Program to check if an array is bitonic or not

• Difficulty Level : Medium
• Last Updated : 22 Dec, 2022

Given an array of N elements. The task is to check if the array is bitonic or not.
An array is said to be bitonic if the elements in the array are first strictly increasing and then strictly decreasing.

Examples

Input: arr[] = {-3, 9, 11, 20, 17, 5, 1}
Output: YES

Input: arr[] = {5, 6, 7, 8, 9, 10, 1, 2, 11};
Output: NO

Approach:

• Start traversing the array and keep checking if the next element is greater than the current element or not.
• If at any point, the next element is not greater than the current element, break the loop.
• Again start traversing from the current element and check if the next element is less than current element or not.
• If at any point before the end of array is reached, if the next element is not less than the current element, break the loop and print NO.
• If the end of array is reached successfully, print YES.

Below is the implementation of above approach:

C++

 `// C++ program to check if an array is bitonic``#include ``using` `namespace` `std;` `// Function to check if the given array is bitonic``int` `checkBitonic(``int` `arr[], ``int` `n)``{``    ``int` `i, j;` `    ``// Check for increasing sequence``    ``for` `(i = 1; i < n; i++) {``        ``if` `(arr[i] > arr[i - 1])``            ``continue``;` `        ``if` `(arr[i] <= arr[i - 1])``            ``break``;``    ``}` `    ``if` `(i == n - 1)``        ``return` `1;` `    ``// Check for decreasing sequence``    ``for` `(j = i + 1; j < n; j++) {``        ``if` `(arr[j] < arr[j - 1])``            ``continue``;` `        ``if` `(arr[j] >= arr[j - 1])``            ``break``;``    ``}` `    ``i = j;` `    ``if` `(i != n)``        ``return` `0;` `    ``return` `1;``}` `// Driver code``int` `main()``{``    ``int` `arr[] = { 1,2,3 };` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]);` `    ``(checkBitonic(arr, n) == 1) ? cout << ``"YES"``                                ``: cout << ``"NO"``;` `    ``return` `0;``}`

Java

 `// Java program to check``// if an array is bitonic``class` `GFG``{``// Function to check if the``// given array is bitonic``static` `int` `checkBitonic(``int` `arr[], ``int` `n)``{``    ``int` `i, j;` `    ``// Check for increasing sequence``    ``for` `(i = ``1``; i < n; i++)``    ``{``        ``if` `(arr[i] > arr[i - ``1``])``            ``continue``;` `        ``if` `(arr[i] <= arr[i - ``1``])``            ``break``;``    ``}` `    ``if` `(i == n - ``1``)``        ``return` `1``;` `    ``// Check for decreasing sequence``    ``for` `(j = i + ``1``; j < n; j++)``    ``{``        ``if` `(arr[j] < arr[j - ``1``])``            ``continue``;` `        ``if` `(arr[j] >= arr[j - ``1``])``            ``break``;``    ``}` `    ``i = j;` `    ``if` `(i != n)``        ``return` `0``;` `    ``return` `1``;``}` `// Driver Code``public` `static` `void` `main(String args[])``{``    ``int` `arr[] = { -``3``, ``9``, ``7``, ``20``, ``17``, ``5``, ``1` `};` `    ``int` `n = arr.length;` `    ``System.out.println((checkBitonic(arr, n) == ``1``) ?``                                             ``"YES"` `: ``"NO"``);``}``}` `// This code is contributed by Bilal`

Python3

 `# Python3 program to check if``# an array is bitonic or not.` `# Function to check if the``# given array is bitonic``def` `checkBitonic(arr, n) :` `    ``# Check for increasing sequence``    ``for` `i ``in` `range``(``1``, n) :``        ``if` `arr[i] > arr[i ``-` `1``] :``            ``continue``        ``else` `:``            ``break` `    ``if` `i ``=``=` `n``-``1` `:``        ``return` `1` `    ``# Check for decreasing sequence``    ``for` `j ``in` `range``(i ``+` `1``, n) :``        ` `        ``if` `arr[j] < arr[j ``-` `1``] :``            ``continue``        ``else` `:``            ``break` `    ``i ``=` `j``    ``if` `i !``=` `n ``-` `1` `:``        ``return` `0` `    ``return` `1` `# Driver Code``if` `__name__ ``=``=` `"__main__"` `:``    ` `    ``arr ``=` `[``-``3``, ``9``, ``7``, ``20``, ``17``, ``5``, ``1``]``    ` `    ``n ``=` `len``(arr)` `    ``if` `checkBitonic(arr, n) ``=``=` `1` `:``        ``print``(``"YES"``)``    ``else` `:``        ``print``(``"NO"``)` `# This code is contributed``# by ANKITRAI1`

C#

 `// C# program to check``// if an array is bitonic``using` `System;` `class` `GFG``{``// Function to check if the``// given array is bitonic``static` `int` `checkBitonic(``int` `[]arr,``                        ``int` `n)``{``    ``int` `i, j;` `    ``// Check for increasing sequence``    ``for` `(i = 1; i < n; i++)``    ``{``        ``if` `(arr[i] > arr[i - 1])``            ``continue``;` `        ``if` `(arr[i] <= arr[i - 1])``            ``break``;``    ``}` `    ``if` `(i == n - 1)``        ``return` `1;` `    ``// Check for decreasing sequence``    ``for` `(j = i + 1; j < n; j++)``    ``{``        ``if` `(arr[j] < arr[j - 1])``            ``continue``;` `        ``if` `(arr[j] >= arr[j - 1])``            ``break``;``    ``}` `    ``i = j;` `    ``if` `(i != n)``        ``return` `0;` `    ``return` `1;``}` `// Driver Code``public` `static` `void` `Main()``{``    ``int` `[]arr = { -3, 9, 7, 20, 17, 5, 1 };` `    ``int` `n = arr.Length;` `    ``Console.WriteLine((``            ``checkBitonic(arr, n) == 1) ?``                                 ``"YES"` `: ``"NO"``);``}``}` `// This code is contributed by Bilal`

PHP

 ` ``\$arr``[``\$i` `- 1])``            ``continue``;` `        ``if` `(``\$arr``[``\$i``] <= ``\$arr``[``\$i` `- 1])``            ``break``;``    ``}` `    ``if` `(``\$i` `== ``\$n` `- 1)``        ``return` `1;` `    ``// Check for decreasing sequence``    ``for` `(``\$j` `= ``\$i` `+ 1; ``\$j` `< ``\$n``; ``\$j``++)``    ``{``        ``if` `(``\$arr``[``\$j``] < ``\$arr``[``\$j` `- 1])``            ``continue``;` `        ``if` `(``\$arr``[``\$j``] >= ``\$arr``[``\$j` `- 1])``            ``break``;``    ``}` `    ``\$i` `= ``\$j``;` `    ``if` `(``\$i` `!= ``\$n``)``        ``return` `0;` `    ``return` `1;``}` `// Driver code``\$arr` `= ``array``( -3, 9, 7, 20, 17, 5, 1 );` `\$n` `= sizeof(``\$arr``);` `checkBitonic(``\$arr``, ``\$n``) == 1 ?``               ``print``(``"YES"``) : ``print``(``"NO"``);` `// This code is contributed by ChitraNayal``?>`

Javascript

 ``

Output

`YES`

Time Complexity: O(n), As we are traversing the array only once.
Auxiliary Space: O(1), As constant extra space is used.

My Personal Notes arrow_drop_up