Sorting

Sorting algorithms aren't usually tested alone. They are often paired with another problem type and the goal is to really see if you're able to think about the algorithms in a wider context

Bubble Sort

$$O(n^2)$$

• Traverse from left to right, swap element with neighbor is neighbor is less than element

• Repeat till no more swaps are needed

• Invariant: after every iteration, the largest elements are arranged at the end in ascending order

Selection Sort

$$O(n^2)$$

• Traverse from left to right, maintaining a “sorted” section in the front of the array

• Every iteration, we seek the next smallest element in the “unsorted” section and insert to the front

• Invariant: after iteration $j$, the front $j$ elements are arranged the smallest elements sorted in ascending order

Insertion Sort

$$O(n^2)$$

• Traverse from left to right

• When encountering a number less than the current element, we move backwards, finding elements before the current element that is greater than the number

• Repeat till a smaller number found, and then swap the elements into position

• Invariant: after iteration $j$, the front $j$ elements are arranged in ascending order (no guarantee to be the smallest $j$

Merge Sort

$$O(n \log n)$$

• $T(n) = 2T(n/2) + O(n)$

• Break up the array into half and merge at the end

• Stop breaking it up once only 1 element is left

• Invariant: the leftmost set of elements will be sorted before the rightmost

# [left, right)
def merge_sort(arr, left, right):
    if left == right:
        return arr[left]
    elif left < right:
        mid = left + (right - left) // 2
        left_merged = merge_sort(arr, left, mid)   # [left, mid)
        right_merged = merge_sort(arr, mid, right) # [mid, right)
        return merge(left_merged, right_merged)
        
def merge(a, b):
    c = []
    i, j = 0, 0
    while i < len(a) and j < len(b):
        if a[i] < b[j]:
            c.append(a[i])
            i += 1
        else:
            c.append(b[j])
            j += 1
    idx, remaining = (i, a) if i < len(a) else (j, b)
    for k in range(idx, len(remaining)):
        c.append(remaining[k])
        
    return c

Quick Sort

$$O(n\log n)$$

• Pick a pivot, partition array around pivot value, and repeat for all halves till 1 element

• Same recurrence as merge sort

• If duplicates allowed, use dutch flag algorithm, otherwise, can refer to the partitioning algorithms used in Quick Select

• Worst case can be $O(n^2)$ if array already sorted

Counting Sort

$$O(n+k)$$

• $k$ is the number of distinct elements

• Count frequency of each element

• Reconstruct the array from the frequency

• Can appear as ways to re-construct an array that has very little elements but many duplicates

def counting_sort(arr):
    offset = min(arr)
    freq = [0] * (max(arr) - offset + 1)
    for num in arr:
        freq[num - offset] += 1
    result = []
    for i in range(freq):
        while freq[i] > 0:
            result.append(i + offset)
    return result
    

Bucket Sort

$$O(n + \frac{n^2}{k} + k)$$

• For every element in the array, create buckets that correspond to some property (such as value and each bucket is a range)

• Sort elements of each individual bucket

• Join all buckets to form result

• Usually best if $k \approx n$ so the overall runtime could be $O(n)$