What is the time complexity of Nth Digit?

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

What data structures are used to solve Nth Digit?

This problem uses Math, Binary Search. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Nth Digit?

Insight 1: We can group numbers by the number of digits they have (1-digit numbers, 2-digit numbers, 3-digit numbers, etc.). This allows us to efficiently determine which number contains the nth digit. Insight 2: We can calculate the number of digits in each group and subtract them from 'n' until 'n' falls within the range of a specific group. This simplifies finding the exact number and digit.

Which companies ask LeetCode 400. Nth Digit in interviews?

This problem is frequently asked at Accenture.

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Nth Digit - LeetCode 400 Solution

Nth Digit - Complete Solution Guide

Nth Digit is LeetCode problem 400, 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 n th digit of the infinite integer sequence [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, ...] . Example 1: Input: n = 3 Output: 3 Example 2: Input: n = 11 Output: 0 Explanation: The 11 th digit of the sequence 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, ... is a 0, which is part of the number 10. Constraints: 1 <= n <= 2 31 - 1

Detailed Explanation

The problem asks us to find the nth digit in the infinite sequence of integers: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12,... Given an integer 'n', we need to return the digit located at the nth position in this sequence. For example, if n=3, the answer is 3. If n=11, the answer is 0, because the 11th digit belongs to the number 10, and it's the '0'. The constraint is that 'n' can be a relatively large number (up to 2<sup>31</sup> - 1), so a naive approach of generating the entire sequence until the nth position would be inefficient.

Solution Approach

The solution first determines the number of digits of the number containing the nth digit. It iteratively subtracts the number of digits contributed by numbers with increasing lengths (1-digit, 2-digit, etc.) from 'n' until 'n' is small enough to fall within a particular range of numbers with the same number of digits. Once we know the digit length, we can determine the number containing the nth digit and then extract the specific digit at the corresponding index.

Step-by-Step Algorithm

Step 1: Initialize `digits` to 1 (number of digits), `count` to 9 (number of integers with `digits` digits), and `start` to 1 (first integer with `digits` digits).
Step 2: While `n` is greater than `digits * count`, subtract `digits * count` from `n`, increment `digits`, multiply `count` by 10, and multiply `start` by 10.
Step 3: After the loop, `digits` represents the number of digits of the number containing the nth digit, `start` is the first number with `digits` digits, and `n` is the index of the digit within the range of numbers with the same number of digits.
Step 4: Calculate the number that contains the nth digit by adding `(n - 1) // digits` to `start`. The integer division `//` is used.
Step 5: Calculate the index of the nth digit within the number by taking the modulo `(n - 1) % digits`.
Step 6: Convert the number to a string and extract the digit at the calculated index. In C, a different approach is taken to extract the digit without string conversion.

Key Insights

Insight 1: We can group numbers by the number of digits they have (1-digit numbers, 2-digit numbers, 3-digit numbers, etc.). This allows us to efficiently determine which number contains the nth digit.
Insight 2: We can calculate the number of digits in each group and subtract them from 'n' until 'n' falls within the range of a specific group. This simplifies finding the exact number and digit.
Insight 3: Integer overflow should be carefully considered, especially when dealing with large 'n' values. Use `long` type in Java and C++ when necessary.

Complexity Analysis

Time Complexity: O(log(n))

Space Complexity: O(1)

Topics

This problem involves: Math, Binary Search.

Companies

Asked at: Accenture.

Nth Digit | LeetCode 400

Medium

Math Binary Search

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

Solution Code

Available in:

class Solution {
    public int findNthDigit(int n) {
        int digits = 1;
        long count = 9;
        long start = 1;
        
        while (n > (long)digits * count) {
            n -= (int)(digits * count);
            digits++;
            count *= 10;
            start *= 10;
        }
        
        long number = start + (n - 1) / digits;
        
        int digit_index = (n - 1) % digits;
        
        return String.valueOf(number).charAt(digit_index) - '0';
    }
}

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Problem Info

Description

Problem 400: Nth Digit - Detailed explanation and solution approaches.

