Number of Digit One - Practice Coding Problems

Number of Digit One - Problem

Given an integer n, you need to count how many times the digit 1 appears in all non-negative integers from 0 to n (inclusive).

This is a classic mathematical problem that requires careful analysis of digit patterns. For example, if n = 13, we need to examine all numbers: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13. The digit 1 appears in: 1 (once), 10 (once), 11 (twice), 12 (once), 13 (once) = 6 times total.

Goal: Return the total count of digit 1 occurrences across all numbers from 0 to n.

Input & Output

example_1.py — Basic Case

$ Input: n = 13

› Output: 6

💡 Note: Numbers from 0-13: 0,1,2,3,4,5,6,7,8,9,10,11,12,13. Digit '1' appears in: 1(1 time), 10(1 time), 11(2 times), 12(1 time), 13(1 time). Total = 1+1+2+1+1 = 6

example_2.py — Single Digit

$ Input: n = 1

› Output: 1

💡 Note: Numbers from 0-1: 0,1. Only the number '1' contains digit '1', appearing once. Total = 1

example_3.py — Edge Case Zero

$ Input: n = 0

› Output: 0

💡 Note: Only number 0 is in range, which contains no digit '1'. Total = 0

Visualization

Tap to expand

Understanding the Visualization

Position Analysis

For each digit position, analyze the pattern of when digit 1 appears

Pattern Formula

Apply mathematical formula based on current digit value (0, 1, or >1)

Count Aggregation

Sum up the counts from all positions to get total occurrences

Key Takeaway

🎯 Key Insight: Instead of checking every number individually, we can mathematically calculate how many times digit 1 appears at each position by analyzing the patterns in the higher and lower digits.

Time & Space Complexity

Time Complexity

⏱️

O(log n)

We only examine each digit position once, and there are log n positions

⚡ Linearithmic

Space Complexity

O(1)

Only using constant extra space for calculations

✓ Linear Space

Constraints

0 ≤ n ≤ 10⁹
Time limit: 1 second
Space limit: O(1) space preferred for optimal solution

Asked in

G Google 45 ⊞ Microsoft 38 a Amazon 32 f Meta 25

The optimal solution uses mathematical pattern analysis to count digit 1 occurrences at each position (ones, tens, hundreds, etc.) without iterating through all numbers. This achieves O(log n) time complexity by recognizing that digit 1 appears in predictable patterns based on higher and lower digits at each position.

Common Approaches

Approach	Time	Space	Notes
✓ Brute Force (Check Every Number)	O(n * log n)	O(1)	Iterate through all numbers from 0 to n and count digit 1 in each number
Mathematical Pattern Analysis (Optimal)	O(log n)	O(1)	Use mathematical analysis to count 1s at each digit position without iterating through all numbers

Brute Force (Check Every Number) — Algorithm Steps

Initialize counter to 0
For each number i from 0 to n:
Convert number i to string
Count occurrences of character '1' in the string
Add count to total counter
Return total counter

Visualization

Tap to expand

Step-by-Step Walkthrough

Start at 0

Begin counting from number 0, no 1s found

Check each number

For each number, convert to string and count '1' characters

Accumulate count

Add the count of 1s from current number to total

Continue until n

Repeat process for all numbers up to and including n

Code -

solution.c — C

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

int countDigitOne(int n) {
    int count = 0;
    for (int i = 0; i <= n; i++) {
        int num = i;
        while (num > 0) {
            if (num % 10 == 1) count++;
            num /= 10;
        }
    }
    return count;
}

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

Time & Space Complexity

Time Complexity

⏱️

O(n * log n)

We iterate through n numbers, and for each number we need O(log n) time to count digits

⚡ Linearithmic

Space Complexity

O(1)

Only using constant extra space for counter variables

✓ Linear Space

Constraints

0 ≤ n ≤ 10⁹
Time limit: 1 second
Space limit: O(1) space preferred for optimal solution

32.7K Views

Medium Frequency

~25 min Avg. Time

1.2K Likes

Ln 1, Col 1

Smart Actions

💡 Explanation

AI Ready

💡 Suggestion Tab to accept Esc to dismiss

// Output will appear here after running code

Code Editor Closed

Click the red button to reopen

Input & Output

Visualization

Time & Space Complexity

Related Problems

Constraints

Common Approaches

Brute Force (Check Every Number) — Algorithm Steps

Visualization

Code -

Time & Space Complexity

Constraints

Select Compiler