Facebook Interview Question for Software Engineer Interns


Country: United States
Interview Type: Phone Interview




Comment hidden because of low score. Click to expand.
9
of 9 vote

c++, implementation

typedef unsigned long long UINT64;

UINT64 getXOROne2N(UINT64 n) {
	switch (n % 4) {
		case 0: return n;
		case 1: return 1;
		case 2: return n + 1;
		case 3: return 0;
	}
	return 0;
}

- kyduke October 07, 2015 | Flag Reply
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0
of 0 votes

Nice solution, Can you please explain the logic ?

- paul October 08, 2015 | Flag
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3
of 3 votes

@Paul:
It's easy to see the cyclicity:
As a start:
1 mod 4 = 1.
1 xor 0 = 1.
Then in the 2nd step (n mod 4 = 2) you xor an even number (ie ending 0 in binary) with 1, thus it will always result in n+1.
In the third step (n mod 4 = 3) you xor a number by itself (the result was n+1 in the previous iteration), which naturally returns 0.
In the last step (n mod 4 = 0) you xor a number by 0, which results in the number itself.
Since that's even you will xor that with n+1, which is odd, the only bit difference will be the least significant, thus will result in 1 and the cycle starts over.

- tamas October 08, 2015 | Flag
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0
of 0 vote

#define FOLD(X)\
	X |= X >> 1;\
	X |= X >> 2;\
	X |= X >> 4;\
	X |= X >> 8; \
	X |= X >> 16; \
	X |= X >> 32


unsigned long long XOR(unsigned long long N)
{
	unsigned long long Res = 0;
	unsigned long long Original = N;

	if (N == 1)
	{
		return (1);
	}

	while (N > 1)
	{
		unsigned long long Closest = N;
		FOLD(Closest);
		Closest ^= Closest >> 1;
		Res |= (Original - Closest + 1) % 2 ? Closest : 0;
		N = N & (Closest - 1);

		if (N == 0)
		{
			Res += (Original % 2) ? 0 : 1;
			break;
		}
	}

	return (Res);
}

- paul October 08, 2015 | Flag Reply
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0
of 0 vote

#include <stdio.h>

int main()
{
int n;
long long value;
printf("Enter number : ");
scanf("%d", &n);
value = n--;
while (n > 0) {
value = value ^ n--;
}
printf("Value = %lld\n", value);
}

- Anonymous December 04, 2015 | Flag Reply
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0
of 0 vote

value = n--;
while (n > 0) {
value = value ^ n--;
}

- Anonymous December 04, 2015 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

unsigned int CalcXorFrom1ToN(int n)
{
    if (n % 2 == 0)
        return n;
    else
        return n ^ (n-1);
}

- Mohammad December 30, 2015 | Flag Reply
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0
of 0 vote

if n%4==0:
	print n
elif n%4==1:
	print 1
elif n%4==2:
	print (n/4)*4+3
else:
	print 0

- Sharad Boni February 22, 2017 | Flag Reply
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-1
of 1 vote

value = n--;
    while (n > 0) {
        value = value ^ n--;
    }

- Sanjeev December 04, 2015 | Flag Reply
Comment hidden because of low score. Click to expand.
-1
of 1 vote

In Ruby (2.0 or above):

def f(n)
  (1..n).inject {|a,b| a^b }
end

or shorter version:

def f(n)
  (1..n).inject(:^)
end

- kenneth.kin.lum January 04, 2016 | Flag Reply
Comment hidden because of low score. Click to expand.


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