PowerMockito: Mocking Private and Public Static Methods

Assume following class with private and public static methods and we want to write junit test which access those methods. We are mocking these static methods.

	//=================================================================================================
	class MockStaticMethod {
	public int getResult(int a, int b, boolean sum) {
	if (sum) {
	return sum(a, b);
	} else {
	return diff(a, b);
	}
	} // getResult
	//-----------------------------------------------------------------------------------------------
	public static int sum(int a, int b) {
	return a + b;
	} // sum
	//-----------------------------------------------------------------------------------------------
	private static int diff(int a, int b) {
	return a - b;
	} // diff
	} // MockStaticMethod

view raw MockStaticMethodTest.java hosted with ❤ by GitHub

Mock Private Static Method

	@Test
	public void testDiff() throws Exception {
	PowerMockito.mockStatic(MockStaticMethod.class);
	PowerMockito.doReturn(3).when(MockStaticMethod.class, "diff", Mockito.anyInt(), Mockito.anyInt());
	MockStaticMethod obj = new MockStaticMethod();
	Assert.assertEquals(3, obj.getResult(0, 0, false));
	Assert.assertEquals(3, obj.getResult(1, 2, false));
	} // testDiff

view raw MockStaticMethodTest.java hosted with ❤ by GitHub

For mocking private static method we need to use

	PowerMockito.mockStatic(MockStaticMethod.class);
	PowerMockito.doReturn(3).when(MockStaticMethod.class, "diff", Mockito.anyInt(), Mockito.anyInt());

view raw MockStaticMethodTest.java hosted with ❤ by GitHub

Mock Public Static Method

When mocking public static method we could use either of following:

	PowerMockito.mockStatic(MockStaticMethod.class);
	PowerMockito.when(MockStaticMethod.sum(Mockito.anyInt(), Mockito.anyInt())).thenReturn(100);

view raw MockStaticMethodTest.java hosted with ❤ by GitHub

OR

	PowerMockito.mockStatic(MockStaticMethod.class);
	PowerMockito.doReturn(100).when(MockStaticMethod.class, "sum", Mockito.anyInt(), Mockito.anyInt());

view raw MockStaticMethodTest.java hosted with ❤ by GitHub

Here is the implementation

	@Test
	public void testSumRight() throws Exception {
	PowerMockito.mockStatic(MockStaticMethod.class);
	PowerMockito.when(MockStaticMethod.sum(Mockito.anyInt(), Mockito.anyInt())).thenReturn(100);
	MockStaticMethod obj = new MockStaticMethod();
	Assert.assertEquals(100, obj.getResult(0, 0, true));
	Assert.assertEquals(100, obj.getResult(1, 2, true));
	} // testSumRight
	//-----------------------------------------------------------------------------------------------
	@Test
	public void testSumRight2() throws Exception {
	PowerMockito.mockStatic(MockStaticMethod.class);
	PowerMockito.doReturn(100).when(MockStaticMethod.class, "sum", Mockito.anyInt(), Mockito.anyInt());
	MockStaticMethod obj = new MockStaticMethod();
	Assert.assertEquals(100, obj.getResult(0, 0, true));
	Assert.assertEquals(100, obj.getResult(1, 2, true));
	} // testSumRight2

view raw MockStaticMethodTest.java hosted with ❤ by GitHub

BE CAREFUL NOT USE FOLLOWING WITH PUBLIC STATIC METHOD

	@Test
	public void testSumWrong() throws Exception {
	PowerMockito.mockStatic(MockStaticMethod.class);
	PowerMockito.doReturn(100).when(MockStaticMethod.sum(Mockito.anyInt(), Mockito.anyInt()));
	MockStaticMethod obj = new MockStaticMethod();
	Assert.assertEquals(100, obj.getResult(0, 0, true));
	Assert.assertEquals(100, obj.getResult(1, 2, true));
	} // testSumWrong

view raw MockStaticMethodTest.java hosted with ❤ by GitHub

This will give ERROR

  org.mockito.exceptions.misusing.UnfinishedStubbingException: 
  Unfinished stubbing detected here:
  -> at org.powermock.api.mockito.internal.PowerMockitoCore.doAnswer(PowerMockitoCore.java:36)

  E.g. thenReturn() may be missing.
  Examples of correct stubbing:
      when(mock.isOk()).thenReturn(true);
      when(mock.isOk()).thenThrow(exception);
      doThrow(exception).when(mock).someVoidMethod();
  Hints:
   1. missing thenReturn()
   2. you are trying to stub a final method, you naughty developer!
   3: you are stubbing the behaviour of another mock inside before 'thenReturn' instruction if completed      
      

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DeRegistration of a service removed health checks on other services?

