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1067. Digit Count in Range πŸ”’

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 <= 9
  • 1 <= low <= high <= 2 * 108

Solutions

Solution 1

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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)

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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;
    }
}

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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