Description
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].
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.
Constraints:
0 <= d <= 91 <= low <= high <= 2 * 108
Solutions
Solution 1
PythonJavaC++Go
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)(code-box)
class Solution {
private int d;
private int[] a = new int[11];
private int[][] dp = new int[11][11];
public int digitsCount(int d, int low, int high) {
this.d = d;
return f(high) - f(low - 1);
}
private int f(int n) {
for (var e : dp) {
Arrays.fill(e, -1);
}
int len = 0;
while (n > 0) {
a[++len] = n % 10;
n /= 10;
}
return dfs(len, 0, true, true);
}
private int dfs(int pos, int cnt, boolean lead, boolean 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 ? 1 : 0), false, limit && i == up);
}
}
if (!lead && !limit) {
dp[pos][cnt] = ans;
}
return ans;
}
}(code-box)
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;
}
};(code-box)
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)
}(code-box)