$\mathtt{REFERENCE}$ @ GitHub Issue

Problem

I had been tyring to integrate Consul and Fabio.

I was able to run both these servers. Also I was able to register services to consul. That was reflected in Fabio as well. I was able to access the registered services through Fabio.

The problem I stuck into was when I tried to Deregister a service from consul, it did deregister but all the Checks from all other services were also gone. Because of the Fabio stopped accessing other services as well.

Solution

Well I really didn't find what was going on with Consul or Fabio.

All I did to fix this issue was Update the Consul version.

I had been using Consul Linux version 1.4.3.

I updated the consul to 1.5.0 and BOOM!! it fixed the issue.

---

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Daily Coding Problem #294: Smallest Weight Strictly Ascending Then Descending Path

Problem Definition

This problem was asked by Square.

A competitive runner would like to create a route that starts and ends at his house, with the condition that the route goes entirely uphill at first, and then entirely downhill.

Given a dictionary of places of the form {location: elevation}, and a dictionary mapping paths between some of these locations to their corresponding distances, find the length of the shortest route satisfying the condition above. Assume the runner's home is location 0.

For example, suppose you are given the following input:

  elevations = {0: 5, 1: 25, 2: 15, 3: 20, 4: 10}
  paths = {
      (0, 1): 10,
      (0, 2): 8,
      (0, 3): 15,
      (1, 3): 12,
      (2, 4): 10,
      (3, 4): 5,
      (3, 0): 17,
      (4, 0): 10
  }
  

In this case, the shortest valid path would be 0 -> 2 -> 4 -> 0, with a distance of 28.

Solution using BackTracking

To solve this problem we need to go through all possible valid paths and determine the minimum weight path.

It can be solved easily with Back Tracking using recursive method.

Solution Overview

• Start from the starting node which is 0.
• For each of the path that can be explored from current node, travel along that path if it is valid i.e. Elevation must be valid since we must travel strictly uphill and then downhill only. Anytime we encounter invalid node, we back track and follow different path. Also if node which has already been visited in given path is encountered we should back track since it will give cycle.
• If we can safely reach final destination node which is 0 again, we store the path summation in a list (OR update miniumum weight)
• After we have found a path, we need to back track again since there may be many possible paths and we need to find the optimal path.

Java Implementation

Following is implementation in Java. This prints lightest weight and the path. (GoLang version solution, below prints all valid paths with their weights.)

