Given an array arr[] consisting of N integers, the task is to count all the subarrays which are Bitonic in nature.
A bitonic subarray is a subarray in which elements are either strictly increasing or strictly decreasing, or are first increasing and then decreasing.
Examples:
Input: arr[] = {2, 1, 4, 5}
Output: 8
Explanation:
All subarray which are bitonic in subarray are : {2}, {2, 1}, {1}, {1, 4}, {1, 4, 5}, {4}, {4, 5} and {5}.
Input: arr[] = {1, 2, 3, 4}
Output:10
Approach:
Follow the steps below to solve the problems:
- Generate all possible subarrays.
- For each subarray, check if it is bitonic or not. If the subarray is Bitonic then increment count for answer.
- Finally return the answer.
Below is the implementation of the above approach :
C++
// C++ program to count the// number of possible// bitonic subarrays#include <bits/stdc++.h>using namespace std;// Function to return the count// of bitonic subarraysvoid countbitonic(int arr[], int n){ int c = 0; // Starting element of subarray for (int i = 0; i < n; i++) { // Ending element of subarray for (int j = i; j < n; j++) { int temp = arr[i], f = 0; // for 1 length if (j == i) { c++; continue; } int k = i + 1; // For increasing sequence while (temp < arr[k] && k <= j) { temp = arr[k]; k++; } // If strictly increasing if (k > j) { c++; f = 2; } // For decreasing sequence while (temp > arr[k] && k <= j && f != 2) { temp = arr[k]; k++; } if (k > j && f != 2) { c++; f = 0; } } } cout << c << endl;}// Driver Codeint main(){ int arr[] = { 1, 2, 4, 3, 6, 5 }; int N = 6; countbitonic(arr, N);} |
Java
// Java program to count the number // of possible bitonic subarraysimport java.io.*; import java.util.*; class GFG{ // Function to return the count // of bitonic subarrays public static void countbitonic(int arr[], int n){ int c = 0; // Starting element of subarray for(int i = 0; i < n; i++) { // Ending element of subarray for(int j = i; j < n; j++) { int temp = arr[i], f = 0; // For 1 length if (j == i) { c++; continue; } int k = i + 1; // For increasing sequence while (temp < arr[k] && k <= j) { temp = arr[k]; k++; } // If strictly increasing if (k > j) { c++; f = 2; } // For decreasing sequence while ( k <= j && temp > arr[k] && f != 2) { temp = arr[k]; k++; } if (k > j && f != 2) { c++; f = 0; } } } System.out.println(c);}// Driver codepublic static void main(String[] args) { int arr[] = { 1, 2, 4, 3, 6, 5 }; int N = 6; countbitonic(arr, N);} } // This code is contributed by grand_master |
Python3
# Python3 program to count the number # of possible bitonic subarrays# Function to return the count # of bitonic subarrays def countbitonic(arr, n): c = 0; # Starting element of subarray for i in range(n): # Ending element of subarray for j in range(i, n): temp = arr[i] f = 0; # For 1 length if (j == i) : c += 1 continue; k = i + 1; # For increasing sequence while (temp < arr[k] and k <= j): temp = arr[k]; k += 1 # If strictly increasing if (k > j) : c += 1 f = 2; # For decreasing sequence while (k <= j and temp > arr[k] and f != 2): temp = arr[k]; k += 1; if (k > j and f != 2): c += 1; f = 0; print(c) # Driver Codearr = [ 1, 2, 4, 3, 6, 5 ];N = 6;countbitonic(arr, N);# This code is contributed by grand_master |
C#
// C# program to count the number // of possible bitonic subarrays using System;class GFG{ // Function to return the count // of bitonic subarrays public static void countbitonic(int []arr, int n){ int c = 0; // Starting element of subarray for(int i = 0; i < n; i++) { // Ending element of subarray for(int j = i; j < n; j++) { int temp = arr[i], f = 0; // for 1 length if (j == i) { c++; continue; } int k = i + 1; // For increasing sequence while (temp < arr[k] && k <= j) { temp = arr[k]; k++; } // If strictly increasing if (k > j) { c++; f = 2; } // For decreasing sequence while ( k <= j && temp > arr[k] && f != 2) { temp = arr[k]; k++; } if (k > j && f != 2) { c++; f = 0; } } } Console.Write(c);}// Driver code public static void Main() { int[] arr = { 1, 2, 4, 3, 6, 5 }; int N = 6; countbitonic(arr, N);} } // This code is contributed by grand_master |
Javascript
<script>// Javascript program to count the// number of possible// bitonic subarrays// Function to return the count// of bitonic subarraysfunction countbitonic(arr, n){ var c = 0; // Starting element of subarray for (var i = 0; i < n; i++) { // Ending element of subarray for (var j = i; j < n; j++) { var temp = arr[i], f = 0; // for 1 length if (j == i) { c++; continue; } var k = i + 1; // For increasing sequence while (temp < arr[k] && k <= j) { temp = arr[k]; k++; } // If strictly increasing if (k > j) { c++; f = 2; } // For decreasing sequence while (temp > arr[k] && k <= j && f != 2) { temp = arr[k]; k++; } if (k > j && f != 2) { c++; f = 0; } } } document.write( c );}// Driver Codevar arr = [1, 2, 4, 3, 6, 5];var N = 6;countbitonic(arr, N);// This code is contributed by rutvik_56.</script> |
Output:
15
Time Complexity: O (N 3)
Auxiliary Space: O (1)
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