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.


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.
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);
            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;



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 or mail your article to See your article appearing on the GeeksforGeeks main page and help other Geeks.

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