This is inspired by reading through the book by Aditya Y. Bhargava
These are based on examples out of the book and either the same examples or ones applied to organizing data (top 100 babies first names, boys).
I did my best to refactor the Python algorithms in the book to Golang for pracice.
The selection sort algorithm sorts an array by repeatedly finding the minimum element (considering ascending order) from unsorted part and putting it at the beginning. The algorithm maintains two subarrays in a given array.
- The subarray which is already sorted.
- Remaining subarray which is unsorted. In every iteration of selection sort, the minimum element (considering ascending order) from the unsorted subarray is picked and moved to the sorted subarray.
Time Complexity: O(n2) as there are two nested loops
Bubble Sort is the simplest sorting algorithm that works by repeatedly swapping the adjacent elements if they are in wrong order.
Example: First Pass: ( 5 1 4 2 8 ) –> ( 1 5 4 2 8 ), Here, algorithm compares the first two elements, and swaps since 5 > 1. ( 1 5 4 2 8 ) –> ( 1 4 5 2 8 ), Swap since 5 > 4 ( 1 4 5 2 8 ) –> ( 1 4 2 5 8 ), Swap since 5 > 2 ( 1 4 2 5 8 ) –> ( 1 4 2 5 8 ), Now, since these elements are already in order (8 > 5), algorithm does not swap them.
Second Pass: ( 1 4 2 5 8 ) –> ( 1 4 2 5 8 ) ( 1 4 2 5 8 ) –> ( 1 2 4 5 8 ), Swap since 4 > 2 ( 1 2 4 5 8 ) –> ( 1 2 4 5 8 ) ( 1 2 4 5 8 ) –> ( 1 2 4 5 8 ) Now, the array is already sorted, but our algorithm does not know if it is completed. The algorithm needs one whole pass without any swap to know it is sorted.
Third Pass: ( 1 2 4 5 8 ) –> ( 1 2 4 5 8 ) ( 1 2 4 5 8 ) –> ( 1 2 4 5 8 ) ( 1 2 4 5 8 ) –> ( 1 2 4 5 8 ) ( 1 2 4 5 8 ) –> ( 1 2 4 5 8 )
Worst and Average Case Time Complexity: O(n*n). Worst case occurs when array is reverse sorted. Best Case Time Complexity: O(n). Best case occurs when array is already sorted.
Heap sort is a comparison-based sorting technique based on Binary Heap data structure. It is similar to selection sort where we first find the minimum element and place the minimum element at the beginning. We repeat the same process for the remaining elements.
A Binary Heap is a Complete Binary Tree where items are stored in a special order such that the value in a parent node is greater(or smaller) than the values in its two children nodes. The former is called max heap and the latter is called min-heap. The heap can be represented by a binary tree or array.
- Build a max heap from the input data.
- At this point, the largest item is stored at the root of the heap. Replace it with the last item of the heap followed by reducing the size of heap by 1. Finally, heapify the root of the tree.
- Repeat step 2 while the size of the heap is greater than 1.
O(nLogn)
Like QuickSort, Merge Sort is a Divide and Conquer algorithm. It divides the input array into two halves, calls itself for the two halves, and then merges the two sorted halves. The merge() function is used for merging two halves. The merge(arr, l, m, r) is a key process that assumes that arr[l..m] and arr[m+1..r] are sorted and merges the two sorted sub-arrays into one. The array is recursively divided into two halves till the size becomes 1. Once the size becomes 1, the merge processes come into action and start merging arrays back till the complete array is merged.
O(nLogn)
Like Merge Sort, QuickSort is a Divide and Conquer algorithm. It picks an element as pivot and partitions the given array around the picked pivot.
The key process in quickSort is partitioning. Target of partitions is, given an array and an element x of array as pivot, put x at its correct position in sorted array and put all smaller elements (smaller than x) before x, and put all greater elements (greater than x) after x. All this should be done in linear time.
Worst Case: O(n^2) Best and Avg Case: O(nLogn)
Search a sorted array by repeatedly dividing the search interval in half. Begin with an interval covering the whole array. If the value of the search key is less than the item in the middle of the interval, narrow the interval to the lower half. Otherwise, narrow it to the upper half. Repeatedly check until the value is found or the interval is empty.
O(Log n)
Breadth First Traversal (or Search) for a graph is similar to Breadth First Traversal of a tree. The only catch here is, unlike trees, graphs may contain cycles, so we may come to the same node again. To avoid processing a node more than once, we use a boolean visited array. For simplicity, it is assumed that all vertices are reachable from the starting vertex.
O(V+E) where V is number of vertices in the graph and E is number of edges in the graph.
Given a graph and a source vertex in the graph, find the shortest paths from the source to all vertices in the given graph. Dijkstra’s algorithm is very similar to Prim’s algorithm for minimum spanning tree. Like Prim’s MST, we generate a SPT (shortest path tree) with a given source as a root. We maintain two sets, one set contains vertices included in the shortest-path tree, other set includes vertices not yet included in the shortest-path tree. At every step of the algorithm, we find a vertex that is in the other set (set of not yet included) and has a minimum distance from the source.
- Create a set sptSet (shortest path tree set) that keeps track of vertices included in the shortest-path tree, i.e., whose minimum distance from the source is calculated and finalized. Initially, this set is empty.
- Assign a distance value to all vertices in the input graph. Initialize all distance values as INFINITE. Assign distance value as 0 for the source vertex so that it is picked first.
- While sptSet doesn’t include all vertices
- Pick a vertex u which is not there in sptSet and has a minimum distance value.
- Include u to sptSet.
- Update distance value of all adjacent vertices of u. To update the distance values, iterate through all adjacent vertices. For every adjacent vertex v, if the sum of distance value of u (from source) and weight of edge u-v, is less than the distance value of v, then update the distance value of v.
Given weights and values of n items, put these items in a knapsack of capacity W to get the maximum total value in the knapsack. In other words, given two integer arrays val[0..n-1] and wt[0..n-1] which represent values and weights associated with n items respectively. Also given an integer W which represents knapsack capacity, find out the maximum value subset of val[] such that sum of the weights of this subset is smaller than or equal to W. You cannot break an item, either pick the complete item or don’t pick it (0-1 property).
Recursion by Brute-Force algorithm OR Exhaustive Search Approach: A simple solution is to consider all subsets of items and calculate the total weight and value of all subsets. Consider the only subsets whose total weight is smaller than W. From all such subsets, pick the maximum value subset. Optimal Sub-structure: To consider all subsets of items, there can be two cases for every item.
- Case 1: The item is included in the optimal subset.
- Case 2: The item is not included in the optimal set. Therefore, the maximum value that can be obtained from ‘n’ items is the max of the following two values.
Maximum value obtained by n-1 items and W weight (excluding nth item). Value of nth item plus maximum value obtained by n-1 items and W minus the weight of the nth item (including nth item). If the weight of ‘nth’ item is greater than ‘W’, then the nth item cannot be included and Case 1 is the only possibility.