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

Solutions to Chapter 20 | Hard
20.4 Write a method to count the number of 2s between 0 and n.
pg 91
SOLUTION
Picture a sequence of numbers:
0 1 2 3 4 5 6 7 8 9
10 11 12 13 14 15 16 17 18 19
20 21 22 23 24 25 26 27 28 29
...
110 111 112 113 114 115 116 117 118 119
The last digit will be repeated every 10 numbers, the last two digits will be repeated every
10^2 numbers, the last 3 digits will be repeated every 10^3 numbers, etc.
So, if there are X 2s between 0 and 99, then we know there are 2x twos between 0 and 199.
Between 0 and 299, we have 3x twos from the last two digits, and another 100 2s from the
first digit.
In other words, we can look at a number like this:
f(513) = 5 * f(99) + f(13) + 100
To break this down individually:
»»
The sequence of the last two digits are repeated 5 times, so add 5 * f(99)
»»
We need to account for the last two digits in 500 -> 513, so add f(13)
»»
We need to account for the first digit being two between 200 -> 299, so add 100
Of course, if n is, say, 279, we’ll need to account for this slightly differently:
f(279) = 2 * f(99) + f(79) + 79 + 1
To break this down individually:
»»
The sequence of the last two digits are repeated 2 times, so add 2 * f(99)
»»
We need to account for the last two digits in 200 -> 279, so add f(79)
»»
We need to account for the first digit being two between 200 -> 279, so add 79 + 1
Recu
rsive Code:
1 public static int count2sR(int n) {
2 // Base case
3 if (n == 0) return 0;
4
5 // 513 into 5 * 100 + 13. [Power = 100; First = 5; Remainder = 13]
6 int power = 1;
7 while (10 * power < n) power *= 10;
8 int first = n / power;
9 int remainder = n % power;
2 8 3
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