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++
#include <iostream>
using namespace std;
void checkInfinite(string s)
{
bool flag = 1;
int N = s.length();
for (int i = 0; i < N - 1; i++) {
if (s[i] == char(int(s[i + 1]) + 1)) {
continue;
}
else if (s[i] == 'a'
&& s[i + 1] == 'z') {
continue;
}
else {
flag = 0;
break;
}
}
if (flag == 0)
cout << "NO";
else
cout << "YES";
}
int main()
{
string s = "ecbaz";
checkInfinite(s);
return 0;
}
|
Java
class GFG{
public static void checkInfinite(String s)
{
boolean flag = true;
int N = s.length();
for(int i = 0; i < N - 1; i++)
{
if (s.charAt(i) == (char)((int)
(s.charAt(i + 1)) + 1))
{
continue;
}
else if (s.charAt(i) == 'a' &&
s.charAt(i + 1) == 'z')
{
continue;
}
else
{
flag = false;
break;
}
}
if (!flag)
System.out.print("NO");
else
System.out.print("YES");
}
public static void main(String[] args)
{
String s = "ecbaz";
checkInfinite(s);
}
}
|
Python3
def checkInfinite(s):
flag = 1
N = len(s)
for i in range(N - 1):
if (s[i] == chr(ord(s[i + 1]) + 1)):
continue
elif (s[i] == 'a' and s[i + 1] == 'z'):
continue
else:
flag = 0
break
if (flag == 0):
print("NO")
else:
print("YES")
if __name__ == '__main__':
s = "ecbaz"
checkInfinite(s)
|
C#
using System;
class GFG{
public static void checkInfinite(String s)
{
bool flag = true;
int N = s.Length;
for(int i = 0; i < N - 1; i++)
{
if (s[i] == (char)((int)
(s[i + 1]) + 1))
{
continue;
}
else if (s[i] == 'a' &&
s[i + 1] == 'z')
{
continue;
}
else
{
flag = false;
break;
}
}
if (!flag)
Console.Write("NO");
else
Console.Write("YES");
}
public static void Main(String[] args)
{
String s = "ecbaz";
checkInfinite(s);
}
}
|
Javascript
<script>
function checkInfinite(s)
{
var flag = 1;
var N = s.length;
for (var i = 0; i < N - 1; i++) {
if (s[i] == String.fromCharCode((s[i + 1].charCodeAt(0)) + 1)) {
continue;
}
else if (s[i] == 'a'
&& s[i + 1] == 'z') {
continue;
}
else {
flag = 0;
break;
}
}
if (flag == 0)
document.write( "NO");
else
document.write( "YES");
}
var s = "ecbaz";
checkInfinite(s);
</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++
#include <iostream>
#include <string>
using namespace std;
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;
}
int main() {
string str = "ywxtuv";
if (isSubstring(str)) {
cout << "YES" << endl;
} else {
cout << "NO" << endl;
}
return 0;
}
|
Java
public class GFG {
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;
}
public static void main(String[] args) {
String str = "ywxtuv";
if (isSubstring(str)) {
System.out.println("YES");
} else {
System.out.println("NO");
}
}
}
|
Python3
def is_substring(s):
infinite_string = "zyxwvutsrqponmlkjihgfedcbazyxwvutsrqponmlkjihgfedcba"
n = len(s)
m = len(infinite_string)
if n > m:
return False
for i in range(m - n + 1):
j = 0
while j < n and infinite_string[i + j] == s[j]:
j += 1
if j == n:
return True
return False
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
{
static bool IsSubstring(string str)
{
string S = "zyxwvutsrqponmlkjihgfedcbazyxwvutsrqponmlkjihgfedcba";
int n = str.Length;
int 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;
}
static void Main(string[] args)
{
string str = "ywxtuv";
if (IsSubstring(str))
{
Console.WriteLine("YES");
}
else
{
Console.WriteLine("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)