233. Number of Digit One

Description

Given an integer n, count the total number of digit 1 appearing in all non-negative integers less than or equal to n.

Example 1:

Input: n = 13
Output: 6

Example 2:

Input: n = 0
Output: 0

Constraints:

0 <= n <= 10⁹

Solutions

Solution 1: Digit DP

This problem essentially asks for the number of times the digit $1$ appears in the given range $[l, ..r]$. The count is related to the number of digits and the value of each digit. We can use the concept of Digit DP to solve this problem. In Digit DP, the size of the number has little impact on the complexity.

For the range $[l, ..r]$ problem, we generally convert it to the problem of $[1, ..r]$ and then subtract the result of $[1, ..l - 1]$, i.e.:

$$ ans = \sum_{i=1}^{r} ans_i - \sum_{i=1}^{l-1} ans_i $$

However, for this problem, we only need to find the value for the range $[1, ..r]$.

Here, we use memoized search to implement Digit DP. We search from the starting point downwards, and at the lowest level, we get the number of solutions. We then return the answers layer by layer upwards, and finally get the final answer from the starting point of the search.

The basic steps are as follows:

First, we convert the number $n$ to a string $s$. Then we design a function $\textit{dfs}(i, \textit{cnt}, \textit{limit})$, where:

The digit $i$ represents the current position being searched, starting from the highest digit, i.e., $i = 0$ represents the highest digit.
The digit $\textit{cnt}$ represents the current count of the digit $1$ in the number.
The boolean $\textit{limit}$ indicates whether the current number is restricted by the upper bound.

The function executes as follows:

If $i$ exceeds the length of the number $n$, it means the search is over, directly return $cnt$. If $\textit{limit}$ is true, $up$ is the $i$-th digit of the current number. Otherwise, $up = 9$. Next, we iterate $j$ from $0$ to $up$. For each $j$:

If $j$ equals $1$, we increment $cnt$ by one.
Recursively call $\textit{dfs}(i + 1, \textit{cnt}, \textit{limit} \land j = up)$.

The answer is $\textit{dfs}(0, 0, \text{True})$.

The time complexity is $O(m^2 \times D)$, and the space complexity is $O(m^2)$. Here, $m$ is the length of the number $n$, and $D = 10$.

Similar Problems:

Here is the translation of the similar problems into English:

Python3JavaC++GoTypeScriptC#

class Solution:
    def countDigitOne(self, n: int) -> int:
        @cache
        def dfs(i: int, cnt: int, limit: bool) -> int:
            if i >= len(s):
                return cnt
            up = int(s[i]) if limit else 9
            ans = 0
            for j in range(up + 1):
                ans += dfs(i + 1, cnt + (j == 1), limit and j == up)
            return ans

        s = str(n)
        return dfs(0, 0, True)

class Solution {
    private int m;
    private char[] s;
    private Integer[][] f;

    public int countDigitOne(int n) {
        s = String.valueOf(n).toCharArray();
        m = s.length;
        f = new Integer[m][m];
        return dfs(0, 0, true);
    }

    private int dfs(int i, int cnt, boolean limit) {
        if (i >= m) {
            return cnt;
        }
        if (!limit && f[i][cnt] != null) {
            return f[i][cnt];
        }
        int up = limit ? s[i] - '0' : 9;
        int ans = 0;
        for (int j = 0; j <= up; ++j) {
            ans += dfs(i + 1, cnt + (j == 1 ? 1 : 0), limit && j == up);
        }
        if (!limit) {
            f[i][cnt] = ans;
        }
        return ans;
    }
}

class Solution {
public:
    int countDigitOne(int n) {
        string s = to_string(n);
        int m = s.size();
        int f[m][m];
        memset(f, -1, sizeof(f));
        auto dfs = [&](auto&& dfs, int i, int cnt, bool limit) -> int {
            if (i >= m) {
                return cnt;
            }
            if (!limit && f[i][cnt] != -1) {
                return f[i][cnt];
            }
            int up = limit ? s[i] - '0' : 9;
            int ans = 0;
            for (int j = 0; j <= up; ++j) {
                ans += dfs(dfs, i + 1, cnt + (j == 1), limit && j == up);
            }
            if (!limit) {
                f[i][cnt] = ans;
            }
            return ans;
        };
        return dfs(dfs, 0, 0, true);
    }
};

func countDigitOne(n int) int {
    s := strconv.Itoa(n)
    m := len(s)
    f := make([][]int, m)
    for i := range f {
        f[i] = make([]int, m)
        for j := range f[i] {
            f[i][j] = -1
        }
    }
    var dfs func(i, cnt int, limit bool) int
    dfs = func(i, cnt int, limit bool) int {
        if i >= m {
            return cnt
        }
        if !limit && f[i][cnt] != -1 {
            return f[i][cnt]
        }
        up := 9
        if limit {
            up = int(s[i] - '0')
        }
        ans := 0
        for j := 0; j <= up; j++ {
            t := 0
            if j == 1 {
                t = 1
            }
            ans += dfs(i+1, cnt+t, limit && j == up)
        }
        if !limit {
            f[i][cnt] = ans
        }
        return ans
    }
    return dfs(0, 0, true)
}

function countDigitOne(n: number): number {
    const s = n.toString();
    const m = s.length;
    const f: number[][] = Array.from({ length: m }, () => Array(m).fill(-1));
    const dfs = (i: number, cnt: number, limit: boolean): number => {
        if (i >= m) {
            return cnt;
        }
        if (!limit && f[i][cnt] !== -1) {
            return f[i][cnt];
        }
        const up = limit ? +s[i] : 9;
        let ans = 0;
        for (let j = 0; j <= up; ++j) {
            ans += dfs(i + 1, cnt + (j === 1 ? 1 : 0), limit && j === up);
        }
        if (!limit) {
            f[i][cnt] = ans;
        }
        return ans;
    };
    return dfs(0, 0, true);
}

public class Solution {
    private int m;
    private char[] s;
    private int?[,] f;

    public int CountDigitOne(int n) {
        s = n.ToString().ToCharArray();
        m = s.Length;
        f = new int?[m, m];
        return Dfs(0, 0, true);
    }

    private int Dfs(int i, int cnt, bool limit) {
        if (i >= m) {
            return cnt;
        }
        if (!limit && f[i, cnt] != null) {
            return f[i, cnt].Value;
        }
        int up = limit ? s[i] - '0' : 9;
        int ans = 0;
        for (int j = 0; j <= up; ++j) {
            ans += Dfs(i + 1, cnt + (j == 1 ? 1 : 0), limit && j == up);
        }
        if (!limit) {
            f[i, cnt] = ans;
        }
        return ans;
    }
}

233. Number of Digit One

Description

Solutions

Solution 1: Digit DP

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