Smallest Number With Given Digit Product

Smallest Number With Given Digit Product - Problem

Math Greedy Medium

Given a positive integer n, return a string representing the smallest positive integer such that the product of its digits is equal to n.

If no such number exists, return "-1".

Note: The returned number should be as small as possible, which means it should have the fewest digits, and among numbers with the same number of digits, it should be lexicographically smallest.

Input & Output

Example 1 — Basic Case

$ Input: n = 12

› Output: 26

💡 Note: We need digits that multiply to 12. Using greedy approach: 12 = 6 × 2. Sort digits: [2,6] → "26"

Example 2 — Single Digit

$ Input: n = 1

› Output: 1

💡 Note: Special case: the smallest number with digit product 1 is 1 itself

Example 3 — Impossible Case

$ Input: n = 11

› Output: -1

💡 Note: 11 is prime and > 9, so it cannot be expressed as a product of single digits (2-9)

Constraints

1 ≤ n ≤ 10⁹

Visualization

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Asked in

G Google 15 a Amazon 12 M Microsoft 8

The key insight is to use greedy factorization with digits 9 down to 2, then sort the result digits in ascending order. This approach ensures we use the fewest digits possible (by using larger digits first) and get the lexicographically smallest result. Time: O(log n), Space: O(log n)

Common Approaches

✓ Brute Force

⏱️ Time: O(10^k) Space: O(1)

Generate numbers incrementally and check if the product of their digits equals the target. This approach guarantees finding the smallest number but is extremely slow for large n.

Greedy Factorization

⏱️ Time: O(log n) Space: O(log n)

Use greedy approach by factorizing n using digits 9 down to 2. Start with the largest digit (9) and use it as many times as possible, then move to 8, 7, etc. This minimizes the total number of digits.

Brute Force — Algorithm Steps

Start from number 1
For each number, calculate product of its digits
If product equals n, return the number
Continue until found or timeout

Visualization

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Step-by-Step Walkthrough

Start from 1

Begin checking numbers from 1

Calculate Product

For each number, multiply all its digits

Check Match

If product equals target, return that number

Code -

solution.c — C

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

char* solution(int n) {
    char* result = (char*)malloc(20);
    
    if (n == 0) {
        strcpy(result, "10");
        return result;
    }
    if (n == 1) {
        strcpy(result, "1");
        return result;
    }
    
    // Try numbers starting from 1
    long long num = 1;
    while (num <= 1000000000LL) {  // Reasonable limit
        // Calculate product of digits
        long long product = 1;
        long long temp = num;
        while (temp > 0) {
            product *= temp % 10;
            temp /= 10;
        }
        
        if (product == n) {
            sprintf(result, "%lld", num);
            return result;
        }
        num++;
    }
    
    strcpy(result, "-1");
    return result;
}

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

Time & Space Complexity

Time Complexity

⏱️

O(10^k)

Where k is the number of digits in result, we may check exponentially many numbers

⚡ Linearithmic

Space Complexity

O(1)

Only using constant extra space for calculations

⚡ Linearithmic Space

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Input & Output

Constraints

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Related Problems

Common Approaches

Brute Force — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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