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:

  1. 0 <= d <= 9
  2. 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.

  1. Loop over i from 1 to n. Each time set i *= 10.
  2. The value n / (i * 10) * i represents the number of digit d at digit i * 10.
  3. The value Math.min(Math.max(n % (i * 10) - d * i + 1, 0), i) represents the extra number of digit d at digit i * 10.
  4. 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.

  1. Loop over i from 1 to n / 10. Each time set i *= 10.
  2. The value (n / (i * 10) - 1) * i represents the number of digit d at digit i * 10.
  3. The value Math.min(Math.max(n % (i * 10) - d * i + 1, 0), i) represents the extra number of digit d at digit i * 10.
  4. 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;
    }
}

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