	/* File: ShortestPath294.java
	* Created: 2019-05-08
	* Author: Sabbir Manandhar
	*
	* Copyright (c) 2019 Hogwart Inc.
	*/
	import java.util.ArrayList;
	import java.util.HashMap;
	import java.util.HashSet;
	import java.util.List;
	import java.util.Map;
	import java.util.Set;
	//=================================================================================================
	public class ShortestPath294 {
	public static void main(String[] args) {
	Map<Integer, Integer> elevations = new HashMap<>();
	elevations.put(0, 5);
	elevations.put(1, 25);
	elevations.put(2, 15);
	elevations.put(3, 20);
	elevations.put(4, 10);
	int[][] weights = {
	{0, 1, 10},
	{0, 2, 8},
	{0, 3, 15},
	{1, 3, 12},
	{2, 4, 10},
	{3, 4, 5},
	{3, 0, 17},
	{4, 0, 10},
	{4, 2, 1}
	};
	findPath(elevations, weights);
	elevations.clear();
	elevations.put(0, 5);
	elevations.put(1, 15);
	elevations.put(2, 10);
	elevations.put(3, 25);
	elevations.put(4, 10);
	elevations.put(5, 20);
	elevations.put(6, 0);
	int[][] weights2 = {
	{0, 1, 3},
	{0, 2, 1},
	{1, 5, 5},
	{2, 4, 1},
	{2, 6, 1},
	{4, 0, 0},
	{4, 5, 1},
	{5, 0, 5},
	{5, 4, 1},
	{6, 0, 2},
	};
	findPath(elevations, weights2);
	} // main
	/**
	* Finds smallest weight path that ascends and then descends
	* @param elevations
	* @param weights
	*/
	public static void findPath(Map<Integer, Integer> elevations, int[][] weights) {
	Map<String, Integer> pathsWeight = new HashMap<>();
	Map<Integer, List<Integer>> neighbors = new HashMap<>();
	for(int[] path : weights) {
	pathsWeight.put(path[0] + "-" + path[1], path[2]);
	if(!neighbors.containsKey(path[0])) {
	neighbors.put(path[0], new ArrayList<>());
	}
	neighbors.get(path[0]).add(path[1]);
	}
	List<Integer> bestPath = new ArrayList<>();
	List<Integer> currentPath = new ArrayList<>();
	currentPath.add(0);
	int bestWeight = findPathHelper(elevations, pathsWeight, neighbors, new HashSet<Integer>(), false, 0, currentPath, 0, bestPath, Integer.MAX_VALUE);
	System.out.println(bestWeight);
	System.out.println(bestPath);
	} // findPath
	//---------------------------------------------------------------------------------------------
	/**
	* Helper method that recursively traverses all valid paths and identifies the lightest path.
	*
	* @param elevations node to elevation map
	* @param pathsWeight path to path weight map
	* @param neighbors node to neighbors map
	* @param visited visited nodes in a path
	* @param descending are we currently descending
	* @param current current node being explored
	* @param currentPath nodes in current path so far
	* @param currentPathSum current path sum so far
	* @param bestPath best path so far
	* @param bestWeight best weight so far
	* @return best weight along current node
	*/
	public static int findPathHelper(Map<Integer, Integer> elevations, Map<String, Integer> pathsWeight, Map<Integer, List<Integer>> neighbors,
	Set<Integer> visited, boolean descending, int current, List<Integer> currentPath, int currentPathSum, List<Integer> bestPath, int bestWeight) {
	int currentBestWeight = bestWeight; // multiple path can exist from a node. So we cannot return as soon as we find a path.
	// explore current node
	for(int neighbor : neighbors.get(current)) {
	String weightKey = current + "-" + neighbor;
	if(neighbor == 0) { // we are back to source i.e Destination
	if(elevations.get(neighbor) < elevations.get(current)) { // path should descend at end
	if(currentPathSum + pathsWeight.get(weightKey) < currentBestWeight) { // is the path best path than already found?
	bestPath.clear();
	bestPath.addAll(currentPath);
	bestPath.add(neighbor);
	currentBestWeight = currentPathSum + pathsWeight.get(weightKey);
	}
	}
	} else {
	boolean isPathValid = elevations.get(neighbor) != elevations.get(current); // Path should ascend or descend
	boolean expectDescend = elevations.get(neighbor) < elevations.get(current);
	isPathValid = isPathValid && (!descending || descending && expectDescend); // If we are ascending next can be either ascend or descend.
	// But if we are descending, it must not ascend again.
	if (isPathValid) {
	visited.add(neighbor);
	currentPath.add(neighbor);
	currentBestWeight = findPathHelper(elevations, pathsWeight, neighbors, visited, expectDescend, neighbor, currentPath, currentPathSum + pathsWeight.get(weightKey), bestPath, currentBestWeight);
	currentPath.remove(currentPath.size() - 1);
	visited.remove(neighbor);
	} else {
	// Path is not valid. IGNORE
	}
	}
	}
	return currentBestWeight;
	} // findPathHelper
	} // ShortestPath294

view raw ShortestPath294.java hosted with ❤ by GitHub

GoLang Implementation

Following is implementation in GoLang. This solution actually prints all valid paths with their path weight. It can be easily modified to return the minimum weight path.

