Check if given string is a substring of string formed by repeated concatenation of z to a
Last Updated :
16 Oct, 2023
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:
- Create a flag variable to mark if a given string is a valid substring or not. Initially set it to true.
- 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.
- 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 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 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# 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
<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>
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 <iostream>
#include <string>
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;
}
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");
}
}
}
# 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()
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");
}
}
}
}
// Function to check if a string is a substring of an infinite length string
function isSubstring(str) {
// Define an infinite-length string 'S'
const S = "zyxwvutsrqponmlkjihgfedcbazyxwvutsrqponmlkjihgfedcba";
// Get the lengths of the input 'str' and 'S'
const n = str.length;
const m = S.length;
// Check if the length of 'str' is greater than 'S'
if (n > m) {
return false; // 'str' cannot be a substring
}
// Iterate through 'S' to find possible substrings
for (let i = 0; i <= m - n; i++) {
let j;
// Check characters of 'str' against characters in 'S'
for (j = 0; j < n; j++) {
if (S[i + j] !== str[j]) {
break; // Mismatch found, move to the next substring
}
}
// If we reach the end of 'str', it is a substring
if (j === n) {
return true;
}
}
return false; // 'str' is not a substring of 'S'
}
// Driver code
const str = "ywxtuv";
if (isSubstring(str)) {
console.log("YES"); // Print "YES" if 'str' is a substring
} else {
console.log("NO"); // Print "NO" if 'str' is not a substring
}
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)
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