Given a string str, the task is to check if that string is a Bitonic String or not. If string str is Bitonic String then print “YES” else print “NO”.
A Bitonic String is a string in which the characters are arranged in increasing order followed by decreasing order of their ASCII values.
Examples:
Input: str = “abcdfgcba”
Output: YES
Explanation:
In the above string, the ASCII values first increases {a, b, c, d, f, g} and then decreases {g, c, b, a}.
Input: str = “abcdwef”
Output: NO
Approach:
To solve the problem, we simply need to traverse the string and check if the ASCII values of the characters of the string follow any of the following patterns:
- Strictly increasing.
- Strictly decreasing.
- Strictly increasing followed by strictly decreasing.
Follow these steps below to solve the problem:
- Start traversing the string and keep checking if the ASCII value of the next character is greater than the ASCII value of the current character or not.
- If at any point, the ASCII value of the next character is not greater than the ASCII value of the current character, break the loop.
- Again start traversing from the above break index and check if the ASCII value of the next character is smaller than the ASCII value of the current character or not.
- If at any point the ASCII value of the next character is not smaller than the ASCII value of the current character before the end of the array is reached, then print “NO” and break the loop.
- If the end of the array is reached successfully, print “YES”.
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Function to check if the given// string is bitonicint checkBitonic(string s){ int i, j; // Check for increasing sequence for (i = 1; i < s.size(); i++) { if (s[i] > s[i - 1]) continue; if (s[i] <= s[i - 1]) break; } // If end of string has been reached if (i == s.size() - 1) return 1; // Check for decreasing sequence for (j = i + 1; j < s.size(); j++) { if (s[j] < s[j - 1]) continue; if (s[j] >= s[j - 1]) break; } i = j; // If the end of string hasn't // been reached if (i != s.size()) return 0; // Return true if bitonic return 1;}// Driver Codeint main(){ // Given string string s = "abcdfgcba"; // Function Call (checkBitonic(s) == 1) ? cout << "YES" : cout << "NO"; return 0;} |
Java
// Java program for the above approachimport java.util.*;class GFG{// Function to check if the given// String is bitonicstatic int checkBitonic(char []s){ int i, j; // Check for increasing sequence for(i = 1; i < s.length; i++) { if (s[i] > s[i - 1]) continue; if (s[i] <= s[i - 1]) break; } // If end of String has been reached if (i == s.length - 1) return 1; // Check for decreasing sequence for(j = i + 1; j < s.length; j++) { if (s[j] < s[j - 1]) continue; if (s[j] >= s[j - 1]) break; } i = j; // If the end of String hasn't // been reached if (i != s.length) return 0; // Return true if bitonic return 1;}// Driver Codepublic static void main(String[] args){ // Given String String s = "abcdfgcba"; // Function Call System.out.print((checkBitonic( s.toCharArray()) == 1) ? "YES" : "NO");}}// This code is contributed by PrinciRaj1992 |
Python3
# Python3 program for the above approach # Function to check if the given # string is bitonic def checkBitonic(s): i = 0 j = 0 # Check for increasing sequence for i in range(1, len(s)): if (s[i] > s[i - 1]): continue; if (s[i] <= s[i - 1]): break; # If end of string has been reached if (i == (len(s) - 1)): return True; # Check for decreasing sequence for j in range(i + 1, len(s)): if (s[j] < s[j - 1]): continue; if (s[j] >= s[j - 1]): break; i = j; # If the end of string hasn't # been reached if (i != len(s) - 1): return False; # Return true if bitonic return True; # Driver code# Given strings = "abcdfgcba"# Function Callif(checkBitonic(s)): print("YES")else: print("NO") # This code is contributed by grand_master |
C#
// C# program for the above approach using System;class GFG{ // Function to check if the given // String is bitonic static int checkBitonic(char []s) { int i, j; // Check for increasing sequence for(i = 1; i < s.Length; i++) { if (s[i] > s[i - 1]) continue; if (s[i] <= s[i - 1]) break; } // If end of String has been reached if (i == s.Length - 1) return 1; // Check for decreasing sequence for(j = i + 1; j < s.Length; j++) { if (s[j] < s[j - 1]) continue; if (s[j] >= s[j - 1]) break; } i = j; // If the end of String hasn't // been reached if (i != s.Length) return 0; // Return true if bitonic return 1; } // Driver Code public static void Main(String[] args) { // Given String String s = "abcdfgcba"; // Function call Console.Write((checkBitonic( s.ToCharArray()) == 1) ? "YES" : "NO"); } } // This code is contributed by PrinciRaj1992 |
Javascript
<script>// Javascript program for the above approach// Function to check if the given// string is bitonicfunction checkBitonic(s){ var i, j; // Check for increasing sequence for (i = 1; i < s.length; i++) { if (s[i] > s[i - 1]) continue; if (s[i] <= s[i - 1]) break; } // If end of string has been reached if (i == s.length - 1) return 1; // Check for decreasing sequence for (j = i + 1; j < s.length; j++) { if (s[j] < s[j - 1]) continue; if (s[j] >= s[j - 1]) break; } i = j; // If the end of string hasn't // been reached if (i != s.length) return 0; // Return true if bitonic return 1;}// Driver Code// Given stringvar s = "abcdfgcba";// Function Call(checkBitonic(s) == 1) ? document.write( "YES") : document.write( "NO");// This code is contributed by itsok.</script> |
Output:
YES
Time Complexity: O(N)
Auxiliary Space: O(1)
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!