Formatted question description: https://leetcode.ca/all/1067.html
1067. Digit Count in Range
Level
Hard
Description
Given an integer d
between 0
and 9
, and two positive integers low
and high
as lower and upper bounds, respectively. Return the number of times that d
occurs as a digit in all integers between low
and high
, including the bounds low
and high
.
Example 1:
Input: d = 1, low = 1, high = 13
Output: 6
Explanation:
The digit d=1 occurs 6 times in 1,10,11,12,13. Note that the digit d=1 occurs twice in the number 11.
Example 2:
Input: d = 3, low = 100, high = 250
Output: 35
Explanation:
The digit d=3 occurs 35 times in 103,113,123,130,131,…,238,239,243.
Note:
0 <= d <= 9
1 <= low <= high <= 2×10^8
Solution
Calculate the number of times that d
occurs as a digit in all integers between 1 and high
, and calculate the number of times that d
occurs as a digit in all integers between 1 and low  1
. Then calculate the difference between the two numbers of times to obtain the result.
If d
is not 0, the number of times is calculated as follows.
 Loop over
i
from 1 ton
. Each time seti *= 10
.  The value
n / (i * 10) * i
represents the number of digitd
at digiti * 10
.  The value
Math.min(Math.max(n % (i * 10)  d * i + 1, 0), i)
represents the extra number of digitd
at digiti * 10
.  Calculate the number of times of
d
by calculating the sum of the values in the previous two steps.
If d
is 0, the number of times is calculated as follows.
 Loop over
i
from 1 ton / 10
. Each time seti *= 10
.  The value
(n / (i * 10)  1) * i
represents the number of digitd
at digiti * 10
.  The value
Math.min(Math.max(n % (i * 10)  d * i + 1, 0), i)
represents the extra number of digitd
at digiti * 10
.  Calculate the number of times of
d
by calculating the sum of the values in the previous two steps.
The difference exists because digit 0 cannot appear at the highest digit, except the number 0. However, in this problem, both low
and high
are positive integers, so 0 is never in the range.

class Solution { public int digitsCount(int d, int low, int high) { long highCount = countDigits(high, d); long lowCount = countDigits(low  1, d); int count = (int) (highCount  lowCount); return count; } public long countDigits(int n, int digit) { if (n <= 0) return 0; long count = 0; if (digit == 0) { for (long i = 1; i <= n / 10; i *= 10) { long unit = i * 10; count += (n / unit  1) * i + Math.min(i, Math.max(0, n % unit  digit * i + 1)); } } else { for (long i = 1; i <= n; i *= 10) { long unit = i * 10; count += n / unit * i + Math.min(i, Math.max(0, n % unit  digit * i + 1)); } } return count; } }

Todo

print("Todo!")