Numbers At Most N Given Digit Set - Practice Coding Problems

Numbers At Most N Given Digit Set - Problem

Array Math String Binary Search Dynamic Programming Hard

Digital Number Building Challenge

Imagine you have a special calculator that only has certain digit buttons working! Given an array of digits sorted in non-decreasing order, you can create numbers by pressing these buttons as many times as you want.

For example, if your working digits are ['1','3','5'], you could create numbers like:
• '1', '3', '5' (single digits)
• '13', '31', '55' (two digits)
• '135', '311', '555' (three digits)
• And so on...

Your Mission: Count how many positive integers you can generate that are less than or equal to a given integer n.

This is a classic combinatorial counting problem that tests your understanding of digit manipulation and mathematical reasoning!

Input & Output

example_1.py — Basic Case

$ Input: digits = ['1','3','5'], n = 100

› Output: 20

💡 Note: We can form: 1-digit (1,3,5), 2-digit (11,13,15,31,33,35,51,53,55), 3-digit (all combinations except those > 100). Total = 3 + 9 + 8 = 20.

example_2.py — Small Limit

$ Input: digits = ['1','4','9'], n = 1000000000

› Output: 29523

💡 Note: With digits [1,4,9] and large n, we count all combinations systematically: 1-digit (3), 2-digit (9), 3-digit (27), ..., up to 10-digit numbers.

example_3.py — Edge Case

$ Input: digits = ['7'], n = 8

› Output: 1

💡 Note: Only digit 7 available. We can only form the number 7, which is ≤ 8. Numbers like 77, 777 are all > 8.

Constraints

1 ≤ digits.length ≤ 9
digits[i].length == 1
digits[i] is a digit from '1' to '9'
All the values in digits are unique
digits is sorted in non-decreasing order
1 ≤ n ≤ 10⁹

Visualization

Tap to expand

Understanding the Visualization

Count Short Wins

All combinations with fewer reels than n automatically win

Same Length Analysis

For same number of reels, check reel by reel from left to right

Constraint Tracking

Once a reel shows a digit less than n's digit, all following reels are free

Mathematical Sum

Add up all valid combinations without spinning every possibility

Key Takeaway

🎯 Key Insight: Mathematical counting avoids generating every number. Count shorter lengths (always valid) + same length (digit-by-digit comparison) = O(log n) solution!

Asked in

G Google 45 a Amazon 38 Apple 22 ⊞ Microsoft 18

The optimal solution uses mathematical counting instead of generating numbers. Key insight: numbers with fewer digits than n are always valid, while same-length numbers require digit-by-digit comparison using dynamic programming. Time complexity: O(log n).

Common Approaches

Approach	Time	Space	Notes
✓ Recursive Generation (Brute Force)	O(k^m)	O(m)	Generate all possible numbers recursively and count those ≤ n
Mathematical Counting (Optimal)	O(log n)	O(log n)	Use mathematical approach to count without generating numbers

Recursive Generation (Brute Force) — Algorithm Steps

Convert n to string to get its length
For each possible length (1 to len(n)), generate all numbers
Use recursion to build numbers digit by digit
Compare each generated number with n
Count valid numbers

Visualization

Tap to expand

Step-by-Step Walkthrough

Start Empty

Begin with empty string, try each digit

Length 1

Generate '1', '3', '5' - all ≤ 100

Length 2

Generate '11', '13', '15', '31', '33', '35', '51', '53', '55' - all ≤ 100

Length 3

Generate '111'... some > 100, count only valid ones

Code -

solution.c — C

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

int count = 0;

void generateNumbers(char* current, int length, char** digits, int digitsSize, int n) {
    if (strlen(current) == length) {
        if (atoi(current) <= n) {
            count++;
        }
        return;
    }
    
    for (int i = 0; i < digitsSize; i++) {
        char next[20];
        strcpy(next, current);
        strcat(next, digits[i]);
        generateNumbers(next, length, digits, digitsSize, n);
    }
}

int numAtMostNGivenDigitSet(char** digits, int digitsSize, int n) {
    char nStr[20];
    sprintf(nStr, "%d", n);
    int nLen = strlen(nStr);
    count = 0;
    
    // Generate numbers of all possible lengths
    for (int length = 1; length <= nLen; length++) {
        char empty[1] = "";
        generateNumbers(empty, length, digits, digitsSize, n);
    }
    
    return count;
}

Time & Space Complexity

Time Complexity

⏱️

O(k^m)

Where k is number of digits and m is length of n. We generate k^m possible numbers.

✓ Linear Growth

Space Complexity

O(m)

Recursion depth is at most m (length of n)

✓ Linear Space

28.4K Views

Medium Frequency

~25 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

Recursive Generation (Brute Force) — Algorithm Steps

Visualization

Code -

Time & Space Complexity

Select Compiler