LeetCode #357 — MEDIUM

Count Numbers with Unique Digits

Move from brute-force thinking to an efficient approach using math strategy.

The Problem

Problem Statement

Given an integer n, return the count of all numbers with unique digits, x, where 0 <= x < 10ⁿ.

Example 1:

Input: n = 2
Output: 91
Explanation: The answer should be the total numbers in the range of 0 ≤ x < 100, excluding 11,22,33,44,55,66,77,88,99

Example 2:

Input: n = 0
Output: 1

Constraints:

0 <= n <= 8

Step 01

Brute Force Baseline

Problem summary: Given an integer n, return the count of all numbers with unique digits, x, where 0 <= x < 10n.

Baseline thinking

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

Pattern signal: Math · Dynamic Programming · Backtracking

Example 1

Example 2

Core Insight

What unlocks the optimal approach

A direct way is to use the backtracking approach.
Backtracking should contains three states which are (the current number, number of steps to get that number and a bitmask which represent which number is marked as visited so far in the current number). Start with state (0,0,0) and count all valid number till we reach number of steps equals to 10<sup>n</sup>.
This problem can also be solved using a dynamic programming approach and some knowledge of combinatorics.
Let f(k) = count of numbers with unique digits with length equals k.
f(1) = 10, ..., f(k) = 9 * 9 * 8 * ... (9 - k + 2) [The first factor is 9 because a number cannot start with 0].

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

Upper-end input sizes

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 #357: Count Numbers with Unique Digits
class Solution {
    private Integer[][] f;

    public int countNumbersWithUniqueDigits(int n) {
        f = new Integer[n][1 << 10];
        return dfs(n - 1, 0, true);
    }

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

// Accepted solution for LeetCode #357: Count Numbers with Unique Digits
func countNumbersWithUniqueDigits(n int) int {
	f := make([][1 << 10]int, n)
	for i := range f {
		for j := range f[i] {
			f[i][j] = -1
		}
	}
	var dfs func(i, mask int, lead bool) int
	dfs = func(i, mask int, lead bool) int {
		if i < 0 {
			return 1
		}
		if !lead && f[i][mask] != -1 {
			return f[i][mask]
		}
		ans := 0
		for j := 0; j < 10; j++ {
			if mask>>j&1 == 1 {
				continue
			}
			if lead && j == 0 {
				ans += dfs(i-1, mask, true)
			} else {
				ans += dfs(i-1, mask|1<<j, false)
			}
		}
		if !lead {
			f[i][mask] = ans
		}
		return ans
	}
	return dfs(n-1, 0, true)
}

# Accepted solution for LeetCode #357: Count Numbers with Unique Digits
class Solution:
    def countNumbersWithUniqueDigits(self, n: int) -> int:
        @cache
        def dfs(i: int, mask: int, lead: bool) -> int:
            if i < 0:
                return 1
            ans = 0
            for j in range(10):
                if mask >> j & 1:
                    continue
                if lead and j == 0:
                    ans += dfs(i - 1, mask, True)
                else:
                    ans += dfs(i - 1, mask | 1 << j, False)
            return ans

        return dfs(n - 1, 0, True)

// Accepted solution for LeetCode #357: Count Numbers with Unique Digits
struct Solution;

impl Solution {
    fn count_numbers_with_unique_digits(n: i32) -> i32 {
        if n < 11 {
            Self::f(n)
        } else {
            Self::f(10)
        }
    }

    fn f(n: i32) -> i32 {
        match n {
            1 => 10,
            2 => 9 * 9 + 10,
            3 => 9 * 9 * 8 + 9 * 9 + 10,
            4 => 9 * 9 * 8 * 7 + 9 * 9 * 8 + 9 * 9 + 10,
            5 => 9 * 9 * 8 * 7 * 6 + 9 * 9 * 8 * 7 + 9 * 9 * 8 + 9 * 9 + 10,
            6 => 9 * 9 * 8 * 7 * 6 * 5 + 9 * 9 * 8 * 7 * 6 + 9 * 9 * 8 * 7 + 9 * 9 * 8 + 9 * 9 + 10,
            7 => {
                9 * 9 * 8 * 7 * 6 * 5 * 4
                    + 9 * 9 * 8 * 7 * 6 * 5
                    + 9 * 9 * 8 * 7 * 6
                    + 9 * 9 * 8 * 7
                    + 9 * 9 * 8
                    + 9 * 9
                    + 10
            }
            8 => {
                9 * 9 * 8 * 7 * 6 * 5 * 4 * 3
                    + 9 * 9 * 8 * 7 * 6 * 5 * 4
                    + 9 * 9 * 8 * 7 * 6 * 5
                    + 9 * 9 * 8 * 7 * 6
                    + 9 * 9 * 8 * 7
                    + 9 * 9 * 8
                    + 9 * 9
                    + 10
            }
            9 => {
                9 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2
                    + 9 * 9 * 8 * 7 * 6 * 5 * 4 * 3
                    + 9 * 9 * 8 * 7 * 6 * 5 * 4
                    + 9 * 9 * 8 * 7 * 6 * 5
                    + 9 * 9 * 8 * 7 * 6
                    + 9 * 9 * 8 * 7
                    + 9 * 9 * 8
                    + 9 * 9
                    + 10
            }
            10 => {
                9 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2
                    + 9 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2
                    + 9 * 9 * 8 * 7 * 6 * 5 * 4 * 3
                    + 9 * 9 * 8 * 7 * 6 * 5 * 4
                    + 9 * 9 * 8 * 7 * 6 * 5
                    + 9 * 9 * 8 * 7 * 6
                    + 9 * 9 * 8 * 7
                    + 9 * 9 * 8
                    + 9 * 9
                    + 10
            }
            0 => 1,
            _ => 0,
        }
    }
}

#[test]
fn test() {
    let n = 2;
    let res = 91;
    assert_eq!(Solution::count_numbers_with_unique_digits(n), res);
}

// Accepted solution for LeetCode #357: Count Numbers with Unique Digits
function countNumbersWithUniqueDigits(n: number): number {
    const f: number[][] = Array.from({ length: n }, () => Array(1 << 10).fill(-1));
    const dfs = (i: number, mask: number, lead: boolean): number => {
        if (i < 0) {
            return 1;
        }
        if (!lead && f[i][mask] !== -1) {
            return f[i][mask];
        }
        let ans = 0;
        for (let j = 0; j < 10; ++j) {
            if ((mask >> j) & 1) {
                continue;
            }
            if (lead && j === 0) {
                ans += dfs(i - 1, mask, true);
            } else {
                ans += dfs(i - 1, mask | (1 << j), false);
            }
        }
        if (!lead) {
            f[i][mask] = ans;
        }
        return ans;
    };
    return dfs(n - 1, 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(n × 2^D × D)

Space

O(n × 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.

Missing undo step on backtrack

Wrong move: Mutable state leaks between branches.

Usually fails on: Later branches inherit selections from earlier branches.

Fix: Always revert state changes immediately after recursive call.