Oursky Developer Pretest - haivicky@gmail.com

answer1.java

/*
  Program to check if an array is a subset of another array
*/

public class answer1{
	public static void main(String args[]){
		char [] a ={'A','B','C','D','E'};
		char [] b ={'A','D','D','Z'};
    
    System.out.println(isSubset(a,b));
	}

	private static boolean isSubset(char[] a, char []b){
		//Hashset is a Set implementation that uses Hashmap and stores values without duplicates
    HashSet <Character> hSet = new HashSet<Character>();
    
    //Store all the elements of major array that is to be searched
		for(char c:a){
			hSet.add(c);
		}
    
    //Loop through the second array and return false, if any one of its elements is not found in the original array
		for(char c:b){
			if(hSet.contains(c)!=true)
				return false;
		}
    //This section will reach only if the second array is the subset of the first array.
		return true;
	}
}

answer2.md

Time complexity is: O(n)

• The first loop takes O(n) to insert all the elements in the Hashset

• The second loop takes O(n) to loop through the query array, to check if the elemets are present in Hashset.

• So O(n) + O(n) = O(n)

The worst case time complexity of my algorithm is still O(n)

answer3.java

import java.util.*;

public class answer3{
	public static void main(String args[]){
		int [] c = {1,9,22};
		nextFibonacci(c);
	}

	private static void nextFibonacci(int[] arr){
		//Copy the initial array.
		int arr2[] = arr.clone(); 
		//HashMap is used with Key as the value in the array and initial value as 0
    HashMap<Integer,Integer> hMap = new HashMap<Integer,Integer>();
    
		// Initialize the HashMap with key as array values and value as 0;
    for(int val:arr){
			hMap.put(val, 0);
		}
    
    //Sort the input array and setup the initial parameters for the Fibonacci series. 
		Arrays.sort(arr);
		int a=0,b=1,c=0;
		int count =0;
		int i=0;
    
    //The below loop runs untill all the required fibonacci numbers for the entire input array is found. 
		do{
			
			while(c<=arr[i]){
				c = a+b;
				a=b;
				b=c;
			}
			hMap.put(arr[i],c);
			i++;
			count++;
			

		}while(count<arr.length);

    //Loop through the backup array and find the corresponding Fibonacci values from the Hashmap.
    for(int key:arr2){
			System.out.println(hMap.get(key));
		}


	}
}

answer4.js

/*
  The Bug in the program is, 
  
  function createArrayOfFunctions(y) {
    var arr = [];
    for(var i = 0; i<y; i++) {
      arr[i] = function(x) { return x + i; }
    }
    return arr;
  }
  
  The value of i is always y. 
  
  lets say, if we createArrayOffunctions(3), when the function is executed, we will be given 
  
  arr(3)
    0 ->function(x) 
    1 ->function(x)
    2 ->function(x)
  
  if you call one of this function eg, arr[0](3) - it will return, 3+3 = 6. 
  but I believe, what they want is 3+0 = 3. 
  
  So the solution is
  *******************
  
  we can write a closure or a self executing function to trap the variable i inside each function. 
*/

function createArrayOfFunctions(y)
{ 
  var arr = []; 
  for(var i = 0; i<y; i++) { 
    arr[i] = (
      function(i){
        return function(x) {
          return x + i; 
        } 
      })(i) 
  } return arr; }