Digit Count in Range
AmazonThe problem
Given a single-digit integer d and two integers low and high, return the number of times that d occurs as a digit in all integers in the inclusive range [low, high].
Input: d = 2, low = 2, high = 25 Output: 9 Explanation: The digit d = 2 occurs 9 times in 2,20,12,21, 22, 23, 24,25.
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.
Input: d = 1, low = 1, high = 13
- 0 <= d <= 9
- 1 <= low <= high <= 2 * 108
cpp
class Solution {
public:
int digitsCount(int d, int low, int high) {
//User code goes here
}
};java
class Solution {
public int digitsCount(int d, int low, int high) {
// User code goes here
}
}python
class Solution:
def digitsCount(self, d: int, low: int, high: int) -> int:
# User code goes herejavascript
class Solution {
digitsCount(d, low, high) {
// User Code goes here
}
}csharp
public class Solution
{
public int DigitsCount(int d, int low, int high)
{
// User code goes here
}
}go
func digitsCount(d int, low int, high int) int {
// User code goes here
}Stuck? Show a way to structure it+
- 01Compute occurrences in `[1,n]` and subtract `count(low-1)`.
- 02For each decimal factor, split n into higher, current, and lower parts.
- 03Use the standard three-case formula for nonzero digit d.
- 04Handle d=0 separately to avoid counting leading zeroes.
Reference answer
Then expect these follow-ups
How would you count all digits 0 through 9 at once?
Tests: digit counting
How would a digit-DP solution handle arbitrary bounds?
Tests: digit DP
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