	package main
	import "fmt"
	// TestShortestPath solves sample examples
	func TestShortestPath() {
	elevations := map[int]int{
	0: 5,
	1: 25,
	2: 15,
	3: 20,
	4: 10,
	}
	paths := [][3]int{
	{0, 1, 10},
	{0, 2, 8},
	{0, 3, 15},
	{1, 3, 12},
	{2, 4, 10},
	{3, 4, 5},
	{3, 0, 17},
	{4, 0, 10},
	{4, 2, 1},
	}
	findPath(elevations, paths)
	elevations = map[int]int{
	0: 5,
	1: 15,
	2: 10,
	3: 25,
	4: 10,
	5: 20,
	6: 0,
	}
	paths = [][3]int{
	{0, 1, 3},
	{0, 2, 1},
	{1, 5, 5},
	{2, 4, 1},
	{2, 6, 1},
	{4, 0, 0},
	{4, 5, 1},
	{5, 0, 5},
	{5, 4, 1},
	{6, 0, 2},
	}
	findPath(elevations, paths)
	} // TestShortestPath
	//-------------------------------------------------------------------------------------------------
	type path struct {
	weight int
	path []int
	}
	func (p *path) String() string {
	return fmt.Sprintf("Weight: %d Path: %v\n", p.weight, p.path)
	}
	//-------------------------------------------------------------------------------------------------
	// findPath initializes necessary data structures to call findPathHelper
	func findPath(elevations map[int]int, paths [][3]int) {
	neighborsMap := make(map[int]*[]int)
	weights := make(map[string]int)
	for _, path := range paths {
	neighbors, ok := neighborsMap[path[0]]
	if !ok {
	newNeighbors := make([]int, 0)
	neighbors = &newNeighbors
	neighborsMap[path[0]] = neighbors
	}
	*neighbors = append(*neighbors, path[1])
	weightKey := fmt.Sprintf("%d-%d", path[0], path[1])
	weights[weightKey] = path[2]
	}
	var resultPaths = make([]*path, 0)
	var currentPath = make([]int, 0, 1)
	currentPath = append(currentPath, 0) // 0 is starting node of path
	var visited = make(map[int]bool)
	findPathHelper(elevations, neighborsMap, weights, visited, &resultPaths, &currentPath, 0, 0, false)
	fmt.Println(resultPaths)
	} // findPath
	//-------------------------------------------------------------------------------------------------
	// findPathHelper recursively finds all the paths and their sum
	func findPathHelper(elevations map[int]int, neighbors map[int]*[]int, weights map[string]int, visited map[int]bool, paths *[]*path, currentPath *[]int, current int, pathSum int, descending bool) {
	for _, neighbor := range *neighbors[current] {
	weightKey := fmt.Sprintf("%d-%d", current, neighbor)
	if neighbor == 0 {
	if elevations[current] > elevations[neighbor] {
	cpath := make([]int, len(*currentPath), len(*currentPath)+1)
	copy(cpath, *currentPath)
	cpath = append(cpath, neighbor)
	*paths = append(*paths, &path{
	weight: pathSum + weights[weightKey],
	path: cpath})
	} else {
	// Path didn't descend. Path is not valid. Do nothing
	}
	} else { // THIS BLOCK HAS REAPEATED CODE BLOCK. CAN BE MERGED AS IN JAVA VERSION. BUT THIG MIGHT CLARIFY THE SOLUTION.
	if !descending {
	if elevations[current] < elevations[neighbor] {
	visited[neighbor] = true
	*currentPath = append(*currentPath, neighbor) // add neighbor to current path
	findPathHelper(elevations, neighbors, weights, visited, paths, currentPath, neighbor, pathSum+weights[weightKey], false)
	*currentPath = (*currentPath)[:len(*currentPath)-1] // remove neighbor from current path
	visited[neighbor] = false
	} else if elevations[current] > elevations[neighbor] {
	visited[neighbor] = true
	*currentPath = append(*currentPath, neighbor) // add neighbor to current path
	findPathHelper(elevations, neighbors, weights, visited, paths, currentPath, neighbor, pathSum+weights[weightKey], true)
	*currentPath = (*currentPath)[:len(*currentPath)-1] // remove neighbor from current path
	visited[neighbor] = false
	} else {
	// Path didn't strictly = ascend or descend. Path not valid. ignore this neighbor
	}
	} else { // descending
	if elevations[current] > elevations[neighbor] {
	visited[neighbor] = true
	*currentPath = append(*currentPath, neighbor) // add neighbor to current path
	findPathHelper(elevations, neighbors, weights, visited, paths, currentPath, neighbor, pathSum+weights[weightKey], true)
	*currentPath = (*currentPath)[:len(*currentPath)-1] // remove neighbor from current path
	visited[neighbor] = false
	} else {
	// Path didn't descend. Path is not valid. Ignore this neighbor
	}
	}
	}
	}
	} // findPathHelper

view raw shortest_path.go hosted with ❤ by GitHub

Complexity

This solves the problem by traversing every possible paths. So the theoretical time complexity is $O(2^n)$.

However since the question has restriction that the path has to strictly ascend and then descend, we are returning path as invalid as soon as we encounter invalid path. Because of this restriction, in practice the time complexity will be $O(n).$

Alternative Solution with Memoization

In the previous solution, we start from first node 0 and keep on exploring all possible paths all the way to destination. During the exploration we keep updating track of path sum for all paths and update the best path.

