Number of Digit One

Number of Digit One - Problem

Given an integer n, count the total number of digit 1 appearing in all non-negative integers less than or equal to n.

For example, if n = 13, the numbers are: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13. The digit 1 appears in: 1, 10, 11, 12, 13. Counting each occurrence: 1 appears 1 time, 10 has 1 occurrence, 11 has 2 occurrences, 12 has 1 occurrence, and 13 has 1 occurrence. Total = 6.

Input & Output

Example 1 — Small Number

$ Input: n = 13

› Output: 6

💡 Note: Numbers 0-13: digit 1 appears in 1(×1), 10(×1), 11(×2), 12(×1), 13(×1) = total 6 times

Example 2 — Single Digit

$ Input: n = 9

› Output: 1

💡 Note: Numbers 0-9: digit 1 appears only in number 1, so total is 1

Example 3 — Larger Number

$ Input: n = 99

› Output: 20

💡 Note: Units place: 1,11,21,31,41,51,61,71,81,91 (10 ones). Tens place: 10-19 (10 ones). Total: 20

Constraints

0 ≤ n ≤ 10⁹

Visualization

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Asked in

G Google 15 M Microsoft 12 a Amazon 8

The mathematical approach is optimal by analyzing digit patterns instead of checking every number. Key insight: Count 1s by digit position using mathematical formulas. Best approach: O(log n) time, O(1) space.

Common Approaches

✓ Brute Force - Check Every Number

⏱️ Time: O(n log n) Space: O(1)

For each number from 1 to n, convert it to string and count how many times '1' appears. Sum all occurrences across all numbers.

Mathematical Pattern Recognition

⏱️ Time: O(log n) Space: O(1)

For each digit position (units, tens, hundreds), calculate how many 1s appear by analyzing the pattern. The key insight is that 1s appear in predictable cycles based on powers of 10.

Brute Force - Check Every Number — Algorithm Steps

Iterate through numbers 1 to n
Convert each number to string
Count '1' characters in each string
Sum all counts

Visualization

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Step-by-Step Walkthrough

Check Each Number

Go through 1, 2, 3, ..., n

Count 1s

Count digit 1s in each number

Sum Total

Add all counts together

Code -

solution.c — C

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

int solution(int n) {
    int count = 0;
    for (int i = 1; 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);
    int result = solution(n);
    printf("%d\n", result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n log n)

Loop n times, each iteration processes log n digits

⚡ Linearithmic

Space Complexity

O(1)

Only using constant extra space for counting

✓ Linear Space

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Constraints

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Related Problems

Common Approaches

Brute Force - Check Every Number — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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