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

Solutions to Chapter 5 | Bit Manipulation
5.7 An array A[1...n] contains all the integers from 0 to n except for one number which is
missing. In this problem, we cannot access an entire integer in A with a single operation.
The elements of A are represented in binary, and the only operation we can use
to access them is “fetch the jth bit of A[i]”, which takes constant time. Write code to
find the missing integer. Can you do it in O(n) time?
SOLUTION
pg 58
Picture a list of binary numbers between 0 to n. What will change when we remove one
number? We’ll get an imbalance of 1s and 0s in the least significant bit. That is, before removing
the number k, we have this list of least significant bits (in some order):
0 0 0 0 0 1 1 1 1 1 OR 0 0 0 0 0 1 1 1 1
Suppose we secretly removed either a 1 or a 0 from this list. Could we tell which one was
removed?
remove(0 from 0 0 0 0 0 1 1 1 1 1) --> 0 0 0 0 1 1 1 1 1
remove(1 from 0 0 0 0 0 1 1 1 1 1) --> 0 0 0 0 0 1 1 1 1
remove(0 from 0 0 0 0 0 1 1 1 1) --> 0 0 0 0 1 1 1 1
remove(1 from 0 0 0 0 0 1 1 1 1) --> 0 0 0 0 0 1 1 1
Note that if 0 is removed, we always wind up with count(1) >= count(0). If 1 is removed, we
wind up with count(1) < count(0). Therefore, we can look at the least significant bit to figure
out in O(N) time whether the missing number has a 0 or a 1 in the least significant bit (LSB).
If LSB(missing) == 0, then we can discard all numbers with LSB = 1. If LSB(missing) == 1, we
can discard all numbers with LSB = 0.
What about the next iteration, with the second least significant bit (SLSB)? We’ve discarded
all the numbers with LSB = 1, so our list looks something like this (if n = 5, and missing = 3):
00000
00001
00010
-----
00100
00101
00110
00111
01000
01001
01010
01011
01100
01101
Our SLSBs now look like 0 0 1 0 1 0. Using the same logic as we applied for LSB, we can figure
out that the missing number must have SLSB = 1. Our number must look like xxxx11.
Third iteration, discarding numbers with SLSB = 0:
00000
00001
00010
-----
00100
00101
00110
00111
01000
01001
01010
01011
01100
01101
We can now compute that count(TLSB = 1) = 1 and count(TLSB = 1) = 1. Therefore, TLSB = 0.
We can recurse repeatedly, building our number bit by bit:
1 4 1
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