Remove 9

Remove 9 - Problem

Math Hard

Start from integer 1, remove any integer that contains the digit 9 such as 9, 19, 29, etc. Now, you will have a new integer sequence [1, 2, 3, 4, 5, 6, 7, 8, 10, 11, 12, 13, 14, 15, 16, 17, 18, 20, 21, ...].

Given an integer n, return the nth (1-indexed) integer in the new sequence.

Note: The sequence is 1-indexed, meaning the first element is at position 1.

Input & Output

Example 1 — Basic Case

$ Input: n = 8

› Output: 8

💡 Note: The sequence without 9s is [1,2,3,4,5,6,7,8,10,11,...]. The 8th number is 8.

Example 2 — Small Position

$ Input: n = 5

› Output: 6

💡 Note: The sequence is [1,2,3,4,5,6,7,8,10,...]. The 5th number is 6.

Example 3 — After Multiple 9s

$ Input: n = 18

› Output: 20

💡 Note: Sequence: [1,2,3,4,5,6,7,8,10,11,12,13,14,15,16,17,18,20,...]. 18th number is 20 (skip 9,19).

Constraints

1 ≤ n ≤ 8 × 10⁸

Visualization

Tap to expand

Asked in

G Google 15 a Amazon 10 f Facebook 8

The key insight is recognizing that numbers without digit 9 form a base-9 system using digits 1-8. Best approach is mathematical conversion: convert position n to base-9, then map each digit d to d+1. Time: O(log n), Space: O(log n)

Common Approaches

✓ Brute Force - Count and Skip

⏱️ Time: O(n × log(result)) Space: O(1)

Iterate through integers starting from 1, check each number to see if it contains digit 9. If not, increment our count. When count reaches n, return the current number.

Mathematical Base-9 Conversion

⏱️ Time: O(log n) Space: O(log n)

The key insight is that numbers without digit 9 form a base-9 system using digits 1-8. Convert the position n to base-9, then map each digit d to d+1 if d < 9, effectively skipping digit 9.

Brute Force - Count and Skip — Algorithm Steps

Start counting from 1
For each number, check if it contains digit 9
If no 9, increment valid count
Return number when count equals n

Visualization

Tap to expand

Step-by-Step Walkthrough

Count Valid Numbers

Check each number from 1 onwards

Skip Numbers with 9

Skip 9, 19, 29, etc.

Return nth Valid

When count reaches n, return current number

Code -

solution.c — C

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

int containsNine(int num) {
    char str[20];
    sprintf(str, "%d", num);
    return strchr(str, '9') != NULL;
}

int solution(int n) {
    int count = 0;
    int num = 1;
    while (1) {
        if (!containsNine(num)) {
            count++;
            if (count == n) {
                return num;
            }
        }
        num++;
    }
}

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(result))

For each of n positions, we check O(log(result)) digits to see if 9 exists

⚡ Linearithmic

Space Complexity

O(1)

Only uses a few variables for counting and checking

⚡ Linearithmic Space

23.0K Views

Medium Frequency

~15 min Avg. Time

892 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

Constraints

Visualization

Related Problems

Common Approaches

Brute Force - Count and Skip — Algorithm Steps

Visualization

Code -

Time & Space Complexity

Select Compiler