In this solution, we compute the best path and best path weight that can be attained from given node. We cache this result and use it later when we encounter the same node in other path.

This solution does not require to pass currentPath, currentPathSum, bestPath and bestWeight to findPathHelper()

Computation is done from the end. Idea is lightest tail path will be part of the solution path.

Following in implementation.

	/* File: ShortestPath294.java
	* Created: 2019-05-08
	* Author: Sabbir Manandhar
	*
	* Copyright (c) 2019 Hogwart Inc.
	*/
	import java.util.ArrayList;
	import java.util.HashMap;
	import java.util.HashSet;
	import java.util.List;
	import java.util.Map;
	import java.util.Set;
	//=================================================================================================
	/**
	* Daily coding problem #294
	*
	* Computes shortest path that starts from 0 and ends in 0 again, strictly ascends and then descends.
	*
	*/
	public class ShortestPath294 {
	public static void main(String[] args) {
	Map<Integer, Integer> elevations = new HashMap<>();
	elevations.put(0, 5);
	elevations.put(1, 25);
	elevations.put(2, 15);
	elevations.put(3, 20);
	elevations.put(4, 10);
	int[][] weights = {
	{0, 1, 10},
	{0, 2, 8},
	{0, 3, 15},
	{1, 3, 12},
	{2, 4, 10},
	{3, 4, 5},
	{3, 0, 17},
	{4, 0, 10},
	{4, 2, 1}
	};
	findPath(elevations, weights);
	elevations.clear();
	elevations.put(0, 5);
	elevations.put(1, 15);
	elevations.put(2, 10);
	elevations.put(3, 25);
	elevations.put(4, 10);
	elevations.put(5, 20);
	elevations.put(6, 0);
	int[][] weights2 = {
	{0, 1, 3},
	{0, 2, 1},
	{1, 5, 5},
	{2, 4, 1},
	{2, 6, 1},
	{4, 0, 0},
	{4, 5, 1},
	{5, 0, 5},
	{5, 4, 1},
	{6, 0, 2},
	};
	findPath(elevations, weights2);
	} // main
	/**
	* Finds smallest weight path that ascends and then descends
	* @param elevations
	* @param weights
	*/
	public static void findPath(Map<Integer, Integer> elevations, int[][] weights) {
	Map<String, Integer> pathsWeight = new HashMap<>();
	Map<Integer, List<Integer>> neighbors = new HashMap<>();
	for(int[] path : weights) {
	pathsWeight.put(path[0] + "-" + path[1], path[2]);
	if(!neighbors.containsKey(path[0])) {
	neighbors.put(path[0], new ArrayList<>());
	}
	neighbors.get(path[0]).add(path[1]);
	}
	Object[] best = findPathHelper(elevations, pathsWeight, neighbors, new HashSet<Integer>(), false, 0, new HashMap<>(), new HashMap<>());
	System.out.println(best[0]);
	System.out.println(best[1]);
	} // findPath
	//---------------------------------------------------------------------------------------------
	/**
	* Helper method that recursively traverses all valid paths and identifies the lightest path. It computes from the end and uses memoization.
	*
	* @param elevations node to elevation map
	* @param pathsWeight path to path weight map
	* @param neighbors node to neighbors map
	* @param visited visited nodes in a path
	* @param descending are we currently descending
	* @param current current node being explored
	* @param weightCache cache for weights holding best weight that can be achieved from a node
	* @param pathCache cache for paths holding best weight that can be achieved from a node
	* @return array of best weight and best path from current node
	*/
	public static Object[] findPathHelper(Map<Integer, Integer> elevations, Map<String, Integer> pathsWeight, Map<Integer, List<Integer>> neighbors,
	Set<Integer> visited, boolean descending, int current, Map<Integer, Integer> weightCache, Map<Integer, List<Integer>> pathCache) {
	if (weightCache.containsKey(current)) {
	return new Object[]{weightCache.get(current), pathCache.get(current)};
	}
	int bestWeightFromCurrent = Integer.MAX_VALUE;
	List<Integer> bestPathFromCurrent = new ArrayList<>();
	// explore current node
	for(int neighbor : neighbors.get(current)) {
	String weightKey = current + "-" + neighbor;
	if(neighbor == 0) { // we are back to source i.e Destination
	if(elevations.get(neighbor) < elevations.get(current)) { // path should descend at end
	if (bestWeightFromCurrent > pathsWeight.get(weightKey)) {
	bestWeightFromCurrent = pathsWeight.get(weightKey);
	bestPathFromCurrent.add(neighbor);
	}
	}
	} else {
	boolean isPathValid = elevations.get(neighbor) != elevations.get(current); // Path should ascend or descend
	boolean expectDescend = elevations.get(neighbor) < elevations.get(current);
	isPathValid = isPathValid && (!descending || descending && expectDescend); // If we are ascending next can be either ascend or descend.
	// But if we are descending, it must not ascend again.
	if (isPathValid) {
	visited.add(neighbor);
	Object[] pathDetails = findPathHelper(elevations, pathsWeight, neighbors, visited, expectDescend, neighbor, weightCache, pathCache);
	int pathWeightFromNeighbor = (int) pathDetails[0];
	if(pathWeightFromNeighbor != Integer.MAX_VALUE) {
	int weightWithNeighbor = pathsWeight.get(weightKey) + pathWeightFromNeighbor;
	if (bestWeightFromCurrent > weightWithNeighbor) {
	bestWeightFromCurrent = weightWithNeighbor;
	bestPathFromCurrent.clear();
	bestPathFromCurrent.addAll((List)pathDetails[1]);
	}
	}
	visited.remove(neighbor);
	} else {
	// Path is not valid. IGNORE
	}
	}
	}
	weightCache.put(current, bestWeightFromCurrent);
	bestPathFromCurrent.add(0, current);
	pathCache.put(current, bestPathFromCurrent);
	return new Object[]{bestWeightFromCurrent, bestPathFromCurrent};
	} // findPathHelper
	} // ShortestPath294

