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Revisit the English version (#1835)

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* Review the English version using Claude-4.5.

* Update mkdocs.yml

* Align the section titles.

* Bug fixes
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Yudong Jin
2025-12-30 17:54:01 +08:00
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### Key review
- Bubble sort works by swapping adjacent elements. By adding a flag to enable early return, we can optimize the best-case time complexity of bubble sort to $O(n)$.
- Insertion sort sorts each round by inserting elements from the unsorted interval into the correct position in the sorted interval. Although the time complexity of insertion sort is $O(n^2)$, it is very popular in sorting small amounts of data due to relatively fewer operations per unit.
- Quick sort is based on sentinel partitioning operations. In sentinel partitioning, it's possible to always pick the worst pivot, leading to a time complexity degradation to $O(n^2)$. Introducing median or random pivots can reduce the probability of such degradation. Tail recursion effectively reduce the recursion depth, optimizing the space complexity to $O(\log n)$.
- Merge sort includes dividing and merging two phases, typically embodying the divide-and-conquer strategy. In merge sort, sorting an array requires creating auxiliary arrays, resulting in a space complexity of $O(n)$; however, the space complexity for sorting a list can be optimized to $O(1)$.
- Bucket sort consists of three steps: distributing data into buckets, sorting within each bucket, and merging results in bucket order. It also embodies the divide-and-conquer strategy, suitable for very large datasets. The key to bucket sort is the even distribution of data.
- Counting sort is a variant of bucket sort, which sorts by counting the occurrences of each data point. Counting sort is suitable for large datasets with a limited range of data and requires data conversion to positive integers.
- Radix sort processes data by sorting it digit by digit, requiring data to be represented as fixed-length numbers.
- Overall, we seek sorting algorithm that has high efficiency, stability, in-place operation, and adaptability. However, like other data structures and algorithms, no sorting algorithm can meet all these conditions simultaneously. In practical applications, we need to choose the appropriate sorting algorithm based on the characteristics of the data.
- The figure below compares mainstream sorting algorithms in terms of efficiency, stability, in-place nature, and adaptability.
- Bubble sort achieves sorting by swapping adjacent elements. By adding a flag to enable early return, we can optimize the best-case time complexity of bubble sort to $O(n)$.
- Insertion sort completes sorting by inserting elements from the unsorted interval into the correct position in the sorted interval each round. Although the time complexity of insertion sort is $O(n^2)$, it is very popular in small data volume sorting tasks because it involves relatively few unit operations.
- Quick sort is implemented based on sentinel partitioning operations. In sentinel partitioning, it is possible to select the worst pivot every time, causing the time complexity to degrade to $O(n^2)$. Introducing median pivot or random pivot can reduce the probability of such degradation. By preferentially recursing on the shorter sub-interval, the recursion depth can be effectively reduced, optimizing the space complexity to $O(\log n)$.
- Merge sort includes two phases: divide and merge, which typically embody the divide-and-conquer strategy. In merge sort, sorting an array requires creating auxiliary arrays, with a space complexity of $O(n)$; however, the space complexity of sorting a linked list can be optimized to $O(1)$.
- Bucket sort consists of three steps: distributing data into buckets, sorting within buckets, and merging results. It also embodies the divide-and-conquer strategy and is suitable for very large data volumes. The key to bucket sort is distributing data evenly.
- Counting sort is a special case of bucket sort, which achieves sorting by counting the number of occurrences of data. Counting sort is suitable for situations where the data volume is large but the data range is limited, and requires that data can be converted to positive integers.
- Radix sort achieves data sorting by sorting digit by digit, requiring that data can be represented as fixed-digit numbers.
- Overall, we hope to find a sorting algorithm that is efficient, stable, in-place, and adaptive, with good versatility. However, just like other data structures and algorithms, no sorting algorithm has been found so far that simultaneously possesses all these characteristics. In practical applications, we need to select the appropriate sorting algorithm based on the specific characteristics of the data.
- The figure below compares mainstream sorting algorithms in terms of efficiency, stability, in-place property, and adaptability.
![Sorting Algorithm Comparison](summary.assets/sorting_algorithms_comparison.png)
![Sorting algorithm comparison](summary.assets/sorting_algorithms_comparison.png)
### Q & A
**Q**: When is the stability of sorting algorithms necessary?
**Q**: In what situations is the stability of sorting algorithms necessary?
In reality, we might sort based on one attribute of an object. For example, students have names and heights as attributes, and we aim to implement multi-level sorting: first by name to get `(A, 180) (B, 185) (C, 170) (D, 170)`; then by height. Because the sorting algorithm is unstable, we might end up with `(D, 170) (C, 170) (A, 180) (B, 185)`.
