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translation: update binary_search _edge.md
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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 $-1$.
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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 $-1$.
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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`**.
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Look back on the method of binary search for an insertion point, after the search is completed, the index $i$ will point to the leftmost occurrence of `target`. Therefore, **searching for the insertion point is essentially the same as finding the index of the leftmost `target`**.
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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:
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We can use the function for finding an insertion point to find the left boundary of `target`. Note that the array might not contain `target`, which could lead to the following two results:
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- The index $i$ of the insertion point is out of bounds.
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- The index $i$ of the insertion point is out of bounds.
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- The element `nums[i]` is not equal to `target`.
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- The element `nums[i]` is not equal to `target`.
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## Find the right boundary
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## Find the right boundary
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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.
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How do we find the rightmost occurrence of `target`? The most straightforward way is to modify the traditional binary search logic by changing how we adjust the search boundaries in the case of `nums[m] == target`. The code is omitted here. If you are interested, try to implement the code on your own.
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Below we introduce two more cunning methods.
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Below we are going to introduce two more ingenious methods.
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### Reusing the search for the left boundary
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### Reuse the left boundary search
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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`**.
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To find the rightmost occurrence of `target`, we can reuse the logic for finding the leftmost occurrence of `target`. Specifically, we can find the leftmost `target`, and then adjust the result to point to the rightmost `target` by simply adding 1 to the index of the leftmost `target`.
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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**.
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As shown in the figure below, after the search is complete, pointer $i$ will point to the the position just after the leftmost `target` (i.e., `target + 1`), and pointer $j$ will point to the rightmost `target`. Therefore, returning $j$ will give us the right boundary.
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![Transforming the search for the right boundary into the search for the left boundary](binary_search_edge.assets/binary_search_right_edge_by_left_edge.png)
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![Transforming the search for the right boundary into the search for the left boundary](binary_search_edge.assets/binary_search_right_edge_by_left_edge.png)
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Please note, the insertion point returned is $i$, therefore, it should be subtracted by $1$ to obtain $j$:
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Note that the insertion point returned is $i$, therefore, it should be subtracted by $1$ to obtain $j$:
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```src
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```src
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[file]{binary_search_edge}-[class]{}-[func]{binary_search_right_edge}
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[file]{binary_search_edge}-[class]{}-[func]{binary_search_right_edge}
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```
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```
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### Transforming into an element search
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### Transform into an element search
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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.
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When the array does not contain `target`, $i$ and $j$ will eventually point to the first element greater and smaller than `target` respectively.
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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.
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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.
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@ -50,7 +50,7 @@ Thus, as shown in the figure below, we can construct an element that does not ex
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![Transforming the search for boundaries into the search for an element](binary_search_edge.assets/binary_search_edge_by_element.png)
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![Transforming the search for boundaries into the search for an element](binary_search_edge.assets/binary_search_edge_by_element.png)
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The code is omitted here, but two points are worth noting.
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The code is omitted here, but here are two important points to note about this approach.
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- The given array does not contain decimals, meaning we do not need to worry about how to handle equal situations.
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- Since the given array does not contain decimals, this means we do not need to worry about handling equal cases.
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- 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).
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- Because this method introduces decimals, the variable `target` in the function needs to be changed to a floating point type (no change needed in Python).
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