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.

1. 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;

}

Output:

NO

Time Complexity: O(N)

Auxiliary Space: O(1)

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