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# Check if a given string is a Reverse Bitonic String or not

• Last Updated : 26 Mar, 2021

Given a string str, the task is to check if that string is a Reverse Bitonic string or not. If the string str is reverse Bitonic string, then print “YES”. Otherwise, print “NO”.

A Reverse Bitonic String is a string in which the characters are arranged in decreasing order followed by increasing order of their ASCII values.

Examples:

Input: str = “zyxbcd”
Output: YES
Explanation:
In the above string, the ASCII values first decreases {z, y, x} and then increases {b, c, d}.

Input: str = “abcdwef”
Output: NO

Approach:
To solve the problem, 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 decreasing followed by strictly increasing.

Follow these steps below to solve the problem:

1. Traverse the string and for each character, check if the ASCII value of the next character is smaller than the ASCII value of the current character or not.
2. If at any point, the ASCII value of the next character is greater than the ASCII value of the current character, break the loop.
3. Now traverse from that index and for each character, check if the ASCII value of the next character is greater than the ASCII value of the current character or not.
4. If at any point, the ASCII value of the next character is 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 entire string is successfully traversed, print “YES”.

Below is the implementation of the above approach:

## C++

 // C++ program to implement// the above approach#include using namespace std; // Function to check if the given// string is reverse bitonicint checkReverseBitonic(string s){    int i, j;     // Check for decreasing 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 increasing 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;     // If the string is    // reverse bitonic    return 1;} // Driver Codeint main(){    string s = "abcdwef";     (checkReverseBitonic(s) == 1)        ? cout << "YES"        : cout << "NO";     return 0;}

## Java

 // Java program to implement// the above approachclass GFG{     // Function to check if the given// string is reverse bitonicstatic int checkReverseBitonic(String s){    int i, j;     // Check for decreasing sequence    for(i = 1; i < s.length(); i++)    {       if (s.charAt(i) < s.charAt(i - 1))           continue;               if (s.charAt(i) >= s.charAt(i - 1))           break;    }         // If end of string has    // been reached    if (i == s.length() - 1)        return 1;     // Check for increasing sequence    for(j = i + 1; j < s.length(); j++)    {       if (s.charAt(j) > s.charAt(j - 1))           continue;       if (s.charAt(j) <= s.charAt(j - 1))           break;    }    i = j;         // If the end of string    // hasn't been reached    if (i != s.length())        return 0;     // If the string is    // reverse bitonic    return 1;} // Driver Codepublic static void main(String []args){    String s = "abcdwef";     if(checkReverseBitonic(s) == 1)        System.out.println("YES");    else        System.out.println("NO");}} // This code is contributed by grand_master

## Python3

 # Python3 program to implement# the above approach # Function to check if the given# string is reverse bitonicdef checkReverseBitonic(s):      i = 0    j = 0     # Check for decreasing sequence    for  i in range(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 1;     # Check for increasing 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)) :        return 0;     # If reverse bitonic    return 1;   # Given strings = "abcdwef"# Function Callif(checkReverseBitonic(s) == 1) :    print("YES")else:    print("NO")

## C#

 // C# program to implement// the above approachusing System; class GFG{     // Function to check if the given// string is reverse bitonicstatic int checkReverseBitonic(String s){    int i, j;     // Check for decreasing 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 increasing 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;     // If the string is    // reverse bitonic    return 1;} // Driver Codepublic static void Main(String []args){    String s = "abcdwef";     if(checkReverseBitonic(s) == 1)        Console.WriteLine("YES");    else        Console.WriteLine("NO");}} // This code is contributed by 29AjayKumar

## Javascript


Output:
NO

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

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