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
// 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 |
NO
Time Complexity: O(N)
Auxiliary Space: O(1)
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