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91 lines
5.6 KiB
Markdown
91 lines
5.6 KiB
Markdown
# Binary search insertion
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Binary search is not only used to search for target elements but also to solve many variant problems, such as searching for the insertion position of target elements.
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## Case with no duplicate elements
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!!! question
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Given an ordered array `nums` of length $n$ and an element `target`, where the array has no duplicate elements. Now insert `target` into the array `nums` while maintaining its order. If the element `target` already exists in the array, insert it to its left side. Please return the index of `target` in the array after insertion. See the example shown in the figure below.
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If you want to reuse the binary search code from the previous section, you need to answer the following two questions.
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**Question one**: When the array contains `target`, is the insertion point index the index of that element?
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The requirement to insert `target` to the left of equal elements means that the newly inserted `target` replaces the original `target` position. Thus, **when the array contains `target`, the insertion point index is the index of that `target`**.
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**Question two**: When the array does not contain `target`, what is the index of the insertion point?
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Further consider the binary search process: when `nums[m] < target`, pointer $i$ moves, meaning that pointer $i$ is approaching an element greater than or equal to `target`. Similarly, pointer $j$ is always approaching an element less than or equal to `target`.
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Therefore, at the end of the binary, it is certain that: $i$ points to the first element greater than `target`, and $j$ points to the first element less than `target`. **It is easy to see that when the array does not contain `target`, the insertion index is $i$**. The code is as follows:
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```src
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[file]{binary_search_insertion}-[class]{}-[func]{binary_search_insertion_simple}
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```
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## Case with duplicate elements
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!!! question
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Based on the previous question, assume the array may contain duplicate elements, all else remains the same.
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Suppose there are multiple `target`s in the array, ordinary binary search can only return the index of one of the `target`s, **and it cannot determine how many `target`s are to the left and right of that element**.
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The task requires inserting the target element to the very left, **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.
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1. Perform a binary search, get an arbitrary index of `target`, denoted as $k$.
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2. Start from index $k$, and perform a linear search to the left until the leftmost `target` is found and return.
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Although this method is feasible, it includes linear search, so its time complexity is $O(n)$. This method is inefficient when the array contains many duplicate `target`s.
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Now consider extending the binary search code. As shown in the figure below, the overall process remains the same, each round first calculates the midpoint index $m$, then judges the size relationship between `target` and `nums[m]`, divided into the following cases.
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- When `nums[m] < target` or `nums[m] > target`, it means `target` has not been found yet, thus use the normal binary search interval reduction operation, **thus making pointers $i$ and $j$ approach `target`**.
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- When `nums[m] == target`, it indicates that the elements less than `target` are in the interval $[i, m - 1]$, therefore use $j = m - 1$ to narrow the interval, **thus making pointer $j$ approach elements less than `target`**.
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After the loop, $i$ points to the leftmost `target`, and $j$ points to the first element less than `target`, **therefore index $i$ is the insertion point**.
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=== "<1>"
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=== "<2>"
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=== "<3>"
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=== "<4>"
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=== "<5>"
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=== "<6>"
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=== "<7>"
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=== "<8>"
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Observe the code, the operations of the branch `nums[m] > target` and `nums[m] == target` are the same, so the two can be combined.
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Even so, we can still keep the conditions expanded, as their logic is clearer and more readable.
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```src
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[file]{binary_search_insertion}-[class]{}-[func]{binary_search_insertion}
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```
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!!! tip
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The code in this section uses "closed intervals". Readers interested can implement the "left-closed right-open" method themselves.
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In summary, binary search is merely about setting search targets for pointers $i$ and $j$, which might be a specific element (like `target`) or a range of elements (like elements less than `target`).
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In the continuous loop of binary search, pointers $i$ and $j$ gradually approach the predefined target. Ultimately, they either find the answer or stop after crossing the boundary.
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