What is the time complexity of Numbers With Repeated Digits?

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

What is the space complexity of Numbers With Repeated Digits?

The space complexity is O(log(N)).

What data structures are used to solve Numbers With Repeated Digits?

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

What are the key insights for solving Numbers With Repeated Digits?

Insight 1: The problem can be solved more efficiently by counting the numbers with *no* repeated digits and subtracting that from n. Insight 2: The number of integers with unique digits and a given length can be calculated using permutations.

Which companies ask LeetCode 1012. Numbers With Repeated Digits in interviews?

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

JavaScript is required for the full interactive experience. However, all content is accessible without JavaScript.

Numbers With Repeated Digits - LeetCode 1012 Solution

Q: How to solve Numbers With Repeated Digits?

The solution counts the number of integers from 1 to n that have *no* repeated digits and subtracts this count from n to get the desired result (numbers with at least one repeated digit). It first counts the number of integers with unique digits that have length less than the length of n+1. Then, it iterates through the digits of n+1, and for each digit, it counts the number of valid numbers with unique digits that have the same prefix as n+1 up to that digit. The permutation function is used to calculate number of such combinations. Finally, the numbers of repeated digits are calculated by n - number of numbers with unique digits.

Numbers With Repeated Digits - Complete Solution Guide

Numbers With Repeated Digits is LeetCode problem 1012, a Hard 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 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

Detailed Explanation

The problem asks us to find the number of positive integers within the range [1, n] that have at least one repeated digit. For example, if n is 20, the only number with repeated digits is 11. If n is 100, the numbers with repeated digits are 11, 22, 33, 44, 55, 66, 77, 88, 99, and 100. The constraint is that 1 <= n <= 10^9.

Solution Approach

The solution counts the number of integers from 1 to n that have *no* repeated digits and subtracts this count from n to get the desired result (numbers with at least one repeated digit). It first counts the number of integers with unique digits that have length less than the length of n+1. Then, it iterates through the digits of n+1, and for each digit, it counts the number of valid numbers with unique digits that have the same prefix as n+1 up to that digit. The permutation function is used to calculate number of such combinations. Finally, the numbers of repeated digits are calculated by n - number of numbers with unique digits.

Step-by-Step Algorithm

Step 1: Convert the input integer `n+1` to a string `s` to easily access individual digits.
Step 2: Calculate the number of integers with unique digits that have a length smaller than the length of s. This is done by iterating from length 1 to length k-1 (where k is the length of s) and using the formula 9 * P(9, i-1). We multiply by 9 because the first digit cannot be 0.
Step 3: Initialize a set `used_digits` to track digits already used in the current number being considered.
Step 4: Iterate through the digits of `s` from left to right. For each digit `d` at index `i`, iterate through possible digits `j` (from 1 if i==0, or 0 otherwise) up to `d-1`. If `j` is not in `used_digits`, it means we can construct a valid number with unique digits. Then, add P(10-(i+1), k-(i+1)) to `count_unique`.
Step 5: If `d` is already in `used_digits`, it means adding `d` would create a number with repeated digits, so stop.
Step 6: Add the digit `d` to `used_digits` for the next iteration.
Step 7: Finally, return n - count_unique, which gives the number of integers with at least one repeated digit.

Key Insights

Insight 1: The problem can be solved more efficiently by counting the numbers with *no* repeated digits and subtracting that from n.
Insight 2: The number of integers with unique digits and a given length can be calculated using permutations.
Insight 3: The solution uses a digit-by-digit approach to account for numbers less than or equal to n, efficiently pruning the search space.

Complexity Analysis

Time Complexity: O(log(N))

Space Complexity: O(log(N))

Topics

This problem involves: Math, Dynamic Programming.

Companies

Asked at: IBM, J.P. Morgan.

Numbers With Repeated Digits | LeetCode 1012

Hard

Math Dynamic Programming

Time: O(log(N))Space: O(log(N))

Solution Code

Available in:

class Solution {
    public int numDupDigitsAtMostN(int n) {
        
        
        
        String s = String.valueOf(n + 1);
        int k = s.length();
        
        int count_unique = 0;
        
        for (int i = 1; i < k; i++) {
            count_unique += 9 * permutations(9, i - 1);
        }
        
        
        java.util.Set<Integer> used_digits = new java.util.HashSet<>();
        for (int i = 0; i < k; i++) {
            int d = Character.getNumericValue(s.charAt(i));
            
            int start = (i == 0) ? 1 : 0;
            for (int j = start; j < d; j++) {
                if (!used_digits.contains(j)) {
                    count_unique += permutations(10 - (i + 1), k - (i + 1));
                }
            }
            
            if (used_digits.contains(d)) {
                break;
            }
            
            used_digits.add(d);
        }
        
        return n - count_unique;
    }
    
    private int permutations(int m, int k) {
            if (k < 0 || k > m) {
                return 0;
            }
            int res = 1;
            for (int i = 0; i < k; i++) {
                res *= (m - i);
            }
            return res;
        }
}

View Full Solution on GitHubGitHub Solution View on LeetCodeLeetCode

Problem Info

Description

Problem 1012: Numbers With Repeated Digits - Detailed explanation and solution approaches.

