LeetCode #233 — HARD

Number of Digit One

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

The Problem

Problem Statement

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

Step 01

Brute Force Baseline

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

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Math · Dynamic Programming

Example 1

Example 2

Core Insight

What unlocks the optimal approach

Beware of overflow.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Largest constraint values

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #233: Number of Digit One
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;
    }
}

// Accepted solution for LeetCode #233: Number of Digit One
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)
}

# Accepted solution for LeetCode #233: Number of Digit One
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)

// Accepted solution for LeetCode #233: Number of Digit One
struct Solution;

impl Solution {
    fn count_digit_one(n: i32) -> i32 {
        let mut m = 1i64;
        let mut res = 0i64;
        let n = n as i64;
        while m <= n {
            let d = 10 * m;
            res += n / d * m + m.min(0.max(n % d - m + 1));
            m *= 10;
        }
        res as i32
    }
}

#[test]
fn test() {
    let n = 13;
    let res = 6;
    assert_eq!(Solution::count_digit_one(n), res);
}

// Accepted solution for LeetCode #233: Number of Digit One
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);
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(m^2 × D)

Space

O(m^2)

Approach Breakdown

RECURSIVE

O(2ⁿ) time

O(n) space

Pure recursion explores every possible choice at each step. With two choices per state (take or skip), the decision tree has 2ⁿ leaves. The recursion stack uses O(n) space. Many subproblems are recomputed exponentially many times.

DYNAMIC PROGRAMMING

O(n × m) time

O(n × m) space

Each cell in the DP table is computed exactly once from previously solved subproblems. The table dimensions determine both time and space. Look for the state variables — each unique combination of state values is one cell. Often a rolling array can reduce space by one dimension.

Shortcut: Count your DP state dimensions → that’s your time. Can you drop one? That’s your space optimization.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.

State misses one required dimension

Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.

Usually fails on: Correctness breaks on cases that differ only in hidden state.

Fix: Define state so each unique subproblem maps to one DP cell.