CS algorithms Flashcards
Linear Search
we simply start at the beginning of our data set and iterate through each item until we either i) find what we were searching for or ii) reach the end and conclude the item being searched for is not in the set.
Issues with Linear Search
it can be very time consuming.
Binary Search
Takes an ordered set of data and starts searching in the middle. If the middle element is what we were looking for, great! That means we’re done. If the middle element is smaller than the target, we can eliminate the entire left-hand side of our data set and repeat this process on the right-hand side.
Issues with Binary Search
can’t use binary searching unless the data is sorted!
Selection Sort
several playing cards you need to put in order. You start off by looking for the lowest card, and when you find it, put it at the front of your hand. Then, you look for the second lowest card and insert it directly behind the first card. You repeat this process, selecting the next lowest card and moving it behind the most recently placed card, until you have moved all cards into the right place.
drawbacks to selection sort
One of the drawbacks of this algorithm is its efficiency. Regardless of the starting order of your elements (random, mostly sorted, reverse sorted, etc.), this algorithm will always have a runtime of O(n²). This is because each pass requires us to check every element of the unsorted part of our list or array.
Is Selection Sort a comparison sorting algorithm?
Yes, since to select the desired element, we compare a current value to the rest of the “unsorted” segment of our list or array.
Why do we end our loop before processing the item in the last index of an array when performing Selection Sort?
By the nature of how this algorithm works, if all other elements have been sorted, the last element will have been moved to the correct index.
When using Selection Sort, what is the difference between the order or arrangement of elements in best case versus worst case?
None, since the same number of operations will be performed regardless of how elements are ordered.
Show how the array [4, 8, 3, 1, 9, 6] changes as it is being sorted using Selection Sort. To do this, write out the contents of the array after each pass of the algorithm. (hint…we should only need n-1 passes to sort the array, where n is the size)
[1, 8, 3, 4, 9, 6] [1, 3, 8, 4, 9, 6] [1, 3, 4, 8, 9, 6] [1, 3, 4, 6, 9, 8] [1, 3, 4, 6, 8, 9]
Insertion Sort
Imagine you lay out several playing cards in a line on a table. Consider the first card to be a sorted subarray of lengh = 1. Pick up each card (in this case, from left to right) and insert it into the correct position in the sorted subarray. Shift other cards to the right to make room for the card you are inserting as you go. O(n²).
main steps of Insertion sort
- Separate the first element from the rest of the array. Consider it a sorted list of one element.
- For all other indices, beginning with [1]:
a. Copy the current item into a temp variableb. Iterate to the left until you find the correct index in the “sorted” part of the array at which this element should be inserted
- Shift items over to the right as you iteratec. When the right index is found, copy temp into this position
Selection Sort steps
- Start with current index = 0
- For all indices EXCEPT last:a. Loop through elements on right-hand-side
of current index and find smallest elementb. Swap element at current index with
smallest element found in above loop
Is Insertion Sort a comparison sorting algorithm?
Yes, since to find the correct index at which we “insert” the current element we are sorting, we compare it to the rest of the “sorted” segment of our list or array.
When using Insertion Sort, what is the difference between the order or arrangement of elements in best case versus worst case?
When performing Insertion Sort, the algorithm runs more efficiently if the elements are already mostly sorted. When the original array is close to sorted order, fewer comparisons are required to sort it. But if the array is in reverse-sorted order, the number of comparisons that need to be done increases significantly. For example: [5, 4, 3, 2, 1] When inserting the 1 into the correct part of the array, we have to compare it with every other number. While this is not such a big deal in an array with 5 elements, it gets very costly as that array becomes bigger.
Show how the array [4, 8, 3, 1, 9, 6] changes as it is being sorted using Insertion Sort.
[4, 8, 3, 1, 9, 6] (comparison done, no swap made)
[3, 4, 8, 1, 9, 6]
[1, 3, 4, 8, 9, 6]
[1, 3, 4, 8, 9, 6] (comparison done, no swap made)
[1, 3, 4, 6, 8, 9]
Merge Sort
- Recursively divide orginal array or list in half until both pieces have a length of 1.
- Starting with these singular subsets, merge sorted pieces back together.
Quick Sort
In this algorithm, we move smaller elements to one side of the array and larger elements to the other side of the array by comparing them to a “pivot” element, which can be chosen in a few different ways. This divides our original array into two subarrays that are closer to being sorted. We then recursively call Quick Sort on each subarray until we are down to singular elements (which by their singular nature, must be sorted) runtime: best case: Ω(nlog(n)) worst: O(n²)
Quick Sort steps
While the data set has a length >= 1:
- Select pivot (typically first, last or middle)
- For each item in data set:
- if item < pivot AND not on LHS of pivot, move to appropriate side of data set
- else if item > pivot AND not on RHS of pivot, move to appropriate side of data set - Recursively Quick Sort LHS and RHS of pivot
