'a' is
        // followed by 4
        else if (s[i] == 'a'
                 && s[i + 1] == 'z') {
            continue;
        }
  
        // Else flip the flag and
        // break from the loop
        else {
            flag = 0;
            break;
        }
    }
  
    // Output according to flag variable
    if (flag == 0)
        cout << "NO";
    else
        cout << "YES";
}
  
// Driver Code
int main()
{
    // Given string
    string s = "ecbaz";
  
    // Function Call
    checkInfinite(s);
  
    return 0;
}

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Java

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// Java program for the above approach
class GFG{
  
// Function checks if a given string is 
// valid or not and prints the output 
public static void checkInfinite(String s) 
      
    // Boolean flag variable to mark 
    // if given string is valid 
    boolean flag = true
  
    int N = s.length(); 
  
    // Traverse the given string 
    for(int i = 0; i < N - 1; i++)
    
          
        // If adjacent character 
        // differ by 1 
        if (s.charAt(i) == (char)((int)
           (s.charAt(i + 1)) + 1)) 
        
            continue
        
  
        // If character 'a' is 
        // followed by 4 
        else if (s.charAt(i) == 'a' && 
                 s.charAt(i + 1) == 'z'
        
            continue
        
  
        // Else flip the flag and 
        // break from the loop 
        else
        
            flag = false
            break
        
    
  
    // Output according to flag variable 
    if (!flag)
        System.out.print("NO");
    else
        System.out.print("YES");
  
// Driver code
public static void main(String[] args)
{
      
    // Given string 
    String s = "ecbaz"
  
    // Function call 
    checkInfinite(s);
}
}
  
// This code is contributed by divyeshrabadiya07

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Python3

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# Python3 program for the above approach
  
# Function checks if a given is
# valid or not and prints the output
def checkInfinite(s):
  
    # Boolean flag variable to mark
    # if given is valid
    flag = 1
  
    N = len(s)
  
    # Traverse the given string
    for i in range(N - 1):
  
        # If adjacent character
        # differ by 1
        if (s[i] == chr(ord(s[i + 1]) + 1)):
            continue
  
        # If character 'a' is
        # followed by 4
        elif (s[i] == 'a' and s[i + 1] == 'z'):
            continue
  
        # Else flip the flag and
        # break from the loop
        else:
            flag = 0
            break
  
    # Output according to flag variable
    if (flag == 0):
        print("NO")
    else:
        print("YES")
  
# Driver Code
if __name__ == '__main__':
  
    # Given string
    s = "ecbaz"
  
    # Function Call
    checkInfinite(s)
  
# This code is contributed by mohit kumar 29

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C#

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// C# program for the above approach
using System;
  
class GFG{
  
// Function checks if a given string is 
// valid or not and prints the output 
public static void checkInfinite(String s) 
      
    // Boolean flag variable to mark 
    // if given string is valid 
    bool flag = true
  
    int N = s.Length; 
  
    // Traverse the given string 
    for(int i = 0; i < N - 1; i++)
    
          
        // If adjacent character 
        // differ by 1 
        if (s[i] == (char)((int)
           (s[i + 1]) + 1)) 
        
            continue
        
  
        // If character 'a' is 
        // followed by 4 
        else if (s[i] == 'a' && 
                 s[i + 1] == 'z'
        
            continue
        
  
        // Else flip the flag and 
        // break from the loop 
        else
        
            flag = false
            break
        
    
  
    // Output according to flag variable 
    if (!flag)
        Console.Write("NO");
    else
        Console.Write("YES");
  
// Driver code
public static void Main(String[] args)
{
      
    // Given string 
    String s = "ecbaz"
  
    // Function call 
    checkInfinite(s);
}
}
  
// This code is contributed by Rajput-Ji

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Output: 

NO

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

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