Nth Digit

Nth Digit - Problem

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

The sequence is formed by concatenating all positive integers: 123456789101112131415...

Note: The digit indexing starts from 1 (1-indexed).

Input & Output

Example 1 — Finding 11th Digit

$ Input: n = 11

› Output: 0

💡 Note: Sequence: 123456789101112... The 11th digit is the second digit of 10, which is 0

Example 2 — Single Digit

$ Input: n = 3

› Output: 3

💡 Note: Sequence: 123456789... The 3rd digit is simply 3

Example 3 — Two-Digit Range

$ Input: n = 15

› Output: 2

💡 Note: Sequence: 123456789101112131415... Position 15 is the first digit of 12, which is 1. Wait, let me recalculate: positions 1-9 are digits 1-9, positions 10-11 are '1','0' from 10, positions 12-13 are '1','1' from 11, positions 14-15 are '1','2' from 12. So 15th digit is '2'

Constraints

1 ≤ n ≤ 2³¹ - 1

Visualization

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Asked in

G Google 12 Apple 8 MS Microsoft 6

The key insight is recognizing the mathematical pattern: 1-digit numbers contribute 9 digits, 2-digit numbers contribute 180 digits, etc. Best approach uses direct mathematical calculation to find the target number and digit position. Time: O(log n), Space: O(1)

Common Approaches

✓ Binary Search on Answer Space

⏱️ Time: O(log² n) Space: O(1)

Use binary search on the number range to find which number contains the nth digit, then extract the specific digit from that number.

Brute Force Counting

⏱️ Time: O(n) Space: O(1)

Iterate through each number starting from 1, count how many digits each number contributes, and stop when we reach the nth digit.

Mathematical Pattern Recognition

⏱️ Time: O(log n) Space: O(1)

Recognize that 1-digit numbers contribute 9 digits total, 2-digit numbers contribute 180 digits, 3-digit numbers contribute 2700 digits, etc. Use this pattern to jump directly to the correct range.

Binary Search on Answer Space — Algorithm Steps

Binary search on numbers 1 to some upper bound
For each mid value, calculate total digits up to that number
Adjust search range based on whether total is less or greater than n
Extract the specific digit from the found number

Visualization

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Step-by-Step Walkthrough

Search Space

Binary search between numbers 1 and n

Count Digits

For each mid point, count total digits up to that number

Narrow Range

Adjust bounds until finding exact number with nth digit

Code -

solution.c — C

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

long long countDigitsUpTo(long long num) {
    if (num <= 0) return 0;
    
    long long total = 0;
    int digits = 1;
    long long start = 1;
    
    while (start <= num) {
        long long end = (start * 10 - 1 < num) ? start * 10 - 1 : num;
        long long count = end - start + 1;
        total += count * digits;
        
        digits++;
        start *= 10;
    }
    
    return total;
}

int solution(int n) {
    long long left = 1, right = n;
    
    while (left < right) {
        long long mid = (left + right) / 2;
        if (countDigitsUpTo(mid) < n) {
            left = mid + 1;
        } else {
            right = mid;
        }
    }
    
    long long targetNum = left;
    long long digitsBefore = countDigitsUpTo(targetNum - 1);
    int positionInNum = n - digitsBefore;
    
    char targetStr[20];
    sprintf(targetStr, "%lld", targetNum);
    
    return targetStr[positionInNum - 1] - '0';
}

int main() {
    int n;
    scanf("%d", &n);
    int result = solution(n);
    printf("%d\n", result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(log² n)

Binary search O(log n) with digit counting O(log n) per iteration

⚡ Linearithmic

Space Complexity

O(1)

Only using constant extra variables for binary search bounds

✓ Linear Space

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Input & Output

Constraints

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Related Problems

Common Approaches

Binary Search on Answer Space — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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