What is the time complexity of Count Numbers with Unique Digits?

The optimal solution has a time complexity of O(n).

What data structures are used to solve Count Numbers with Unique Digits?

This problem uses Math, Dynamic Programming, Backtracking. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Count Numbers with Unique Digits?

Insight 1: The number of unique digits available decreases as the length of the number increases. We start with 9 available digits (1-9) for the first digit (since it can't be 0), then 9 for the second digit (0-9 except the first digit), then 8, and so on. Insight 2: The problem can be solved iteratively or recursively. The number of unique digit numbers of length `i` depends on the number of unique digit numbers of length `i-1`.

Which companies ask LeetCode 357. Count Numbers with Unique Digits in interviews?

This problem is frequently asked at J.P. Morgan.

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Count Numbers with Unique Digits - LeetCode 357 Solution

Count Numbers with Unique Digits - Complete Solution Guide

Count Numbers with Unique Digits is LeetCode problem 357, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

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

Detailed Explanation

The problem asks us to count the number of integers with unique digits within the range [0, 10<sup>n</sup>). For instance, if n=2, we need to count the numbers from 0 to 99, excluding numbers like 11, 22, 33, ..., 99 which have repeating digits. If n=0, we only have the number 0, so the count is 1.

Solution Approach

The solution uses a dynamic programming approach (although implemented iteratively) to calculate the number of integers with unique digits for each length from 1 to `n`. It builds upon the counts from smaller lengths to derive the count for larger lengths. It avoids generating each number and explicitly checking if digits are unique, instead leveraging mathematical principles to compute the result directly.

Step-by-Step Algorithm

Step 1: Handle the base case: if n is 0, return 1 (only the number 0 is considered).
Step 2: Initialize `total` to 10. This accounts for the numbers 0-9 when n >= 1.
Step 3: Initialize `unique_count` to 9, which represents the number of unique one-digit numbers (1-9).
Step 4: Initialize `available_digits` to 9, representing the digits available for the second position onward.
Step 5: Iterate from i = 2 to n (inclusive). In each iteration, update `unique_count` by multiplying it with `available_digits`. This computes the count of unique digit numbers of length `i`. Then add `unique_count` to the `total`. Reduce available_digits by 1 for the next iteration, as we have one less unique digit to choose from.
Step 6: Return the `total`, which is the sum of all unique digit numbers from length 0 to n.

Key Insights

Insight 1: The number of unique digits available decreases as the length of the number increases. We start with 9 available digits (1-9) for the first digit (since it can't be 0), then 9 for the second digit (0-9 except the first digit), then 8, and so on.
Insight 2: The problem can be solved iteratively or recursively. The number of unique digit numbers of length `i` depends on the number of unique digit numbers of length `i-1`.
Insight 3: The constraint n <= 8 is important because if n > 10, there cannot be a number with unique digits (since we only have 10 digits from 0-9).

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

This problem involves: Math, Dynamic Programming, Backtracking.

Companies

Asked at: J.P. Morgan.

Count Numbers with Unique Digits | LeetCode 357

Medium

Math Dynamic Programming Backtracking

Time: O(n)Space: O(1)

Solution Code

Available in:

class Solution {
    public int countNumbersWithUniqueDigits(int n) {
        if (n == 0) {
            return 1;
        }

        int total = 10;
        int unique_count = 9;
        int available_digits = 9;

        for (int i = 2; i <= n; i++) {
            unique_count *= available_digits;
            total += unique_count;
            available_digits -= 1;
        }

        return total;
    }
}

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Problem Info

Description

Problem 357: Count Numbers with Unique Digits - Detailed explanation and solution approaches.

Topics

Math Dynamic Programming Backtracking

Companies

J.P. Morgan

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:357

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

6 steps

1
Step 1: Handle the base case: if n is 0, return 1 (only the number 0 is considered).
2
Step 2: Initialize `total` to 10. This accounts for the numbers 0-9 when n >= 1.
3
Step 3: Initialize `unique_count` to 9, which represents the number of unique one-digit numbers (1-9).
4
Step 4: Initialize `available_digits` to 9, representing the digits available for the second position onward.
5
Step 5: Iterate from i = 2 to n (inclusive). In each iteration, update `unique_count` by multiplying it with `available_digits`. This computes the count of unique digit numbers of length `i`. Then add `unique_count` to the `total`. Reduce available_digits by 1 for the next iteration, as we have one less unique digit to choose from.
6
Step 6: Return the `total`, which is the sum of all unique digit numbers from length 0 to n.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The time complexity is O(n) because the code iterates from 2 to n in the for loop. The space complexity is O(1) because we are only using a fixed number of variables, regardless of the input size `n`.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Backtracking

We can use backtracking to generate all possible numbers with unique digits and count them. We start with an empty number and recursively add digits to it, making sure that each digit is unique. We stop when the length of the number reaches `n`.

Time: O(10<sup>n</sup>)Space: O(n)

Trade-offs: Backtracking is less efficient because it explores all possible combinations. The given solution is significantly faster, especially for larger values of `n`, as it uses a direct formula derived from combinatorics principles. Backtracking has a exponential time complexity compared to the linear time complexity of the iterative solution.

✨

Key Insights

1Insight 1: The number of unique digits available decreases as the length of the number increases. We start with 9 available digits (1-9) for the first digit (since it can't be 0), then 9 for the second digit (0-9 except the first digit), then 8, and so on.
2Insight 2: The problem can be solved iteratively or recursively. The number of unique digit numbers of length `i` depends on the number of unique digit numbers of length `i-1`.
3Insight 3: The constraint n <= 8 is important because if n > 10, there cannot be a number with unique digits (since we only have 10 digits from 0-9).

📚

Key Terms Explained

For Students

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

Edge Cases

Edge case 1: n = 0. The code explicitly handles this base case by returning 1.
Edge case 2: n > 10. Although the constraint specifies n <= 8, if n were greater than 10, the `available_digits` would eventually become 0, resulting in `unique_count` becoming 0, and effectively terminating the calculation of additional unique numbers. The loop correctly handles the case where n exceeds the number of available digits.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not handling the case when n = 0 correctly. The output should be 1 in this case.

✗Common mistake 2: Incorrectly calculating `available_digits` by not decrementing it correctly in each iteration.

✗Common mistake 3: Forgetting to consider the numbers of length less than `n`. We need to sum the numbers with unique digits of all lengths from 1 to `n` along with the number 0.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Generating unique identifiers or serial numbers. This algorithm helps determine the possible number of unique identifiers of a certain length given a set of available characters/digits.

🏦Real-world scenario 2: Password generation or verification. Understanding how many unique passwords can be created with certain restrictions (e.g., no repeating characters) is crucial for security.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Medium problems are the bread and butter of technical interviews. Practice explaining your approach before coding.

⚠️ Watch Out

Always test your solution with edge cases: empty input, single element, and boundary values.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: November 25, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 13

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 6

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJun 29

❓

Frequently Asked Questions

Should I use top-down (memoization) or bottom-up (tabulation)?

Start with top-down (recursion + memoization) as it directly maps to the recursive thinking and handles only needed subproblems. Convert to bottom-up for better space efficiency and to avoid stack overflow. Bottom-up is often faster due to no recursion overhead.

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Count Numbers with Unique Digits"?

For this medium-level Math problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether dynamic programming techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?