Bitonic string

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++ program for the above approach

#include <bits/stdc++.h>
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 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 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# 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

<script>

// Javascript program for the above approach

// Function to check if the given
// string is bitonic
function 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 string
var s = "abcdfgcba";

// Function Call
(checkBitonic(s) == 1)
    ? document.write( "YES")
    : document.write( "NO");

// This code is contributed by itsok.
</script>

YES

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