view raw ShortestPath294.java hosted with ❤ by GitHub

Complexity

The time complexity will be $O(n).$

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Daily Coding Problem #291: Minimum Number of Boats to save population

Problem Definition

An imminent hurricane threatens the coastal town of Codeville. If at most two people can fit in a rescue boat, and the maximum weight limit for a given boat is k, determine how many boats will be needed to save everyone.

For example, given a population with weights [100, 200, 150, 80] and a boat limit of 200, the smallest number of boats required will be 3.

Solution

To be able to determine the smallest number of boats, we need to maximize the number of pairing. To maximize the number of pairing, for given weight we must pair with the maximum value possible. If we pair we smaller weight we may not get the minimum number.

For example, given a population with weights [30, 150, 170, 50] and a boat limit of 200. Here if we pair 30 with 50, then 150 and 170 are left alone hence resulting into the result 3. However, it is possible to pair 30 with 170 and 50 with 150 satisfying the constraint. This pairing will give the result 2.

Therefore, for each weight, we must pair it with the maximum weight possible. We might be tempted to use following solution:

1. For each weight, iterate through rest of the weights to find the maximum weight with which it can be paired. Make sure to avoid already used weight.

This solution will work, but it will have time complexity $O(n^2)$

Solution Using Sorting

Following is implementation is using sorting. This method is still $O(n^2)$. It takes $O(n)$ time find pair. I am trying to figure out if the pair could be found in $O(logn)$ time instead so that the overall time complexity will be $O(nlogn)$

Implementation

	/* File: Boats.java
	* Created: 7/24/2019
	* Author: Sabbir Manandhar
	*
	* Copyright (c) 2019 Hogwart Inc.
	*/
	import java.util.Arrays;
	import java.util.Collections;
	//=======================================================================================
	/**
	* @author Sabbir Manandhar
	* @version 1.0
	*/
	public class Boats {
	/**
	* Driver method
	*/
	public static void main(String[] args) {
	System.out.println(compute(new Integer[]{100, 200, 150, 80}, 200) + " == " + 3);
	System.out.println(compute(new Integer[]{30, 40, 70, 100, 200, 150, 80}, 200) + " == " + 4);
	System.out.println(compute(new Integer[]{}, 1) + " == " + 0);
	System.out.println(compute(new Integer[]{100}, 200) + " == " + 1);
	System.out.println(compute(new Integer[]{200, 10, 200, 200}, 200) + " == " + 4);
	System.out.println(compute(new Integer[]{200, 200, 10, 200, 200}, 200) + " == " + 5);
	System.out.println(compute(new Integer[]{170, 160, 30, 90}, 200) + " == " + 3);
	System.out.println(compute(new Integer[]{150, 150, 150, 30}, 200) + " == " + 3);
	} // main
	//-----------------------------------------------------------------------------------------------
	/**
	* Computes number of boats required
	* @param weights array of weights of population
	* @param limit weight limit of each boat
	* @return number of boats needed
	*
	*/
	public static int compute(Integer[] weights, int limit) {
	if(weights.length == 0) return 0;
	Arrays.sort(weights, Collections.reverseOrder());
	boolean[] used = new boolean[weights.length];
	int boats = 0;
	for(int idx = 0; idx < weights.length; idx++) {
	if (used[idx]) continue;
	if (weights[idx] == limit) {
	boats++;
	continue;
	}
	int pair = findPair(weights, used, limit - weights[idx], idx+1, weights.length-1);
	if (pair > -1) {
	used[pair] = true;
	}
	boats++;
	}
	return boats;
	} // compute
	/**
	* Find pairing for the limit provided
	* @param weights
	* @param used
	* @param limit
	* @param lo
	* @param hi
	* @return
	*/
	private static int findPair(Integer[] weights, boolean[] used, int limit, int lo, int hi) {
	int lastValid = -1;
	for (int i = hi; i >= lo; i--) {
	if (!used[i]) {
	if (weights[i] == limit) {
	return i;
	} else if (weights[i] < limit) {
	lastValid = i;
	} else {
	return lastValid;
	}
	} else {
	continue;
	}
	}
	return lastValid;
	} // findPair
	} // Boats

