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# 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++

 `// C++ program for the above approach``#include ``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

 ``

Output

```NO

```

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

## Approach: Sliding Window

`The sliding window approach involves generating a window of size equal to the length of the given string and slide the window over the infinite string S. At each position of the window, check if the substring of S starting from that position and of length equal to the length of the given string matches the given string. If there is a match, return "YES", otherwise continue sliding the window.`

Below is the code implementation :

## C++

 `// C++ implementation for the problem` `#include ``#include ``using` `namespace` `std;` `// Function to check if a string is a substring of an infinite length string``bool` `isSubstring(string str) {``    ``string S = ``"zyxwvutsrqponmlkjihgfedcbazyxwvutsrqponmlkjihgfedcba"``;``    ``int` `n = str.length(), m = S.length();``    ``if` `(n > m) {``        ``return` `false``;``    ``}``    ``for` `(``int` `i = 0; i <= m - n; i++) {``        ``int` `j;``        ``for` `(j = 0; j < n; j++) {``            ``if` `(S[i+j] != str[j]) {``                ``break``;``            ``}``        ``}``        ``if` `(j == n) {``            ``return` `true``;``        ``}``    ``}``    ``return` `false``;``}` `//Driver Code``int` `main() {``    ``string str = ``"ywxtuv"``;``    ``if` `(isSubstring(str)) {``        ``cout << ``"YES"` `<< endl;``    ``} ``else` `{``        ``cout << ``"NO"` `<< endl;``    ``}``    ``return` `0;``}`

## Java

 `public` `class` `GFG {` `    ``// Function to check if a string is a substring of an infinite length string``    ``public` `static` `boolean` `isSubstring(String str) {``        ``String S = ``"zyxwvutsrqponmlkjihgfedcbazyxwvutsrqponmlkjihgfedcba"``;``        ``int` `n = str.length(), m = S.length();``        ``if` `(n > m) {``            ``return` `false``;``        ``}``        ``for` `(``int` `i = ``0``; i <= m - n; i++) {``            ``int` `j;``            ``for` `(j = ``0``; j < n; j++) {``                ``if` `(S.charAt(i + j) != str.charAt(j)) {``                    ``break``;``                ``}``            ``}``            ``if` `(j == n) {``                ``return` `true``;``            ``}``        ``}``        ``return` `false``;``    ``}``//Driver code``    ``public` `static` `void` `main(String[] args) {``        ``String str = ``"ywxtuv"``;``        ``if` `(isSubstring(str)) {``            ``System.out.println(``"YES"``);``        ``} ``else` `{``            ``System.out.println(``"NO"``);``        ``}``    ``}``}`

## Python3

 `# Function to check if a string is a substring of an infinite length string``def` `is_substring(s):``    ``infinite_string ``=` `"zyxwvutsrqponmlkjihgfedcbazyxwvutsrqponmlkjihgfedcba"``    ``n ``=` `len``(s)``    ``m ``=` `len``(infinite_string)``    ` `    ``# If the input string is longer than the infinite string, it cannot be a substring``    ``if` `n > m:``        ``return` `False``    ` `    ``# Iterate through the infinite string to check for substring``    ``for` `i ``in` `range``(m ``-` `n ``+` `1``):``        ``j ``=` `0``        ``# Compare each character of the substring with the infinite string``        ``while` `j < n ``and` `infinite_string[i ``+` `j] ``=``=` `s[j]:``            ``j ``+``=` `1``        ` `        ``# If j reaches the length of the substring, it means all characters match``        ``if` `j ``=``=` `n:``            ``return` `True``    ` `    ``# If the loop completes without finding a match, the input string is not a substring``    ``return` `False` `# Driver code``def` `main():``    ``input_str ``=` `"ywxtuv"``    ``if` `is_substring(input_str):``        ``print``(``"YES"``)``    ``else``:``        ``print``(``"NO"``)` `if` `__name__ ``=``=` `"__main__"``:``    ``main()`

## C#

 `using` `System;` `namespace` `SubstringCheckExample``{``    ``class` `Program``    ``{``        ``// Function to check if a string is a substring of an infinite length string``        ``static` `bool` `IsSubstring(``string` `str)``        ``{``            ``// The infinite length string``            ``string` `S = ``"zyxwvutsrqponmlkjihgfedcbazyxwvutsrqponmlkjihgfedcba"``;``            ` `            ``int` `n = str.Length;  ``// Length of the input string``            ``int` `m = S.Length;    ``// Length of the infinite length string``            ` `            ``if` `(n > m)``            ``{``                ``return` `false``;  ``// If the input string is longer than the``                               ``// infinite string, it can't be a substring``            ``}``            ` `            ``for` `(``int` `i = 0; i <= m - n; i++)``            ``{``                ``int` `j;``                ``for` `(j = 0; j < n; j++)``                ``{``                    ``if` `(S[i + j] != str[j])``                    ``{``                        ``break``;  ``// If characters don't match, exit the inner loop``                    ``}``                ``}``                ``if` `(j == n)``                ``{``                    ``return` `true``;  ``// If all characters matched, the input string is a substring``                ``}``            ``}``            ``return` `false``;  ``// If no match was found, the input string is not a substring``        ``}` `        ``// Driver Code``        ``static` `void` `Main(``string``[] args)``        ``{``            ``string` `str = ``"ywxtuv"``;``            ``if` `(IsSubstring(str))``            ``{``                ``Console.WriteLine(``"YES"``);``            ``}``            ``else``            ``{``                ``Console.WriteLine(``"NO"``);``            ``}``        ``}``    ``}``}`

Output

```NO

```

Time Complexity: O((m-n+1)*n), where m is the length of the infinite string S and n is the length of the input string str.
Auxiliary Space: O(1)