AMD Interview Question for Associates


Country: India
Interview Type: In-Person




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1
of 1 vote

n = 4

def gen_bin(n):
    l2 = []
    l1 = [bin(i)[2:] for i in range(1, 2**n, 1)]
    for x in l1:
        y = x[::-1]
        if x not in l2 and x == y:
            l2.append(x)
    return l2[:n]

print gen_bin(n)

- Indraja.Punna June 13, 2016 | Flag Reply
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1
of 1 vote

// ZoomBA
def bin_palindrom( n ){
   tot = 0 
   for ( [1: 2 ** (n - 1) : 2] ){
     bin_s = str($,2)
     continue ( bin_s ** -1 != bin_s )
     println( bin_s )
     tot += 1
     break  ( tot == n ) 
   }
}
bin_palindrom ( 10 )

- NoOne October 06, 2016 | Flag Reply
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0
of 0 vote

public static void printPalindrom(int N)
        {
            List<int> list = new List<int>();

            while(N>0)
            {
                list.Add(1);

                if (list.Count <= 2)
                {
                    Console.WriteLine();
                    foreach (int n in list)
                    {
                        Console.Write(n);
                    }
                    N--;
                    continue;
                } 

                int i = 0;
                int j = list.Count - 1;

                while (i <= j - 2)
                {
                    i++;
                    j--;
                    list[i] = 0;
                    list[j] = 0;
                }

                while (i >= 0 && N > 0)
                {
                    Console.WriteLine();
                    foreach (int n in list)
                    {
                        Console.Write(n);
                    }
                    list[i] = 1;
                    list[j] = 1;
                    i--;
                    j++;
                    N--;
                }
            }
        }

- Anonymous June 09, 2016 | Flag Reply
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0
of 0 vote

My idea is we use an ArrayList to store all Palindromes. And
F(2i) = "1" + F(2i-2) + "1";
F(2i +1) = "1" + F(2i-2) + "1";
With F(j) is all values in this ArrayList. Note that before we add new value to ArrayList, we ensure that F(2i-2) and F(2i-1) have exact length by adding "0" at 2 sizes of F. (I use method format(String s, int length) to do this)

import java.util.ArrayList;

public class FirstNPalindromes {

	public static void main(String[] args) {
		print(10);
	}

	private static void print(int n) {
		ArrayList<String> list = new ArrayList<>();
		list.add("0");
		list.add("1");
		list.add("11");
		if (n == 0) {
			return;
		}
		if (n >= 1) {
			System.out.println("1");
		}
		if (n >= 2) {
			System.out.println("11");
		}
		int length = 3;// length of current palindromes
		int current = 3; // count size
		int i = 3;
		int j = 0;
		while (i < n) {
			j = 0;
			current = list.size();
			while (j < current && i <= n) {

				String obj = list.get(j);

				if ((length - obj.length()) % 2 != 0) {
					if (j == 0) {
						obj = "1" + format("", length - 2) + "1";
						list.add(obj);
						System.out.println(obj);
					}
					j++;
					continue;
				}
				obj = "1" + format(obj, length - 2) + "1";
				list.add(obj);
				System.out.println(obj);
				j++;
				i++;
			}
			length++;
		}
	}

	private static String format(String s, int length) {
		StringBuilder builder = new StringBuilder();
		int need = (length - s.length()) / 2;
		for (int i = 0; i < need; i++) {
			builder.append("0");
		}
		StringBuilder builderTotal = new StringBuilder();
		builderTotal.append(builder).append(s).append(builder);
		return builderTotal.toString();
	}
}

- ToanVu June 10, 2016 | Flag Reply
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0
of 0 vote

Converting int to binary and checking if its a palindrome.

public class PrintBinaryPalindome {

	public static void main(String[] args) {
		//Print first N binary palindrome numbers
		binaryPalindome(20);
	}

	////////////////////////////////////////
	//Method to Print N binary palindrome///
	////////////////////////////////////////
	public static void binaryPalindome(int n) {
		int i = 1;
		while (n > 0) {
			int binaryValue = decimalToBinary(i);
			if (isPalindrome(String.valueOf(binaryValue))) {
				System.out.println(binaryValue);
				n--;
			}
			i++;
		}//End of while
	} //End of binaryPalindome() method