view raw Boats.java hosted with ❤ by GitHub

Complexity

Sorting can be done in $O(n logn)$ time.
Finding a pair will take $O(n)$ time.

$\therefore$ The Time Complexity of this method will be $\cal{O(n^2)}$.

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Kruskal's MST Algorithm Using Union-Find Data Structure

$\mathtt{REFERENCE}$ @ Source Class Notes

Kruskal's Algorithm

1. Like Prim's Algorithm, it is also a greedy algorithm.
2. While Prim's Algorithm starts from single node and keep growing to reach a final tree, this algorithm starts with $|V|$ trees and keep on combining to finally reach the MST.

Main Features

1. It starts with a forest of size $n = |V|$ trees.
2. At each step, it combines two trees to obtain a bigger tree and reduces the number of tress in the forest by 1.
3. When the forest has a single tree, it is a Minimum Spanning Tree (MST).

Use Union-Find data structure to make it faster.

Formal Sketch of Kruskal's MST Algorithm

1. $A = \phi$ // A is an empty forest
2. Sort the edges in the graph in increasing order of the weight.
3. for each edge $(u,v) \in G.E$ in increasing order of weight {
   1. if $u$ and $v$ are not already connected (find by using Union-Find) {
      1. $A = A \cup \{(u,v)\}$
      2. UNION(u,v) // Union-Find operation
      }
   }
4. return A

Implementation

Following is the implementation of above explanation.

