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en/docs/chapter_searching/binary_search_edge.md
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comments: true
---
# 10.3   Binary Search Edge Cases
# 10.3   Binary Search Boundaries
## 10.3.1   Finding the Left Boundary
!!! question
Given a sorted array `nums` of length $n$ that may contain duplicate elements, return the index of the leftmost element `target` in the array. If the array does not contain the element, return $-1$.
Given a sorted array `nums` of length $n$ that may contain duplicate elements, return the index of the leftmost occurrence of `target`. If the array does not contain `target`, return $-1$.
Recall the method for finding the insertion point with binary search. After the search completes, $i$ points to the leftmost `target`, **so finding the insertion point is essentially finding the index of the leftmost `target`**.
@@ -232,9 +232,9 @@ Below we introduce two more clever methods.
### 1.   Reusing Left Boundary Search
In fact, we can use the function for finding the leftmost element to find the rightmost element. The specific method is: **Convert finding the rightmost `target` into finding the leftmost `target + 1`**.
In fact, we can use the function for finding the leftmost `target` to find the rightmost `target`. The specific method is: **convert finding the rightmost `target` into finding the leftmost `target + 1`**.
As shown in Figure 10-7, after the search completes, pointer $i$ points to the leftmost `target + 1` (if it exists), while $j$ points to the rightmost `target`, **so we can simply return $j$**.
As shown in Figure 10-7, after the search completes, the pointer $i$ points to the leftmost `target + 1` (if it exists), while $j$ points to the rightmost `target`, **so we can return $j$**.
![Converting right boundary search to left boundary search](binary_search_edge.assets/binary_search_right_edge_by_left_edge.png){ class="animation-figure" }
@@ -480,8 +480,8 @@ We know that when the array does not contain `target`, $i$ and $j$ will eventual
Therefore, as shown in Figure 10-8, we can construct an element that does not exist in the array to find the left and right boundaries.
- Finding the leftmost `target`: Can be converted to finding `target - 0.5` and returning pointer $i$.
- Finding the rightmost `target`: Can be converted to finding `target + 0.5` and returning pointer $j$.
- Finding the leftmost `target`: This can be converted to finding `target - 0.5` and returning the pointer $i$.
- Finding the rightmost `target`: This can be converted to finding `target + 0.5` and returning the pointer $j$.
![Converting boundary search to element search](binary_search_edge.assets/binary_search_edge_by_element.png){ class="animation-figure" }
@@ -489,5 +489,5 @@ Therefore, as shown in Figure 10-8, we can construct an element that does not ex
The code is omitted here, but the following two points are worth noting:
- Since the given array does not contain decimals, we don't need to worry about how to handle equal cases.
- Since the given array does not contain decimal values, we do not need to worry about how to handle equality.
- Because this method introduces decimals, the variable `target` in the function needs to be changed to a floating-point type (Python does not require this change).