Topics

Math Dynamic Programming

Companies

IBMJ.P. Morgan

Stats

Difficulty:Hard

Languages:4

Companies:2

Problem #:1012

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

7 steps

1
Step 1: Convert the input integer `n+1` to a string `s` to easily access individual digits.
2
Step 2: Calculate the number of integers with unique digits that have a length smaller than the length of s. This is done by iterating from length 1 to length k-1 (where k is the length of s) and using the formula 9 * P(9, i-1). We multiply by 9 because the first digit cannot be 0.
3
Step 3: Initialize a set `used_digits` to track digits already used in the current number being considered.
4
Step 4: Iterate through the digits of `s` from left to right. For each digit `d` at index `i`, iterate through possible digits `j` (from 1 if i==0, or 0 otherwise) up to `d-1`. If `j` is not in `used_digits`, it means we can construct a valid number with unique digits. Then, add P(10-(i+1), k-(i+1)) to `count_unique`.
5
Step 5: If `d` is already in `used_digits`, it means adding `d` would create a number with repeated digits, so stop.
6
Step 6: Add the digit `d` to `used_digits` for the next iteration.
7
Step 7: Finally, return n - count_unique, which gives the number of integers with at least one repeated digit.

💡

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(log(N))

🚀 Logarithmic - Very efficient! Halves the problem each step.

💾

Space Complexity

O(log(N))

🎯 Logarithmic space - Very memory efficient!

📝

Why?

The time complexity is determined by the number of digits in `n`, which is log(n). The loops iterate a number of times proportional to the number of digits. Similarly, the space complexity is proportional to the number of digits, which is log(n), because we store the digits in a set.

📖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

Brute Force

Iterate through all numbers from 1 to n and check if each number has repeated digits. This can be done by converting the number to a string and checking if any character appears more than once.

Time: O(N * log(N)), where N is the input integer `n`. For each number, we iterate through its digits, which takes O(log(N)) time.Space: O(log(N)), for storing the digits of the number.

Trade-offs: This is much slower than the given solution, but it's easier to understand and implement. It is suitable for smaller values of `n`.

✨

Key Insights

1Insight 1: The problem can be solved more efficiently by counting the numbers with *no* repeated digits and subtracting that from n.
2Insight 2: The number of integers with unique digits and a given length can be calculated using permutations.
3Insight 3: The solution uses a digit-by-digit approach to account for numbers less than or equal to n, efficiently pruning the search space.

📚

Key Terms Explained

For Students

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

Edge Cases

Edge case 1: n is a single-digit number. In this case, the number of integers with repeated digits is 0, because all numbers from 1 to n have unique digits. The solution correctly handles this case because the main loop is only executed when k > 1.
Edge case 2: n is a number like 99, 999, etc. In these cases, the solution needs to calculate permutations for smaller lengths and consider the digits up to the point where a repeating digit is found (or not found at all). The algorithm iterates over all the digits.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to handle the case where the first digit cannot be 0 when calculating permutations for lengths less than k.

✗Common mistake 2: Not considering the case where the number n itself has no repeating digits, leading to an incorrect subtraction.

✅

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: Password generation. You might want to generate passwords with unique characters, and this problem's concepts relate to understanding combinations of unique elements.

🏦Real-world scenario 2: Number plate design. When creating license plates, you might have restrictions on repeated digits or letters, and algorithms similar to this can help determine the number of possible valid license plates.

💡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

It's okay to not finish hard problems in interviews. Show your problem-solving process.

⚠️ 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: September 21, 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 14

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

@CodeNinjaJul 7

❓

Frequently Asked Questions

How do I identify if a problem needs dynamic programming?

DP problems typically have: 1) Overlapping subproblems (same calculation repeated), 2) Optimal substructure (optimal solution uses optimal solutions to subproblems). Look for keywords like "minimum/maximum," "count ways," "is it possible," or "longest/shortest."

How do I approach hard algorithm problems?

Hard problems often require: 1) Breaking down into subproblems, 2) Combining multiple advanced techniques, 3) Careful optimization of brute force. Start with a working solution (even O(n²) or worse), then optimize. Draw examples, identify patterns, and think about what makes this problem unique.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Numbers With Repeated Digits"?

For this hard-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?