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# Bitonic string

• Difficulty Level : Hard
• Last Updated : 20 May, 2021

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:

1. 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.
2. 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.
3. 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.
4. 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.
5. 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 ``using` `namespace` `std;` `// Function to check if the given``// string is bitonic``int` `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 Code``int` `main()``{``    ``// Given string``    ``string s = ``"abcdfgcba"``;` `    ``// Function Call``    ``(checkBitonic(s) == 1)``        ``? cout << ``"YES"``        ``: cout << ``"NO"``;` `    ``return` `0;``}`

## Java

 `// Java program for the above approach``import` `java.util.*;` `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``    ``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 string``s ``=` `"abcdfgcba"` `# Function Call``if``(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

 ``
Output:
`YES`

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

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