Question:
Given an array whose order of sorting is unknown and a key. You need to check if the key is present or not using binary search.
Solution:
As the question stated, we dont know the order of the array.
Our first step is to get to know the order of the array.
How do we do that?
Simple: We check the first and last element of the array.
If the first element is smaller than the last element, then it is in ascending order, else it is in descnding order.
Solution in C++
#include <iostream>
using namespace std;
int binary_search(int arr[], int start, int end, int key)
{
int ascending = 0;
if (arr[start] < arr[end])
ascending = 1;
while (start <= end)
{
int mid = start + (end - start) / 2;
// Check if key is present at mid
if (arr[mid] == key)
return mid;
if(ascending == 1)
{
// If key greater, ignore left half
if (arr[mid] < key)
start = mid + 1;
// If key is smaller, ignore right half
else
end = mid - 1;
} else {
// If key greater, ignore right half
if (arr[mid] < key)
end = mid - 1;
// If key is smaller, ignore left half
else
start = mid + 1;
}
}
// if we reach here, then element was
// not present
return -1;
}
int main(void)
{
//descending order
int arr[] = { 10, 9, 8, 7, 6, 5, 4, 3, 2, 1};
int key = 2;
int size = sizeof(arr) / sizeof(arr[0]);
int result = binary_search(arr, 0, size - 1, key);
if(result == -1)
{
cout << "Key is not present in array"<<endl;
} else {
cout << "Key is present at index " << result<<endl ;
}
//ascending order
int arr_1[] = { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10};
int key_1 = 2;
int size_1 = sizeof(arr) / sizeof(arr[0]);
int result_1 = binary_search(arr_1, 0, size_1 - 1, key_1);
if(result_1 == -1)
{
cout << "Key is not present in array"<<endl;
} else {
cout << "Key is present at index " << result_1<<endl ;
}
return 0;
}
Output:
Key is present at index 8
Key is present at index 1