LeetCode #1012 — HARD

Numbers With Repeated Digits

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

The Problem

Problem Statement

Given an integer n, return the number of positive integers in the range [1, n] that have at least one repeated digit.

Example 1:

Input: n = 20
Output: 1
Explanation: The only positive number (<= 20) with at least 1 repeated digit is 11.

Example 2:

Input: n = 100
Output: 10
Explanation: The positive numbers (<= 100) with atleast 1 repeated digit are 11, 22, 33, 44, 55, 66, 77, 88, 99, and 100.

Example 3:

Input: n = 1000
Output: 262

Constraints:

1 <= n <= 10⁹

Step 01

Brute Force Baseline

Problem summary: Given an integer n, return the number of positive integers in the range [1, n] that have at least one repeated digit.

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

How many numbers with no duplicate digits? How many numbers with K digits and no duplicates?
How many numbers with same length as N? How many numbers with same prefix as N?

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 #1012: Numbers With Repeated Digits
class Solution {
    private char[] s;
    private Integer[][] f;

    public int numDupDigitsAtMostN(int n) {
        s = String.valueOf(n).toCharArray();
        f = new Integer[s.length][1 << 10];
        return n - 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 (!lead && !limit && 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 (lead && j == 0) {
                ans += dfs(i + 1, mask, true, false);
            } else if ((mask >> j & 1) == 0) {
                ans += dfs(i + 1, mask | 1 << j, false, limit && j == up);
            }
        }
        if (!lead && !limit) {
            f[i][mask] = ans;
        }
        return ans;
    }
}

// Accepted solution for LeetCode #1012: Numbers With Repeated Digits
func numDupDigitsAtMostN(n int) int {
	s := []byte(strconv.Itoa(n))
	m := len(s)
	f := make([][]int, m)
	for i := range f {
		f[i] = make([]int, 1<<10)
		for j := range f[i] {
			f[i][j] = -1
		}
	}

	var dfs func(i, mask int, lead, limit bool) int
	dfs = func(i, mask int, lead, limit bool) int {
		if i >= m {
			if lead {
				return 0
			}
			return 1
		}
		if !lead && !limit && 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 lead && j == 0 {
				ans += dfs(i+1, mask, true, limit && j == up)
			} else if mask>>j&1 == 0 {
				ans += dfs(i+1, mask|(1<<j), false, limit && j == up)
			}
		}
		if !lead && !limit {
			f[i][mask] = ans
		}
		return ans
	}
	return n - dfs(0, 0, true, true)
}

# Accepted solution for LeetCode #1012: Numbers With Repeated Digits
class Solution:
    def numDupDigitsAtMostN(self, n: int) -> int:
        @cache
        def dfs(i: int, mask: int, lead: bool, limit: bool) -> int:
            if i >= len(s):
                return lead ^ 1
            up = int(s[i]) if limit else 9
            ans = 0
            for j in range(up + 1):
                if lead and j == 0:
                    ans += dfs(i + 1, mask, True, False)
                elif mask >> j & 1 ^ 1:
                    ans += dfs(i + 1, mask | 1 << j, False, limit and j == up)
            return ans

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

// Accepted solution for LeetCode #1012: Numbers With Repeated Digits
/**
 * [1012] Numbers With Repeated Digits
 *
 * Given an integer n, return the number of positive integers in the range [1, n] that have at least one repeated digit.
 *  
 * Example 1:
 *
 * Input: n = 20
 * Output: 1
 * Explanation: The only positive number (<= 20) with at least 1 repeated digit is 11.
 *
 * Example 2:
 *
 * Input: n = 100
 * Output: 10
 * Explanation: The positive numbers (<= 100) with atleast 1 repeated digit are 11, 22, 33, 44, 55, 66, 77, 88, 99, and 100.
 *
 * Example 3:
 *
 * Input: n = 1000
 * Output: 262
 *
 *  
 * Constraints:
 *
 * 	1 <= n <= 10^9
 *
 */
pub struct Solution {}

// problem: https://leetcode.com/problems/numbers-with-repeated-digits/
// discuss: https://leetcode.com/problems/numbers-with-repeated-digits/discuss/?currentPage=1&orderBy=most_votes&query=

// submission codes start here

impl Solution {
    pub fn num_dup_digits_at_most_n(n: i32) -> i32 {
        n - Self::num_not_dup_digits_at_most_n(n)
    }

    fn num_not_dup_digits_at_most_n(n: i32) -> i32 {
        let mut n = n;
        let mut digits = Vec::new();

        while n > 0 {
            digits.push(n % 10);
            n /= 10;
        }

        let k = digits.len();

        let mut used: [i32; 10] = [0; 10];
        let mut total = 0;

        for i in 1..k {
            total += 9 * Self::permutation(9, i as i32 - 1);
        }

        for i in 0..k {
            let i = k - 1 - i;
            let num = digits[i];

            for j in (if i == k - 1 { 1 } else { 0 })..num {
                if used[j as usize] != 0 {
                    continue;
                }
                total += Self::permutation((10 - k + i) as i32, i as i32);
            }

            used[num as usize] += 1;
            if used[num as usize] > 1 {
                break;
            }

            if i == 0 {
                total += 1;
            }
        }

        total
    }

    fn permutation(n: i32, k: i32) -> i32 {
        Self::factorial(n) / Self::factorial(n - k)
    }

    fn factorial(n: i32) -> i32 {
        match n {
            0 | 1 => 1,
            n @ _ => n * Self::factorial(n - 1),
        }
    }
}

// submission codes end

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

    #[test]
    fn test_1012_example_1() {
        let n = 20;
        let result = 1;

        assert_eq!(Solution::num_dup_digits_at_most_n(n), result);
    }

    #[test]
    fn test_1012_example_2() {
        let n = 100;
        let result = 10;

        assert_eq!(Solution::num_dup_digits_at_most_n(n), result);
    }

    #[test]
    fn test_1012_example_3() {
        let n = 1000;
        let result = 262;

        assert_eq!(Solution::num_dup_digits_at_most_n(n), result);
    }
}

// Accepted solution for LeetCode #1012: Numbers With Repeated Digits
function numDupDigitsAtMostN(n: number): number {
    const s = n.toString();
    const m = s.length;
    const f = 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 (!lead && !limit && f[i][mask] !== -1) {
            return f[i][mask];
        }
        const up = limit ? parseInt(s[i]) : 9;
        let ans = 0;
        for (let j = 0; j <= up; j++) {
            if (lead && j === 0) {
                ans += dfs(i + 1, mask, true, limit && j === up);
            } else if (((mask >> j) & 1) === 0) {
                ans += dfs(i + 1, mask | (1 << j), false, limit && j === up);
            }
        }
        if (!lead && !limit) {
            f[i][mask] = ans;
        }
        return ans;
    };

    return n - 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(log n × 2^D × D)

Space

O(log 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.