902. Numbers At Most N Given Digit Set

Description

Given an array of digits which is sorted in non-decreasing order. You can write numbers using each digits[i] as many times as we want. For example, if digits = ['1','3','5'], we may write numbers such as '13', '551', and '1351315'.

Return the number of positive integers that can be generated that are less than or equal to a given integer n.

Example 1:

Input: digits = ["1","3","5","7"], n = 100
Output: 20
Explanation:
The 20 numbers that can be written are:
1, 3, 5, 7, 11, 13, 15, 17, 31, 33, 35, 37, 51, 53, 55, 57, 71, 73, 75, 77.


Example 2:

Input: digits = ["1","4","9"], n = 1000000000
Output: 29523
Explanation:
We can write 3 one digit numbers, 9 two digit numbers, 27 three digit numbers,
81 four digit numbers, 243 five digit numbers, 729 six digit numbers,
2187 seven digit numbers, 6561 eight digit numbers, and 19683 nine digit numbers.
In total, this is 29523 integers that can be written using the digits array.


Example 3:

Input: digits = ["7"], n = 8
Output: 1


Constraints:

• 1 <= digits.length <= 9
• digits[i].length == 1
• digits[i] is a digit from '1' to '9'.
• All the values in digits are unique.
• digits is sorted in non-decreasing order.
• 1 <= n <= 109

Solutions

• class Solution {
private int[] a = new int[12];
private int[][] dp = new int[12][2];
private Set<Integer> s = new HashSet<>();

public int atMostNGivenDigitSet(String[] digits, int n) {
for (var e : dp) {
Arrays.fill(e, -1);
}
for (String d : digits) {
}
int len = 0;
while (n > 0) {
a[++len] = n % 10;
n /= 10;
}
return dfs(len, 1, true);
}

private int dfs(int pos, int lead, boolean limit) {
if (pos <= 0) {
return lead ^ 1;
}
if (!limit && lead != 1 && dp[pos][lead] != -1) {
}
int ans = 0;
int up = limit ? a[pos] : 9;
for (int i = 0; i <= up; ++i) {
if (i == 0 && lead == 1) {
ans += dfs(pos - 1, lead, limit && i == up);
} else if (s.contains(i)) {
ans += dfs(pos - 1, 0, limit && i == up);
}
}
if (!limit && lead == 0) {
}
return ans;
}
}

• class Solution {
public:
int a[12];
int dp[12][2];
unordered_set<int> s;

int atMostNGivenDigitSet(vector<string>& digits, int n) {
memset(dp, -1, sizeof dp);
for (auto& d : digits) {
s.insert(stoi(d));
}
int len = 0;
while (n) {
a[++len] = n % 10;
n /= 10;
}
return dfs(len, 1, true);
}

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

• class Solution:
def atMostNGivenDigitSet(self, digits: List[str], n: int) -> int:
@cache
def dfs(pos, lead, limit):
if pos <= 0:
return lead == False
up = a[pos] if limit else 9
ans = 0
for i in range(up + 1):
if i == 0 and lead:
ans += dfs(pos - 1, lead, limit and i == up)
elif i in s:
ans += dfs(pos - 1, False, limit and i == up)
return ans

l = 0
a = [0] * 12
s = {int(d) for d in digits}
while n:
l += 1
a[l] = n % 10
n //= 10
return dfs(l, True, True)


• func atMostNGivenDigitSet(digits []string, n int) int {
s := map[int]bool{}
for _, d := range digits {
i, _ := strconv.Atoi(d)
s[i] = true
}
a := make([]int, 12)
dp := make([][2]int, 12)
for i := range a {
dp[i] = [2]int{-1, -1}
}
l := 0
for n > 0 {
l++
a[l] = n % 10
n /= 10
}
var dfs func(int, int, bool) int
dfs = func(pos, lead int, limit bool) int {
if pos <= 0 {
return lead ^ 1
}
if !limit && lead == 0 && dp[pos][lead] != -1 {
}
up := 9
if limit {
up = a[pos]
}
ans := 0
for i := 0; i <= up; i++ {
if i == 0 && lead == 1 {
ans += dfs(pos-1, lead, limit && i == up)
} else if s[i] {
ans += dfs(pos-1, 0, limit && i == up)
}
}
if !limit {