Facebook Interview Question for SDE1s


Country: United States




Comment hidden because of low score. Click to expand.
2
of 2 vote

Solution in python.
Cost O(n). (We update the variables once in the loop constant time computed n times => O(n))

def fib():
     a = 0; b = 1
     while True:
             yield a
             temp = a
             a += b
             b = temp

- Fernando June 09, 2017 | Flag Reply
Comment hidden because of low score. Click to expand.
1
of 1 vote

I think the interviewer is looking for a Iterator class with functions like hasNext() and next(). Otherwise it is too simple for an FB interview. You can google search Iterator definitions .., the code snippet
is for Java

//class variables
	long current = 1;
	long prev = 0;

	public void reset () {
        	current = 1;
        	prev = 0;
    	}

	@Override
 	public boolean hasNext() {
		//will always have a next 
		//alternatively you could get some brownie points by returning false when you 
		//reach the largest positive 'long' number
        	return true;
    	}

	@Override
	public Long next() {
        	long temp = current;
       		current = prev + current;
        	prev = temp;
        	return temp;
    	}

- nomadicdude July 06, 2017 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

#include <stdio.h>
 
 int main(void) {
         int i, v1 = 1, v2 = 0;
         int element;
 
         scanf("%d", &element);
 
         for (i = 0; i < element; ++i) {
                 v2 += v1;
                 v1 = v2 - v1;
                 printf("%d ", v2);
         }
 
         return 0;
 }

- CRSJ June 09, 2017 | Flag Reply
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0
of 0 vote

import java.util.ArrayList;

public class Fibonacci {
	
	public static void main(String[] args) {
		
		ArrayList<Integer> fibo=new ArrayList<>();
		int end=10;
		fibo.add(0);
		fibo.add(1);
		for(int i=2;i<end;i++){
			fibo.add(fibo.get(i-2)+fibo.get(i-1));
		}
		
		for(int i:fibo)
			System.out.print(i+",");
	}
}

- ea June 09, 2017 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

public class Fibo {
int n;

int fiboSum(int n){
if(n <= 2)
return 1;
else
return fiboSum(n-2) + fiboSum(n-1);
}

public static void main(String[] args) {
Fibo fb = new Fibo();
int num = 9;
int result = fb.fiboSum(num);
System.out.println("num " + num + " is : " + result);
}
}

- Jin June 14, 2017 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

public class Fibo {
	int n;

	int fiboSum(int n){
		if(n <= 2)
			return 1;
		else
			return fiboSum(n-2) + fiboSum(n-1);
	}
	
	public static void main(String[] args) {
		Fibo fb = new Fibo();
		int num = 9;
		int result = fb.fiboSum(num);
		System.out.println("num " + num + " is : " + result);
	}

}

- Jin June 14, 2017 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

using recursion

var act = 1;
  var limit = 8;

  function fib (x, prev) {
  	console.log(act);
  	act = x+prev;
  	prev = x;
    if(act <= limit) {
    	fib(act, prev);
    } 
  }

  fib(1,0);

- jaylu June 26, 2017 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

Assuming we have to implement next() and hasNext() methods (just like traditional iterators), I propose the following solution. I am using a deque since removing/adding elements to the front or back of the structure is O(1). I also assume in the implementation that the user specifies the limit or the first 'n' fibonacci numbers that he wishes to see.

Solution:

from collections import deque
class FibonacciIterator:
  def __init__(self, limit):
    self.results = deque([0, 1])
    self.curPos = 0
    self.limit = limit

  def hasNext(self):
    return self.curPos < self.limit

  def next(self):
    firstNum = self.results.popleft()
    self.curPos += 1
    self.results.append(self.results[0] + firstNum)
    return firstNum

Test Code:

f = FibonacciIterator(15) # Want the first 15 fib numbers
while f.hasNext():
  print(f.next(),end=', ')
print()
# Output: 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377,

- prudent_programmer March 18, 2018 | Flag Reply


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