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Facebook Interview Question for Software Engineer / Developers


Country: United States
Interview Type: Phone Interview
More Questions from This Interview



Comment hidden because of low score. Click to expand.
1
of 1 vote

boolean isPalindrome(String s) {
    if (s == null || s.length() == 0) {
        return false; 
    }
    for (int s = 0, e = s.length() - 1; s < e; ++s, --e) {
       if (s.charSt(s) != s.charAt(e)) {
            return false;
        }
    }
    return true;
}

- Anonymous April 12, 2012 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 votes

e = s.length() - 1 <--- computing s.length() is a O(n) complexity. if you put that in loop, then you will end up making an O(n^2) solution. Not many pic this detail but pointing this out just in case.

- A March 25, 2013 | Flag
Comment hidden because of low score. Click to expand.
0
of 0 votes

No, s.length() is O(1) in java. However, it's true for strlen(s) in C

- Kevin August 15, 2013 | Flag
Comment hidden because of low score. Click to expand.
1
of 1 vote

Is it that simple at Facebook ?

- Orkut April 13, 2012 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 votes

You would think so would you. Are these "official" Facebook questions? Presumptuous at best!

- FBNerd February 26, 2013 | Flag
Comment hidden because of low score. Click to expand.
0
of 0 vote

boolean palinCheck(String s)
{
if(s.length()<=1)return true;
else if(s.charAt(0)==s.charAt(s.length()-1)))
return true&&palincheck(s.substring(1,s.length()-1));
else return false;



}

- Shashi April 12, 2012 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

Palindrome for what? string / array(int etc...) / linked list?

- Anonymous April 12, 2012 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

its easy when you go with string. But if provided with integer, then here comes the trick.
boolean prime(int n){
int length = 0;
int temp=n;
while( temp!=0 ) {temp/=10;length++;}
int i=0;
while(n!=0 ){
int lsd = n%10;
int msd = (int)(n/Math.pow(10,length-1));
if( lsd == msd ){
n %= Math.pow(10,length-1);
n /= 10;
length -= 2;
}else
return false;
}
return true;
}

- virat April 13, 2012 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

For number just reverse the number and check if they are equal.
rev = 0;
while ( num != 0 ) {
rev = rev*10 + num%10;
num=num/10;
}

- Raj May 11, 2012 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

It must work for numbers:

for (i=n;i>9;i=n)
    {
                    a= n/10;
                    b= n%10;
                    printf("%d",b);
                    n=a;
                    }
                    printf("%d",n);
                    
    getch();
}

- Bihan Sen May 23, 2012 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 votes

Of course, n is the number.<integer> :)

- Bihan Sen May 23, 2012 | Flag
Comment hidden because of low score. Click to expand.
0
of 0 votes

You'll get the reverse number by this.. print the original number before it to create the palindrome......

- Bihan Sen May 23, 2012 | Flag
Comment hidden because of low score. Click to expand.
0
of 0 vote

#!/usr/bin/python

def isPalindrome(s):
	for i in range(len(s) / 2):
		if s[i] != s[len(s)-i-1]:
			return False
	return True

s = raw_input("Enter string: ")
print(isPalindrome(s))

- took_two_seconds August 15, 2012 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 votes

I'd like to add a little thing to make this case insensitive:

#!/usr/bin/python

def isPalindrome(s):
	for i in range(len(s) / 2):
		front_char = s[i].lower()
		back_char = s[len(s)-i-1].lower()
		if front_char != back_char:
			return False
	return True

s = raw_input("Enter string: ")
print(isPalindrome(s))

- took_two_more August 15, 2012 | Flag
Comment hidden because of low score. Click to expand.
0
of 0 vote

#include <iostream>

int main()
{
	std::cout << "1234321" << std::endl;
	return 0;
}

- clow July 21, 2013 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

public static void main(String[] args) {
    assert isPalindrome("abbbcbbba");
    assert isPalindrome("aba");
    assert isPalindrome("aa");
    assert isPalindrome("a");
    assert isPalindrome("");
    assert !isPalindrome("ab");
    assert !isPalindrome("abc");
}

private static boolean isPalindrome(String str) {
    int l = 0, r = str.length() - 1;
    while (l < r) {
        if (str.charAt(l++) != str.charAt(r--)) {
            return false;
        }
    }
    return true;
}

- Safi December 11, 2014 | Flag Reply


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