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Formatted question description: https://leetcode.ca/all/1067.html

# 1067. Digit Count in Range

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;
}
}

• class Solution:
def digitsCount(self, d: int, low: int, high: int) -> int:
return self.f(high, d) - self.f(low - 1, d)

def f(self, n, d):
@cache
def dfs(pos, cnt, lead, limit):
if pos <= 0:
return cnt
up = a[pos] if limit else 9
ans = 0
for i in range(up + 1):
if i == 0 and lead:
ans += dfs(pos - 1, cnt, lead, limit and i == up)
else:
ans += dfs(pos - 1, cnt + (i == d), False, limit and i == up)
return ans

a = [0] * 11
l = 0
while n:
l += 1
a[l] = n % 10
n //= 10
return dfs(l, 0, True, True)


• class Solution {
public:
int d;
int a[11];
int dp[11][11];

int digitsCount(int d, int low, int high) {
this->d = d;
return f(high) - f(low - 1);
}

int f(int n) {
memset(dp, -1, sizeof dp);
int len = 0;
while (n) {
a[++len] = n % 10;
n /= 10;
}
return dfs(len, 0, true, true);
}

int dfs(int pos, int cnt, bool lead, bool limit) {
if (pos <= 0) {
return cnt;
}
if (!lead && !limit && dp[pos][cnt] != -1) {
return dp[pos][cnt];
}
int up = limit ? a[pos] : 9;
int ans = 0;
for (int i = 0; i <= up; ++i) {
if (i == 0 && lead) {
ans += dfs(pos - 1, cnt, lead, limit && i == up);
} else {
ans += dfs(pos - 1, cnt + (i == d), false, limit && i == up);
}
}
if (!lead && !limit) {
dp[pos][cnt] = ans;
}
return ans;
}
};

• func digitsCount(d int, low int, high int) int {
f := func(n int) int {
a := make([]int, 11)
dp := make([][]int, 11)
for i := range dp {
dp[i] = make([]int, 11)
for j := range dp[i] {
dp[i][j] = -1
}
}
l := 0
for n > 0 {
l++
a[l] = n % 10
n /= 10
}

var dfs func(int, int, bool, bool) int
dfs = func(pos, cnt int, lead, limit bool) int {
if pos <= 0 {
return cnt
}
if !lead && !limit && dp[pos][cnt] != -1 {
return dp[pos][cnt]
}
up := 9
if limit {
up = a[pos]
}
ans := 0
for i := 0; i <= up; i++ {
if i == 0 && lead {
ans += dfs(pos-1, cnt, lead, limit && i == up)
} else {
t := cnt
if d == i {
t++
}
ans += dfs(pos-1, t, false, limit && i == up)
}
}
if !lead && !limit {
dp[pos][cnt] = ans
}
return ans
}

return dfs(l, 0, true, true)
}
return f(high) - f(low-1)
}