	import java.util.Arrays;
	import java.util.Comparator;
	public class Kruskal {
	/**
	* Driver method
	*/
	public static void main(String[] args) {
	Graph graph = new Graph(7, 13);
	graph.addEdge(0, 1, 6);
	graph.addEdge(0, 2, 6);
	graph.addEdge(0, 3, 5);
	graph.addEdge(1, 3, 8);
	graph.addEdge(1, 4, 5);
	graph.addEdge(2, 3, 8);
	graph.addEdge(2, 5, 3);
	graph.addEdge(3, 4, 9);
	graph.addEdge(3, 5, 4);
	graph.addEdge(5, 6, 1);
	graph.addEdge(4, 6, 1);
	graph.addEdge(3, 6, 2);
	graph.addEdge(2, 6, 1);
	Kruskal k = new Kruskal();
	int[][] result = k.MST(graph);
	for(int[] res : result) {
	System.out.println(res[0] + " " + res[1] + " " + res[2]);
	}
	} // main
	//-----------------------------------------------------------------------------------------------
	/**
	* Computes the MST of given connected graph.
	* Returns array of N-1 edges that belongs to the MST.
	* Assumes that the input graph is connected.
	*
	* @param Graph graph whose MST is to be computed.
	* @return Array of edges belonging to the MST
	*
	* Algorithm:
	* 1. Create a sorted queue of Edges sorted by their weight
	* 2. Greedily Pick lightest edge and include it in the result if it does not create cycle
	* 3. When there are N-1 edges return the edges
	*
	*/
	public int[][] MST(Graph g) {
	// Initialize container to hold result edges. There will be N-1 edges
	int[][] result = new int[g.N - 1][3];
	int cursor = 0; // cursor to point to the index to insert an Edge
	// Initialize Union-Find data structure, that will assist to know
	// if two vertex are already connected or not
	WQuickUnion uf = new WQuickUnion(g.N);
	// Sort Graph Edges in non-decreasing order of the edge weight
	g.sortEdges();
	int[] edge = g.getNextEdge();
	while(edge != null) {
	if (!uf.find(edge[0], edge[1])) { // check if source and target nodes are connected or not
	result[cursor++] = edge;
	uf.union(edge[0], edge[1]);
	if (cursor == g.N) break; // we already have N-1 Tree edges
	}
	edge = g.getNextEdge();
	}
	return result;
	} // MST
	} // Kruskal
	//=================================================================================================
	/**
	* Class to represent a graph
	*/
	class Graph {
	int N; // Number of vertices. Vertices will be numbered 0, 1, 2, 3, ....
	int[][] E; // Edges in Graph. Each edge will be 3 element array representing source vertex, target vertex and the weight.
	int cursor = 0; // cursor for inserting and retrieving edges
	//-----------------------------------------------------------------------------------------------
	/**
	* Constructor
	* @param N number of vertices in the graph
	* @param E number of edges in the graph
	*
	*/
	public Graph(int N, int E) {
	this.N = N;
	this.E = new int[E][3];
	} // Graph
	//-----------------------------------------------------------------------------------------------
	/**
	* Adds an edge to the graph
	*
	* @param s source node
	* @param t target node
	* @param w weight of edge
	*
	*/
	public void addEdge(int s, int t, int w) {
	this.E[cursor++] = new int[]{s, t, w};
	} // addEdge
	//-----------------------------------------------------------------------------------------------
	/**
	* Sort the edges of the graph in increasing order of weights.
	*
	*/
	public void sortEdges() {
	Arrays.sort(this.E, new Comparator<int[]>() {
	@Override
	public int compare(int[] o1, int[] o2) {
	return o1[2] - o2[2];
	} // compare
	});
	this.cursor = 0;
	} // sortEdges
	//-----------------------------------------------------------------------------------------------
	/**
	* Get next edge as pointed by cursor.
	* Cursor will be reset if sortEdges() is called.
	* @return returns current edge pointed by cursor
	*/
	public int[] getNextEdge() {
	if(cursor == E.length) return null;
	return E[cursor++];
	} // getNextEdge
	} // Graph
	//=================================================================================================
	/**
	* Class to represent Union-Find Data structure
	* This is Weighted Quick Union version of the data structure.
	*
	*/
	class WQuickUnion {
	private int[] id; // id[i] represent parent of the element at index i
	private int[] size; //size[i] represent number of nodes element at index has
	//-----------------------------------------------------------------------------------------------
	/**
	* Constructor to initilize the data structure.
	* @param n number of elements.
	*
	*/
	public WQuickUnion(int n) {
	id = new int[n];
	size = new int[n];
	for (int i = 0; i < n; i++) {
	id[i] = i;
	size[i] = 1;
	}
	} // WQuickUnion
	//-----------------------------------------------------------------------------------------------
	/**
	* Identifies whether two nodes p and q are connected or not
	* @param p first node
	* @param q second node
	* @return True if the two nodes are connected otherwise False
	*/
	public boolean find(int p, int q) {
	return root(p) == root(q);
	} // find
	//-----------------------------------------------------------------------------------------------
	/**
	* Connects two nodes in the network
	* @param p first node
	* @param q second node
	*/
	public void union(int p, int q) {
	int i = root(p);
	int j = root(q);
	if (size[i] < size[j]) {
	id[i] = j;
	size[j] += size[i];
	} else {
	id[j] = i;
	size[i] += size[j];
	}
	} // union
	//-----------------------------------------------------------------------------------------------
	/**
	* Finds the root of current element
	* @param p current element index
	* @return index of the root of input index
	*/
	private int root(int p) {
	while(p != id[p]) {
	id[p] = id[id[p]]; // Path Compression. Keeps tree almost completely flat. Will still work if this line is removed.
	p = id[p];
	}
	return p;
	} // root
	} // WQuickUnion

view raw Kruskal.java hosted with ❤ by GitHub

Complexity

• Sorting takes $O(|E|log|E|)$ times
• Union-Find operations can be done in $O(log|V|)$ time if properly implemented

$\therefore$ The Time Complexity of this method will be $O(|E|log|E|)$ time.

for dense graph, $|E| \approx |V|^2 \Rightarrow$ SLOW
for sparse graph, $|E| \approx |V| \Rightarrow$ FAST

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