	////////////////////////////////////////
	//Method to convert Decimal to Binary///
	////////////////////////////////////////
	private static int decimalToBinary(int input) {
		// Special case
		if (input == 1 || input == 0)
			return input;

		int result = 0;
		int p = 0;
		int base = 2;
		int finalBase = 10;
		while (Math.abs(input) > 0) { // 5
			int reminder = input % base;
			result += reminder * Math.pow(finalBase, p);
			input /= base;
			p++;
		} // End of while
		return result;

	}// End of decimalToBinary() Method

	//////////////////////////////////////////////////////
	//Method to Check if a number is palindrome or not?///
	//////////////////////////////////////////////////////
	private static boolean isPalindrome(String str){
		if(str==null) 
			return false;
		StringBuffer sb = new StringBuffer(str);
		sb.reverse();
		return sb.toString().equals(str);
	}//End of isPalindrome() method
}

- hitansu June 10, 2016 | Flag Reply
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0
of 0 vote

kok= lambda(n):True if str(n)==str(n)[::-1] else False 
>>> kok(101)
True

>>> kok(100)
False

- Sasha June 11, 2016 | Flag Reply
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0
of 0 vote

Restrict only to odd numbers for binary palindromes for positive numbers

- Aish June 15, 2016 | Flag Reply
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0
of 0 vote

Restrict only to odd numbers for binary palindromes for positive numbers

- Aish June 15, 2016 | Flag Reply
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0
of 0 vote

{
//if n is less than 3 ,then its a trivial palindrome
private static String [] trivialPalin ={"1","11"};

//main method which prints the palindromes
public static void printBinaryPalin(int n){
String arr[]=new String[n];

for(int i=0;i<((n>2)?2:n);i++)
arr[i]=trivialPalin[i];

int digitCount=1;
//for n>2 ,we use a recursive sub method to fetch substrings which are palindromes
//involving zeroes and ones
for(int i=3;i<=n;){
//call submethod which returns substrings of digit count
//for i=3 we call the sub method to return substring of digitcount i-2 ,that is 1 and
//soon
List<String> subStr = fetchSubStrPalin(digitCount);
digitCount++;
for(String str:subStr){
//we append 1 at the beginning to increase digits and a correspoding 1 at
//the end to make it a palindrome
arr[i-1]="1"+str+"1";
i++;
if(i>n)
break;
}
}
//print the accumulated binary palindromes
for(String str:arr){
System.out.println(str);
}

}

//recursive method to return palindromes involving 0 and 1
public static List<String> fetchSubStrPalin(int digits){
List<String> subStrList = new ArrayList<String>();
//trivial case 1
if(digits==1){
subStrList.add("0");
subStrList.add("1");
return subStrList;
}
else if(digits==2){//trivial case 2
subStrList.add("00");
subStrList.add("11");
return subStrList;
}
//if it is not a trivial case call recursively for digits -2
List<String> recursiveRes= (fetchSubStrPalin(digits-2));
//append 0 at beginning and end to make it a palindrome starting and ending with 0
for(String str:recursiveRes){
subStrList.add("0"+str+"0");
}
//similarly append 1 at beginning and end to make it a palindrome with
//begiining and trailing 1's
for(String str:recursiveRes){
subStrList.add("1"+str+"1");
}
return subStrList;
}
}

- sreenurao77 June 15, 2016 | Flag Reply
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0
of 0 vote

{{
import sys


def gen_bin(n):
pal = ['1']
if n < 1:
return []
if n < 2:
return pal
# How many palindromes I've encountered so far
count = 1
# Lenght of the next binary string
bin_len = 2
while True:
# Odd binary strings need a middle filler
middle = [''] if not bin_len % 2 else ['0', '1']
# Each binary number of lenght L will count up to
# 2**L. I'm only interested in odd counts to avoid
# leading zeros in the solution.
for x in xrange(1, 2**(bin_len/2), 2):
# Get binary string
s = bin(x)[2:]
# Pad with leading zeros
s = '0' * (bin_len/2 - len(s)) + s
for mid in middle:
pal.append(s[::-1] + mid + s)
count += 1
if count == n:
# Short cirtuit here to avoid extra computation.
return pal
bin_len += 1


print gen_bin(int(sys.argv[1])
}}

- DoorHolder June 16, 2016 | Flag Reply
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0
of 0 vote

We can count up from 0 and reverse and add every number to itself which will give us the correct ordering. The important trick is to hold off with the odd length palindromes until we hit a new power of two.

