LeetCode #2376 — HARD

Count Special Integers

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

The Problem

Problem Statement

We call a positive integer special if all of its digits are distinct.

Given a positive integer n, return the number of special integers that belong to the interval [1, n].

Example 1:

Input: n = 20
Output: 19
Explanation: All the integers from 1 to 20, except 11, are special. Thus, there are 19 special integers.

Example 2:

Input: n = 5
Output: 5
Explanation: All the integers from 1 to 5 are special.

Example 3:

Input: n = 135
Output: 110
Explanation: There are 110 integers from 1 to 135 that are special.
Some of the integers that are not special are: 22, 114, and 131.

Constraints:

1 <= n <= 2 * 10⁹

Step 01

Brute Force Baseline

Problem summary: We call a positive integer special if all of its digits are distinct. Given a positive integer n, return the number of special integers that belong to the interval [1, 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

Example 3

Core Insight

What unlocks the optimal approach

Try to think of dynamic programming.
Use the idea of digit dynamic programming to build the numbers, in addition to a bitmask that will tell which digits you have used so far on the number that you are building.

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 #2376: Count Special Integers
class Solution {
    private char[] s;
    private Integer[][] f;

    public int countSpecialNumbers(int n) {
        s = String.valueOf(n).toCharArray();
        f = new Integer[s.length][1 << 10];
        return dfs(0, 0, true, true);
    }

    private int dfs(int i, int mask, boolean lead, boolean limit) {
        if (i >= s.length) {
            return lead ? 0 : 1;
        }
        if (!limit && !lead && f[i][mask] != null) {
            return f[i][mask];
        }
        int up = limit ? s[i] - '0' : 9;
        int ans = 0;
        for (int j = 0; j <= up; ++j) {
            if ((mask >> j & 1) == 1) {
                continue;
            }
            if (lead && j == 0) {
                ans += dfs(i + 1, mask, true, limit && j == up);
            } else {
                ans += dfs(i + 1, mask | (1 << j), false, limit && j == up);
            }
        }
        if (!limit && !lead) {
            f[i][mask] = ans;
        }
        return ans;
    }
}

// Accepted solution for LeetCode #2376: Count Special Integers
func countSpecialNumbers(n int) int {
	s := strconv.Itoa(n)
	m := len(s)
	f := make([][1 << 10]int, m+1)
	for i := range f {
		for j := range f[i] {
			f[i][j] = -1
		}
	}
	var dfs func(int, int, bool, bool) int
	dfs = func(i, mask int, lead, limit bool) int {
		if i >= m {
			if lead {
				return 0
			}
			return 1
		}
		if !limit && !lead && f[i][mask] != -1 {
			return f[i][mask]
		}
		up := 9
		if limit {
			up = int(s[i] - '0')
		}
		ans := 0
		for j := 0; j <= up; j++ {
			if mask>>j&1 == 1 {
				continue
			}
			if lead && j == 0 {
				ans += dfs(i+1, mask, true, limit && j == up)
			} else {
				ans += dfs(i+1, mask|1<<j, false, limit && j == up)
			}
		}
		if !limit && !lead {
			f[i][mask] = ans
		}
		return ans
	}
	return dfs(0, 0, true, true)
}

# Accepted solution for LeetCode #2376: Count Special Integers
class Solution:
    def countSpecialNumbers(self, n: int) -> int:
        @cache
        def dfs(i: int, mask: int, lead: bool, limit: bool) -> int:
            if i >= len(s):
                return int(lead ^ 1)
            up = int(s[i]) if limit else 9
            ans = 0
            for j in range(up + 1):
                if mask >> j & 1:
                    continue
                if lead and j == 0:
                    ans += dfs(i + 1, mask, True, limit and j == up)
                else:
                    ans += dfs(i + 1, mask | 1 << j, False, limit and j == up)
            return ans

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

// Accepted solution for LeetCode #2376: Count Special Integers
/**
 * [2376] Count Special Integers
 */
pub struct Solution {}

// submission codes start here
use std::collections::HashMap;

impl Solution {
    pub fn count_special_numbers(n: i32) -> i32 {
        let n: Vec<i32> = n
            .to_string()
            .chars()
            .map(|c| c.to_digit(10).unwrap() as i32)
            .collect();

        let mut result = 0;
        let mut candidate = 9;

        // 计算位数小于n的特殊整数
        for i in 0..n.len() as i32 - 1 {
            result += candidate;
            candidate *= 9 - i
        }

        let mut memory: HashMap<i32, i32> = HashMap::new();

        result += Self::dp(0, false, &mut memory, &n);

        result
    }

    // mask 表示前缀中使用过的数字 使用二进制表示
    // prefix_smaller 当前的前缀是否小于n的前缀
    fn dp(
        mask: i32,
        prefix_smaller: bool,
        memory: &mut HashMap<i32, i32>,
        number: &Vec<i32>,
    ) -> i32 {
        let used_bits = mask.count_ones() as usize;
        if used_bits == number.len() {
            return 1;
        }

        let key = mask * 2 + if prefix_smaller { 1 } else { 0 };

        if !memory.contains_key(&key) {
            let mut result = 0;

            let lower_bound = if mask == 0 { 1 } else { 0 };

            let upper_bound = if prefix_smaller { 9 } else { number[used_bits] };

            for i in lower_bound..=upper_bound {
                if mask >> i & 1 == 0 {
                    result += Self::dp(
                        mask | 1 << i,
                        prefix_smaller || i < upper_bound,
                        memory,
                        number,
                    );
                }
            }

            memory.insert(key, result);
        }

        *memory.get(&key).unwrap()
    }
}

// submission codes end

#[cfg(test)]
mod tests {
    use super::*;

    #[test]
    fn test_2376() {
        assert_eq!(19, Solution::count_special_numbers(20));
        assert_eq!(5, Solution::count_special_numbers(5));
        assert_eq!(110, Solution::count_special_numbers(135));
    }
}

// Accepted solution for LeetCode #2376: Count Special Integers
function countSpecialNumbers(n: number): number {
    const s = n.toString();
    const m = s.length;
    const f: number[][] = Array.from({ length: m }, () => Array(1 << 10).fill(-1));
    const dfs = (i: number, mask: number, lead: boolean, limit: boolean): number => {
        if (i >= m) {
            return lead ? 0 : 1;
        }
        if (!limit && !lead && f[i][mask] !== -1) {
            return f[i][mask];
        }
        const up = limit ? +s[i] : 9;
        let ans = 0;
        for (let j = 0; j <= up; ++j) {
            if ((mask >> j) & 1) {
                continue;
            }
            if (lead && j === 0) {
                ans += dfs(i + 1, mask, true, limit && j === up);
            } else {
                ans += dfs(i + 1, mask | (1 << j), false, limit && j === up);
            }
        }
        if (!limit && !lead) {
            f[i][mask] = ans;
        }
        return ans;
    };
    return dfs(0, 0, true, 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 × D)

Space

O(m × 2^D)

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.