Topics

Math Binary Search

Companies

Accenture

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:400

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: Initialize `digits` to 1 (number of digits), `count` to 9 (number of integers with `digits` digits), and `start` to 1 (first integer with `digits` digits).
2
Step 2: While `n` is greater than `digits * count`, subtract `digits * count` from `n`, increment `digits`, multiply `count` by 10, and multiply `start` by 10.
3
Step 3: After the loop, `digits` represents the number of digits of the number containing the nth digit, `start` is the first number with `digits` digits, and `n` is the index of the digit within the range of numbers with the same number of digits.
4
Step 4: Calculate the number that contains the nth digit by adding `(n - 1) // digits` to `start`. The integer division `//` is used.
5
Step 5: Calculate the index of the nth digit within the number by taking the modulo `(n - 1) % digits`.
6
Step 6: Convert the number to a string and extract the digit at the calculated index. In C, a different approach is taken to extract the digit without string conversion.

💡

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(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The time complexity is O(log(n)) because the number of digits increases logarithmically with the value of n. The while loop iterates a number of times proportional to the number of digits in the number containing the nth digit. The space complexity is O(1) because we only use a few variables to store the number of digits, count, start, and the result, regardless of the input '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

Brute Force Approach

Generate the sequence of numbers until the nth digit is reached. This involves constructing the numbers sequentially and concatenating them into a string or tracking digits.

Time: O(10^(n/9)) in the worst case (generating all numbers until reaching the nth digit). Can be approximated to O(n)Space: O(10^(n/9)) in the worst case as we need to store the generated numbers as a sequence. Can be approximated to O(n)

Trade-offs: While conceptually simple, this approach is highly inefficient for large values of 'n' due to the need to generate potentially long sequences of digits. The main solution is more efficient and should always be prefered.

✨

Key Insights

1Insight 1: We can group numbers by the number of digits they have (1-digit numbers, 2-digit numbers, 3-digit numbers, etc.). This allows us to efficiently determine which number contains the nth digit.
2Insight 2: We can calculate the number of digits in each group and subtract them from 'n' until 'n' falls within the range of a specific group. This simplifies finding the exact number and digit.
3Insight 3: Integer overflow should be carefully considered, especially when dealing with large 'n' values. Use `long` type in Java and C++ when necessary.

📚

Key Terms Explained

For Students

Binary Search

Halving the search space each step. Like finding a word in a dictionary!

Edge Cases

Edge case 1: n = 1. The code correctly identifies the 1st digit as 1.
Edge case 2: n = 10. The code correctly identifies the 10th digit as 1, which is the first digit of the number 10 after sequence 1,2,3,4,5,6,7,8,9.
Edge case 3: n is a large number such as 2147483647. The use of `long` (in Java and C++) helps to prevent integer overflow during the calculations, ensuring the correctness of the algorithm for large values of n.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Integer overflow. Using `int` instead of `long` can lead to incorrect results when dealing with large values of 'n'.

✗Common mistake 2: Off-by-one errors when calculating the index of the digit. Remember to subtract 1 from n when finding the number and the index within the number.

✗Common mistake 3: Not correctly handling the case when n falls exactly on a transition point between numbers with different digit lengths (e.g., when n = 9, 10, 99, 100). The calculation `(n - 1) // digits` and `(n - 1) % digits` are crucial for handling these transition points.

✅

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: Data compression algorithms might utilize similar digit sequence patterns for encoding purposes.

🏦Real-world scenario 2: Generating unique identifiers or serial numbers where a specific digit at a certain position has significance.

💡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

If stuck, try working through a small example by hand. The pattern often reveals itself.

⚠️ Watch Out

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

👨‍💻 From Experience

This problem pattern appears frequently in real codebases. I've used similar logic when building production systems.

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: July 11, 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

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 14

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 7

❓

Frequently Asked Questions

How do I avoid off-by-one errors in binary search?

Use the pattern: left = 0, right = n-1, while (left <= right), mid = left + (right - left) / 2. Decide if you need to find exact match, first occurrence, or last occurrence, and adjust your conditions accordingly. Test with arrays of size 0, 1, and 2.

What techniques are commonly tested in medium-difficulty problems?

Medium problems often require combining multiple techniques: two-pointer with hash maps, BFS/DFS with state tracking, dynamic programming with optimization, or divide-and-conquer approaches. They test your ability to recognize patterns and choose the right combination of techniques.

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 "Nth Digit"?

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 binary search techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?