In reality, we may sort based on a certain attribute of objects. For example, students have two attributes: name and height. We want to implement multi-level sorting: first sort by name to get `(A, 180) (B, 185) (C, 170) (D, 170)`; then sort by height. Because the sorting algorithm is unstable, we may get `(D, 170) (C, 170) (A, 180) (B, 185)`.
It can be seen that the positions of students D and C have been swapped, disrupting the orderliness of the names, which is undesirable.
It can be seen that the positions of students D and C have been swapped, and the orderliness of names has been disrupted, which is something we don't want to see.
**Q**: Can the order of "searching from right to left" and "searching from left to right" in sentinel partitioning be swapped?
No, when using the leftmost element as the pivot, we must first "search from right to left" then "search from left to right". This conclusion is somewhat counterintuitive, so let's analyze the reason.
No. When we use the leftmost element as the pivot, we must first "search from right to left" and then "search from left to right". This conclusion is somewhat counterintuitive; let's analyze the reason.
The last step of the sentinel partition `partition()` is to swap `nums[left]` and `nums[i]`. After the swap, the elements to the left of the pivot are all `<=` the pivot, **which requires that `nums[left] >= nums[i]` must hold before the last swap**. Suppose we "search from left to right" first, and if no element larger than the pivot is found, **we will exit the loop when `i == j`, possibly with `nums[j] == nums[i] > nums[left]`**. In other words, the final swap operation will exchange an element larger than the pivot to the left end of the array, causing the sentinel partition to fail.
The last step of sentinel partitioning `partition()` is to swap `nums[left]` and `nums[i]`. After the swap is complete, the elements to the left of the pivot are all `<=` the pivot, **which requires that `nums[left] >= nums[i]` must hold before the last swap**. Suppose we first "search from left to right", then if we cannot find an element larger than the pivot, **we will exit the loop when `i == j`, at which point it may be that `nums[j] == nums[i] > nums[left]`**. In other words, the last swap operation will swap an element larger than the pivot to the leftmost end of the array, causing sentinel partitioning to fail.
For example, given the array `[0, 0, 0, 0, 1]`, if we first "search from left to right", the array after the sentinel partition is `[1, 0, 0, 0, 0]`, which is incorrect.
For example, given the array `[0, 0, 0, 0, 1]`, if we first "search from left to right", the array after sentinel partitioning is `[1, 0, 0, 0, 0]`, which is incorrect.
Upon further consideration, if we choose `nums[right]` as the pivot, then exactly the opposite, we must first "search from left to right".
Thinking deeper, if we select `nums[right]` as the pivot, then it's exactly the opposite - we must first "search from left to right".
**Q**: Regarding tail recursion optimization, why does choosing the shorter array ensure that the recursion depth does not exceed $\log n$?
**Q**: Regarding the optimization of recursion depth in quick sort, why can selecting the shorter array ensure that the recursion depth does not exceed $\log n$?
The recursion depth is the number of currently unreturned recursive methods. Each round of sentinel partition divides the original array into two subarrays. With tail recursion optimization, the length of the subarray to be recursively followed is at most half of the original array length. Assuming the worst case always halves the length, the final recursion depth will be $\log n$.
The recursion depth is the number of currently unreturned recursive methods. Each round of sentinel partitioning divides the original array into two sub-arrays. After recursion depth optimization, the length of the sub-array to be recursively processed is at most half of the original array length. Assuming the worst case is always half the length, the final recursion depth will be $\log n$.
Reviewing the original quicksort, we might continuously recursively process larger arrays, in the worst case from $n$, $n - 1$, ..., $2$, $1$, with a recursion depth of $n$. Tail recursion optimization can avoid this scenario.
Reviewing the original quick sort, we may continuously recurse on the longer array. In the worst case, it would be $n$, $n - 1$, $\dots$, $2$, $1$, with a recursion depth of $n$. Recursion depth optimization can avoid this situation.
**Q**: When all elements in the array are equal, is the time complexity of quicksort $O(n^2)$? How should this degenerate case be handled?
**Q**: When all elements in the array are equal, is the time complexity of quick sort $O(n^2)$? How should this degenerate case be handled?
Yes. For this situation, consider using sentinel partitioning to divide the array into three parts: less than, equal to, and greater than the pivot. Only recursively proceed with the less than and greater than parts. In this method, an array where all input elements are equal can be sorted in just one round of sentinel partitioning.
Yes. For this situation, consider partitioning the array into three parts through sentinel partitioning: less than, equal to, and greater than the pivot. Only recursively process the less than and greater than parts. Under this method, an array where all input elements are equal can complete sorting in just one round of sentinel partitioning.
**Q**: Why is the worst-case time complexity of bucket sort $O(n^2)$?
In the worst case, all elements are placed in the same bucket. If we use an $O(n^2)$ algorithm to sort these elements, the time complexity will be $O(n^2)$.
In the worst case, all elements are distributed into the same bucket. If we use an $O(n^2)$ algorithm to sort these elements, the time complexity will be $O(n^2)$.