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

  • Last Updated : 02 Jun, 2021

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++




// 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;
}
Java



// 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
Python3



# 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
C#



// 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
Javascript



<script>
 
// Javascript program for the above approach
 
// Function checks if a given string is
// valid or not and prints the output
function checkInfinite(s)
{
 
    // Boolean flag variable to mark
    // if given string is valid
    var flag = 1;
 
    var N = s.length;
 
    // Traverse the given string
    for (var i = 0; i < N - 1; i++) {
 
        // If adjacent character
        // differ by 1
        if (s[i] == String.fromCharCode((s[i + 1].charCodeAt(0)) + 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)
        document.write( "NO");
    else
        document.write( "YES");
}
 
// Driver Code
// Given string
var s = "ecbaz";
 
// Function Call
checkInfinite(s);
 
// This code is contributed by famously.
</script>
Output: 
NO
Time Complexity: O(N)
Auxiliary Space: O(1)
 
Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.
In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.



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Article Contributed By :
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pradyumanagarwal
@pradyumanagarwal
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