private List<String> getBitPalindromes(int count) {
    List<String> palindromes = new ArrayList<>(), oddLengths = new ArrayList<>();
    palindromes.add("1");
    int current = 1, powOfTwo = 0;
    while (palindromes.size() < count) {
      if (powOfTwo != Integer.numberOfLeadingZeros(current)) {
        palindromes.addAll(oddLengths);
        oddLengths.clear();
        powOfTwo = Integer.numberOfLeadingZeros(current);
      }
      String left = Integer.toBinaryString(current);
      String right = new StringBuilder(left).reverse().toString();
      palindromes.add(left + right);
      oddLengths.add(left + "0" + right);
      oddLengths.add(left + "1" + right);
      current++;
    }
    return palindromes.subList(0, count);
  }

- Johnie June 18, 2016 | Flag Reply
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of 0 vote
{{{ {{{ public class BinStrPalindrome { public static void generateBinStrings(int n){ /* * Helper function to set up the recursive call to generateBinStrings */ //Check for valid input if (n<1){ return; } // These 3 cases can produce all binary string palindromes. It is easy to see that they will not // produce any overlapping solutions. generateBinStrings("", n); // This will generate all even length palindromes generateBinStrings("1", n); //This will generate all odd length palindromes where '1' is middle bit generateBinStrings("0", n); //This will gererate all odd length palindromes where '0' is middle bit } private static void generateBinStrings(String bin_string, int n){ /* * prints all binary string palindromes up to length n */ //We don't want to print any palindromes that start (and end) in '0' also don't print the empty string if (bin_string.length() > 0 && bin_string.charAt(0) != '0'){ System.out.println(bin_string); } // Make sure we stop at n if (bin_string.length() + 2 <= n){ //Recursively generate the next palindrome by either appending '1' to both sides or '0' generateBinStrings("1" + bin_string + "1", n); generateBinStrings("0" + bin_string + "0", n); } } public static void main(String[] args){ generateBinStrings(12); generateBinStrings(3); generateBinStrings(0); } } }}} }}} - Anonymous June 21, 2016 | Flag Reply
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of 0 vote

public class BinStrPalindrome {  	public static void generateBinStrings(int n){ 		/* 		 * Helper function to set up the recursive call to generateBinStrings 		 */ 		 		//Check for valid input 		if (n<1){ 			return; 		} 		 		// These 3 cases can produce all binary string palindromes. It is easy to see that they will not 		// 		produce any overlapping solutions.  		generateBinStrings("", n);	// This will generate all even length palindromes 		generateBinStrings("1", n);		//This will generate all odd length palindromes where '1' is middle bit 		generateBinStrings("0", n);		//This will gererate all odd length palindromes where '0' is middle bit 	} 	 	private static void generateBinStrings(String bin_string, int n){ 		/* 		 * prints all binary string palindromes up to length n  		 */ 		 		//We don't want to print any palindromes that start (and end) in '0' also don't print the empty string 		if (bin_string.length() > 0 && bin_string.charAt(0) != '0'){ 			System.out.println(bin_string); 		} 		 		// Make sure we stop at n 		if (bin_string.length() + 2 <= n){ 			 			//Recursively generate the next palindrome by either appending '1' to both sides or '0' 			generateBinStrings("1" + bin_string + "1", n); 			generateBinStrings("0" + bin_string + "0", n); 		} 	} 	 	public static void main(String[] args){ 		generateBinStrings(12); 		generateBinStrings(3); 		generateBinStrings(0); 	} 	 }

- Jordan Casoli June 21, 2016 | Flag Reply
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-1
of 1 vote

import java.util.LinkedList;
import java.util.Queue;


public class BinaryPal {
	
	public static void printNum(int n){
		Queue q = new LinkedList();
		q.add("1");
		while(n!=0){
			n--;
			String s1=(String) q.peek();//retrives the node's value but do not removes it 
			q.poll();//gets the first node and removes it from the queue
			//System.out.println(s1);
			String temp=new StringBuffer(s1).reverse().toString();
			if(temp.equals(s1)){
				System.out.println(s1);
			}
			String s2=s1;
			q.add(s1.concat("0"));
			q.add(s2.concat("1"));
			
			
		}
		
		
	}
	/**
	 * @param args
	 */
	public static void main(String[] args) {
		// TODO Auto-generated method stub
		int n=10;
		printNum(100);
		
		

	}

}

- nandagirisaipranay June 10, 2016 | Flag Reply


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