Binary Search Insertion Point

Binary search can not only be used to search for target elements but also to solve many variant problems, such as searching for the insertion position of a target element.

Case Without Duplicate Elements

!!! question

Given a sorted array `nums` of length $n$ and an element `target`, where the array contains no duplicate elements. Insert `target` into the array `nums` while maintaining its sorted order. If the array already contains the element `target`, insert it to its left. Return the index of `target` in the array after insertion. An example is shown in the figure below.

If we want to reuse the binary search code from the previous section, we need to answer the following two questions.

Question 1: When the array contains target, is the insertion point index the same as that element's index?

The problem requires inserting target to the left of equal elements, which means the newly inserted target replaces the position of the original target. In other words, when the array contains target, the insertion point index is the index of that target.

Question 2: When the array does not contain target, what is the insertion point index?

Further consider the binary search process: When nums[m] < target, $i$ moves, which means pointer $i$ is approaching elements greater than or equal to target. Similarly, pointer $j$ is always approaching elements less than or equal to target.

Therefore, when the binary search ends, we must have: $i$ points to the first element greater than target, and $j$ points to the first element less than target. It's easy to see that when the array does not contain target, the insertion index is $i$. The code is shown below:

[file]{binary_search_insertion}-[class]{}-[func]{binary_search_insertion_simple}

Case with Duplicate Elements

!!! question

Based on the previous problem, assume the array may contain duplicate elements, with everything else remaining the same.

Suppose there are multiple target elements in the array. Ordinary binary search can only return the index of one target, and cannot determine how many target elements are to the left and right of that element.

The problem requires inserting the target element at the leftmost position, so we need to find the index of the leftmost target in the array. Initially, consider implementing this through the steps shown in the figure below:

1. Perform binary search to obtain the index of any target, denoted as $k$.
2. Starting from index $k$, perform linear traversal to the left, and return when the leftmost target is found.

Although this method works, it includes linear search, resulting in a time complexity of $O(n)$. When the array contains many duplicate target elements, this method is very inefficient.

Now consider extending the binary search code. As shown in the figure below, the overall process remains unchanged: calculate the midpoint index $m$ in each round, then compare target with nums[m], divided into the following cases:

• When nums[m] < target or nums[m] > target, it means target has not been found yet, so use the ordinary binary search interval narrowing operation to make pointers $i$ and $j$ approach target.
• When nums[m] == target, it means elements less than target are in the interval $[i, m - 1]$, so use $j = m - 1$ to narrow the interval, thereby making pointer $j$ approach elements less than target.

After the loop completes, $i$ points to the leftmost target, and $j$ points to the first element less than target, so index $i$ is the insertion point.

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Observe the following code: the operations for branches nums[m] > target and nums[m] == target are the same, so the two can be merged.

Even so, we can still keep the conditional branches expanded, as the logic is clearer and more readable.

[file]{binary_search_insertion}-[class]{}-[func]{binary_search_insertion}

!!! tip

The code in this section all uses the "closed interval" approach. Interested readers can implement the "left-closed right-open" approach themselves.

Overall, binary search is simply about setting search targets for pointers $i$ and $j$ separately. The target could be a specific element (such as target) or a range of elements (such as elements less than target).

Through continuous binary iterations, both pointers $i$ and $j$ gradually approach their preset targets. Ultimately, they either successfully find the answer or stop after crossing the boundaries.