Cracking the Coding Interview, 4 Edition - 150 Programming Interview Questions and Solutions

Solutions to Chapter 5 | Bit Manipulation
5.2 Given a (decimal - e.g. 3.72) number that is passed in as a string, print the binary representation.
If the number can not be represented accurately in binary, print “ERROR”
SOLUTION
pg 58
First, let’s start off by asking ourselves what a non-integer number in binary looks like. By
analogy to a decimal number, the number n = 0.101 = 1 * (1/2^1) + 0 * (1/2^2) + 1 * (1/2^3).
Printing the int part of n is straight-forward (see below). To print the decimal part, we can
multiply by 2 and check if the 2*n is greater than or equal to one. This is essentially “shifting”
the fractional sum. That is:
r = 2*n = 2*0.101 = 1*(1 / 2^0) + 0*(1 / 2^1) + 1*(1 / 2^2) = 1.01
If r >= 1, then we know that n had a 1 right after the decimal point. By doing this continuously,
we can check every digit.
1 public static String printBinary(String n) {
2 int intPart = Integer.parseInt(n.substring(0, n.indexOf(‘.’)));
3 double decPart = Double.parseDouble(
4 n.substring(n.indexOf(‘.’), n.length()));
5 String int_string = “”;
6 while (intPart > 0) {
7 int r = intPart % 2;
8 intPart >>= 1;
9 int_string = r + int_string;
10 }
11 StringBuffer dec_string = new StringBuffer();
12 while (decPart > 0) {
13 if (dec_string.length() > 32) return “ERROR”;
14 if (decPart == 1) {
15 dec_string.append((int)decPart);
16 break;
17 }
18 double r = decPart * 2;
19 if (r >= 1) {
20 dec_string.append(1);
21 decPart = r - 1;
22 } else {
23 dec_string.append(0);
24 decPart = r;
25 }
26 }
27 return int_string + “.” + dec_string.toString();
28 }
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