Find bitonic point in given bitonic sequence

You are given a bitonic sequence, the task is to find Bitonic Point in it. A Bitonic Sequence is a sequence of numbers which is first strictly increasing then after a point strictly decreasing.

A Bitonic Point is a point in bitonic sequence before which elements are strictly increasing and after which elements are strictly decreasing. A Bitonic point doesn’t exist if array is only decreasing or only increasing.

Examples:

Input : arr[] = {6, 7, 8, 11, 9, 5, 2, 1}
Output: 11
All elements before 11 are smaller and all
elements after 11 are greater.

Input : arr[] = {-3, -2, 4, 6, 10, 8, 7, 1}
Output: 10

A simple solution for this problem is to use linear search. Element arr[i] is bitonic point if both i-1’th and i+1’th both elements are less than i’th element. Time complexity for this approach is O(n).

An efficient solution for this problem is to use modified binary search.

  • If arr[mid-1] < arr[mid] > arr[mid+1] then we are done with bitonic point.
  • If arr[mid] < arr[mid+1] then search in right sub-array, else search in left sub-array.
// C++ program to find bitonic point in a bitonic array.
#include<bits/stdc++.h>
using namespace std;

// Function to find bitonic point using binary search
int binarySearch(int arr[], int left, int right)
{
    if (left <= right)
    {
        int mid = (left+right)/2;

        // base condition to check if arr[mid] is
        // bitonic point or not
        if (arr[mid-1]<arr[mid] && arr[mid]>arr[mid+1])
            return mid;

        // We assume that sequence is bitonic. We go to
        // right subarray if middle point is part of
        // increasing subsequence. Else we go to left
        // subarray.
        if (arr[mid] < arr[mid+1])
            return binarySearch(arr, mid+1,right);
        else
            return binarySearch(arr, left, mid-1);
    }

    return -1;
}

// Driver program to run the case
int main()
{
    int arr[] = {6, 7, 8, 11, 9, 5, 2, 1};
    int n = sizeof(arr)/sizeof(arr[0]);
    int index = binarySearch(arr, 1, n-2);
    if (index != -1)
       cout << arr[index];
    return 0;
}

Output:

11

Time complexity : O(Log n)

This article is contributed by Shashank Mishra ( Gullu ). If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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