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
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
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 , the front elements are arranged the smallest elements sorted in ascending order
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 , the front elements are arranged in ascending order (no guarantee to be the smallest
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 cPick 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 if array already sorted
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
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 so the overall runtime could be
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