Nth Digit - Practice Coding Problems

Nth Digit - Problem

Imagine an infinite sequence of positive integers written without any separators: 123456789101112131415.... Your task is to find the nth digit in this sequence.

Given an integer n, return the nth digit of the infinite integer sequence [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, ...].

Key insights:

Single-digit numbers (1-9) contribute 9 digits
Two-digit numbers (10-99) contribute 90 × 2 = 180 digits
Three-digit numbers (100-999) contribute 900 × 3 = 2700 digits
And so on...

The challenge is to efficiently determine which number contains the nth digit and which position within that number.

Input & Output

example_1.py — Basic Case

$ Input: n = 3

› Output: 3

💡 Note: The sequence is 123456789101112... The 3rd digit is 3.

example_2.py — Two-Digit Range

$ Input: n = 11

› Output: 0

💡 Note: The sequence is 123456789101112... Positions 1-9 are single digits (123456789), position 10 is '1' from '10', position 11 is '0' from '10'.

example_3.py — Edge Case

$ Input: n = 15

› Output: 2

💡 Note: The sequence is 123456789101112131415... Position 15 falls on the second digit of number 12, which is '2'.

Constraints

1 ≤ n ≤ 2³¹ - 1
n is guaranteed to be valid
The sequence starts from 1 (not 0)

Visualization

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Understanding the Visualization

Pattern Recognition

1-digit numbers (1-9) contribute 9 digits total

Range Expansion

2-digit numbers (10-99) contribute 90×2=180 digits

Mathematical Jump

Skip entire ranges to find target number quickly

Precise Extraction

Calculate exact position within the target number

Key Takeaway

🎯 Key Insight: Instead of counting digit by digit, use mathematical patterns to calculate which number contains the nth digit and extract it directly in O(log n) time.

Asked in

G Google 45 f Meta 32 a Amazon 28 ⊞ Microsoft 25

The optimal solution uses mathematical pattern recognition to identify digit ranges (1-digit: 9 digits, 2-digit: 180 digits, etc.) and jumps directly to the target number in O(log n) time, avoiding the need to count every digit sequentially.

Common Approaches

Approach	Time	Space	Notes
✓ Mathematical Pattern Recognition (Optimal)	O(log n)	O(1)	Use mathematical patterns to jump directly to the correct range and number
Brute Force (Count Every Digit)	O(n)	O(1)	Iterate through each number and count digits until reaching position n

Mathematical Pattern Recognition (Optimal) — Algorithm Steps

Determine how many digits the target number has (1, 2, 3, etc.)
Calculate which specific number in that range contains the nth digit
Find the exact position within that number
Extract and return the digit at that position

Visualization

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Step-by-Step Walkthrough

Identify Range

Determine if target is in 1-digit, 2-digit, 3-digit numbers, etc.

Calculate Number

Find exact number containing the nth digit

Find Position

Determine position within that number

Extract Digit

Return the digit at calculated position

Code -

solution.c — C

#include <stdio.h>
#include <string.h>

int findNthDigit(int n) {
    // Step 1: Find how many digits the target number has
    long digits = 1;
    long count = 9;
    long start = 1;
    
    while (n > digits * count) {
        n -= digits * count;
        digits++;
        count *= 10;
        start *= 10;
    }
    
    // Step 2: Find which number contains the nth digit
    long number = start + (n - 1) / digits;
    
    // Step 3: Find which digit in that number
    int digitIndex = (n - 1) % digits;
    
    // Step 4: Return the digit
    char numberStr[20];
    sprintf(numberStr, "%ld", number);
    return numberStr[digitIndex] - '0';
}

Time & Space Complexity

Time Complexity

⏱️

O(log n)

We only iterate through digit lengths (1, 2, 3, ...) which is logarithmic

⚡ Linearithmic

Space Complexity

O(1)

Only using constant extra space for calculations

✓ Linear Space

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Medium Frequency

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Smart Actions

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Mathematical Pattern Recognition (Optimal) — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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