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

# Binary Search Insertion Point

Binary search can be used not only to search for target elements, but also to solve many variant problems, such as finding 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 `nums` while maintaining its sorted order. If `target` already exists in the array, insert it to its left. Return the index of `target` after insertion. An example is shown below.

![Binary search insertion point example data](binary_search_insertion.assets/binary_search_insertion_example.png)

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?

To analyze this further, consider the binary search process: when `nums[m] < target`, $i$ moves, meaning that 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, $i$ must point to the first element greater than `target`, and $j$ must point to the first element less than `target`. **It follows that when the array does not contain `target`, the insertion index is $i$**. The code is shown below:

```src
[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**. A straightforward initial approach is to follow 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.

![Linear search for insertion point of duplicate elements](binary_search_insertion.assets/binary_search_insertion_naive.png)

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: in each iteration, we first compute the midpoint index $m$, then compare `target` with `nums[m]`, leading to the following cases:

- When `nums[m] < target` or `nums[m] > target`, it means `target` has not been found yet, so use the standard interval-shrinking operation of binary search to **move pointers $i$ and $j$ closer to `target`**.
- When `nums[m] == target`, it means elements less than `target` are in the interval $[i, m - 1]$, so use $j = m - 1$ to shrink the interval, thereby **moving pointer $j$ closer to 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**.

=== "<1>"
    ![Steps for binary search insertion point of duplicate elements](binary_search_insertion.assets/binary_search_insertion_step1.png)

=== "<2>"
    ![binary_search_insertion_step2](binary_search_insertion.assets/binary_search_insertion_step2.png)

=== "<3>"
    ![binary_search_insertion_step3](binary_search_insertion.assets/binary_search_insertion_step3.png)

=== "<4>"
    ![binary_search_insertion_step4](binary_search_insertion.assets/binary_search_insertion_step4.png)

=== "<5>"
    ![binary_search_insertion_step5](binary_search_insertion.assets/binary_search_insertion_step5.png)

=== "<6>"
    ![binary_search_insertion_step6](binary_search_insertion.assets/binary_search_insertion_step6.png)

=== "<7>"
    ![binary_search_insertion_step7](binary_search_insertion.assets/binary_search_insertion_step7.png)

=== "<8>"
    ![binary_search_insertion_step8](binary_search_insertion.assets/binary_search_insertion_step8.png)

Observe the following code: the branches `nums[m] > target` and `nums[m] == target` perform the same operation, so they can be merged.

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

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

!!! tip

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

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

With each iteration of binary search, pointers $i$ and $j$ gradually approach their preset targets. Ultimately, they either find the answer or stop after crossing the boundary.