Oursky Developer Pre-Test

answers.md

Question: 1 Write a function that takes two arrays as input, each array contains a list of A-Z; Your program should return True if the 2nd array is a subset of 1st array, or False if not.

Answer:

 import java.util.Arrays;
 import java.util.HashSet;

public class SubsetExample {

public static boolean isSubset_simple(String arr1[],String arr2[]) {
    int i = 0;
    int j = 0;
    int m = arr1.length;
    int n = arr2.length;
    for (i = 0; i < n; i++) {
        for (j = 0; j < m; j++) {
            if (arr2[i].equals(arr1[j])) {
                break;
            }
        }
        
        if (j == m) {
            System.out.println("Value of j "+j);
            System.out.println("Value of m "+m);
            return false;
        }
    }
    return true;
}


public static boolean isSubset_hash(String arr1[], String arr2[]){
    int m = arr1.length;
    int n =  arr2.length;
    HashSet<String> hset= new HashSet<>(); 
      
    for(int i = 0; i < m; i++){ 
        if(!hset.contains(arr1[i])) 
            hset.add(arr1[i]); 
    } 
          
    for(int i = 0; i < n; i++) { 
        if(!hset.contains(arr2[i])) 
            return false; 
    } 
    return true; 
}



public static void main(String[] args) {
    String arr1[] = {"A", "B", "C", "D", "E"};
    String arr2[] = {"A", "D", "Z"};

    if (isSubset_simple(arr1, arr2)) {
        System.out.println("True");
    } else {
        System.out.println("False");
    }
    
}

}

Question: 2 What is the computational complexity of your answer in Question 1? Can you explain why?

Answer I'm using O(m*n) time complexity and hashing method to getting a result. I thought that was much easier for me.

Question: 3 Write a function that takes an array of integers as input. For each integer, output the next fibonacci number. Solution that work both cpu and memory efficient are appreciated.

Answer

import java.util.HashMap;
import java.util.Map;

public class FibonacciExample {

//Check the number is perfect square or not.
static boolean isPerfectSquare(int x) {
    int s = (int) Math.sqrt(x);
    return (s * s == x);
}

// Returns true if n is a Fibonacci Number, else false 
static boolean isFibonacci(int n) {
    return isPerfectSquare(5 * n * n + 4)
            || isPerfectSquare(5 * n * n - 4);
}

// find the index of fibonacci number
static int findIndex(int n) {
    if (n <= 1) {
        return n;
    }
    int a = 0, b = 1, c = 1;
    int res = 1;
    while (c < n) {
        c = a + b;
        res++;
        a = b;
        b = c;
    }
    return res;
}

// find the fibonacci number by Index
static int fib(int n) {
    if (n <= 1) {
        return n;
    }
    return fib(n - 1) + fib(n - 2);
}

//    Find the next fibonacci number
public static int getNextFib(int n) {
    int[] fibs = new int[100];
    int i = 0;
    fibs[0] = 0;
    fibs[1] = 1;
    for (i = 2; i <= n + 1; i++) {
        fibs[i] = fibs[i - 1] + fibs[i - 2];
        if (fibs[i] > n) {
            return fibs[i];
        }
    }
            
    if (fibs[i - 1] < n) {
        return fibs[i - 1] + n;
    }
    
    if(fibs[i-1]== n){
         return fibs[i-1]+ fibs[i-2];
    }

    if (fibs[i] == n - 1 || fibs[i] == n) {
        if (n != 0) {
            if (n == 1) {
                return 2;
            } else {
                return fibs[i + 1] + fibs[i + 2];
            }
        } else {
            return 1;
        }
    }

    if (fibs[i - 1] == 1) {
        return fibs[n + n];
    }
    return n;
}

    //Use hashmap
public static Map<String, String> nextFibonacci(int[] num) {
    int len = num.length;
    Map<String, String> fibNum = new HashMap<>();
    for (int i = 0; i < len; i++) {
        if (isFibonacci(num[i])) {
            // Using a index of given number and find the next fibonacci number.
//                int index = findIndex(num[i])+1;
//                int nextFibNum = fib(index);
            int nextFibNum = getNextFib(num[i]);
            fibNum.put(String.valueOf(i), String.valueOf(nextFibNum));
        } else {
            fibNum.put(String.valueOf(i), String.valueOf(num[i]) + " is not Fibonacci Number");
        }
    }
    return fibNum;
}

public static void main(String[] args) {
    int[] numbers = {1, 21, 8};
    for(String nums: nextFibonacci(numbers).values()){
        System.out.println(nums);
    }
  }
}

Question: 4 Please explain what is the bug in the following Javascript function, and how to correct it.

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

Answer:

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

To use const / let instead of var, which makes each closure pointing to the block scope variable, i.