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

Solutions to Chapter 20 | Hard
10
11 // Counts 2s from first digit
12 int nTwosFirst = 0;
13 if (first > 2) nTwosFirst += power;
14 else if (first == 2) nTwosFirst += remainder + 1;
15
16 // Count 2s from all other digits
17 int nTwosOther = first * count2sR(power - 1) + count2sR(remainder);
18
19 return nTwosFirst + nTwosOther;
20 }
We can also implement this algorithm iteratively:
1 public static int count2sI(int num) {
2 int countof2s = 0, digit = 0;
3 int j = num, seendigits=0, position=0, pow10_pos = 1;
4 /* maintaining this value instead of calling pow() is an 6x perf
5 * gain (48s -> 8s) pow10_posMinus1. maintaining this value
6 * instead of calling Numof2s is an 2x perf gain (8s -> 4s).
7 * overall > 10x speedup */
8 while (j > 0) {
9 digit = j % 10;
10 int pow10_posMinus1 = pow10_pos / 10;
11 countof2s += digit * position * pow10_posMinus1;
12 /* we do this if digit , or = 2
13 * Digit < 2 implies there are no 2s contributed by this
14 * digit.
15 * Digit == 2 implies there are 2 * numof2s contributed by
16 * the previous position + num of 2s contributed by the
17 * presence of this 2 */
18 if (digit == 2) {
19 countof2s += seendigits + 1;
20 }
21 /* Digit > 2 implies there are digit * num of 2s by the prev.
22 * position + 10^position */
23 else if(digit > 2) {
24 countof2s += pow10_pos;
25 }
26 seendigits = seendigits + pow10_pos * digit;
27 pow10_pos *= 10;
28 position++;
29 j = j / 10;
30 }
31 return(countof2s);
32 }
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