Most Generalized Binary Search

 

Suppose we have a search space. It could be an array, a range, etc. Usually it's sorted in ascend order. For most tasks, we can transform the requirement into the following generalized form:

Minimize k , s.t. condition(k) is True

The following code is the most generalized binary search template:

def binary_search(array) -> int:
    def condition(value) -> bool:
        pass

    left, right = 0, len(array)
    while left < right:
        mid = left + (right - left) // 2
        if condition(mid):
            right = mid
        else:
            left = mid + 1
    return left

What's really nice of this template is that, for most of the binary search problems, we only need to modify three parts after copy-pasting this template, and never need to worry about corner cases and bugs in code any more:

• Correctly initialize the boundary variables left and right. Only one rule: set up the boundary to include all possible elements;
• Decide return value. Is it return left or return left - 1? Remember this: after exiting the while loop, left is the minimal k​ satisfying the condition function;
• Design the condition function. This is the most difficult and most beautiful part. Needs lots of practice.

Binary Search

Given a sorted array arr[] of n elements, write a function to search a given element x in arr[].
A simple approach is to do linear search.The time complexity of above algorithm is O(n). Another approach to perform the same task is using Binary Search.
Binary Search: Search a sorted array by repeatedly dividing the search interval in half. Begin with an interval covering the whole array. If the value of the search key is less than the item in the middle of the interval, narrow the interval to the lower half. Otherwise narrow it to the upper half. Repeatedly check until the value is found or the interval is empty.
Example:
Image Source : http://www.nyckidd.com/bob/Linear%20Search%20and%20Binary%20Search_WorkingCopy.pdf
The idea of binary search is to use the information that the array is sorted and reduce the time complexity to O(Logn).

Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

We basically ignore half of the elements just after one comparison.

1. Compare x with the middle element.
2. If x matches with middle element, we return the mid index.
3. Else If x is greater than the mid element, then x can only lie in right half subarray after the mid element. So we recur for right half.
4. Else (x is smaller) recur for the left half.

Recursive implementation of Binary Search

// Java implementation of recursive Binary Search class BinarySearch {     // Returns index of x if it is present in arr[l..r], else     // return -1     int binarySearch(int arr[], int l, int r, int x)     {         if (r>=l)         {             int mid = l + (r - l)/2;               // If the element is present at the middle itself             if (arr[mid] == x)                return mid;               // If element is smaller than mid, then it can only             // be present in left subarray             if (arr[mid] > x)                return binarySearch(arr, l, mid-1, x);               // Else the element can only be present in right             // subarray             return binarySearch(arr, mid+1, r, x);         }           // We reach here when element is not present in array         return -1;     }       // Driver method to test above     public static void main(String args[])     {         BinarySearch ob = new BinarySearch();         int arr[] = {2,3,4,10,40};         int n = arr.length;         int x = 10;         int result = ob.binarySearch(arr,0,n-1,x);         if (result == -1)             System.out.println("Element not present");         else             System.out.println("Element found at index "+result);     } } /* This code is contributed by Rajat Mishra */

Output:

Element is present at index 3

Iterative implementation of Binary Search

// Java implementation of iterative Binary Search class BinarySearch {     // Returns index of x if it is present in arr[], else     // return -1     int binarySearch(int arr[], int x)     {         int l = 0, r = arr.length - 1;         while (l <= r)         {             int m = l + (r-l)/2;               // Check if x is present at mid             if (arr[m] == x)                 return m;               // If x greater, ignore left half             if (arr[m] < x)                 l = m + 1;               // If x is smaller, ignore right half             else                 r = m - 1;         }           // if we reach here, then element was not present         return -1;     }       // Driver method to test above     public static void main(String args[])     {         BinarySearch ob = new BinarySearch();         int arr[] = {2, 3, 4, 10, 40};         int n = arr.length;         int x = 10;         int result = ob.binarySearch(arr, x);         if (result == -1)             System.out.println("Element not present");         else             System.out.println("Element found at index "+result);     } }

Output:

Element is present at index 3

Time Complexity:
The time complexity of Binary Search can be written as

T(n) = T(n/2) + c 

The above recurrence can be solved either using Recurrence T ree method or Master method. It falls in case II of Master Method and solution of the recurrence is .
Auxiliary Space: O(1) in case of iterative implementation. In case of recursive implementation, O(Logn) recursion call stack space.
Algorithmic Paradigm: Divide and Conquer
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