hello-algo/en/docs/chapter_searching/binary_search_edge.md

Binary search boundaries

Find the left boundary

!!! question

Given a sorted array `nums` of length $n$, which may contain duplicate elements, return the index of the leftmost element `target`. If the element is not present in the array, return $-1$.

Recall the method of binary search for an insertion point, after the search is completed, i points to the leftmost target, thus searching for the insertion point is essentially searching for the index of the leftmost target.

Consider implementing the search for the left boundary using the function for finding an insertion point. Note that the array might not contain target, which could lead to the following two results:

• The index i of the insertion point is out of bounds.
• The element nums[i] is not equal to target.

In these cases, simply return -1. The code is as follows:

[file]{binary_search_edge}-[class]{}-[func]{binary_search_left_edge}

Find the right boundary

So how do we find the rightmost target? The most straightforward way is to modify the code, replacing the pointer contraction operation in the case of nums[m] == target. The code is omitted here, but interested readers can implement it on their own.

Below we introduce two more cunning methods.

Reusing the search for the left boundary

In fact, we can use the function for finding the leftmost element to find the rightmost element, specifically by transforming the search for the rightmost target into a search for the leftmost target + 1.

As shown in the figure below, after the search is completed, the pointer i points to the leftmost target + 1 (if it exists), while j points to the rightmost target, thus returning j is sufficient.

Please note, the insertion point returned is i, therefore, it should be subtracted by 1 to obtain j:

[file]{binary_search_edge}-[class]{}-[func]{binary_search_right_edge}

Transforming into an element search

We know that when the array does not contain target, i and j will eventually point to the first element greater and smaller than target respectively.

Thus, as shown in the figure below, we can construct an element that does not exist in the array, to search for the left and right boundaries.

• To find the leftmost target: it can be transformed into searching for target - 0.5, and return the pointer i.
• To find the rightmost target: it can be transformed into searching for target + 0.5, and return the pointer j.

The code is omitted here, but two points are worth noting.

• The given array does not contain decimals, meaning we do not need to worry about how to handle equal situations.
• Since this method introduces decimals, the variable target in the function needs to be changed to a floating point type (no change needed in Python).