LeetCode 625: Minimum Factorization Solution in Python – A Step-by-Step Guide

What Is LeetCode 625: Minimum Factorization?

In LeetCode 625: Minimum Factorization, you’re given a positive integer num, and your task is to find the smallest positive integer whose digits, when multiplied together, equal num. The result must fit within a 32-bit signed integer (≤ 2³¹ - 1 = 2,147,483,647), and you return 0 if no such integer exists. For example, for num = 48, the smallest integer is 68 (6 * 8 = 48), but for num = 100, no valid solution fits within bounds, so you return 0. This problem tests your ability to factor numbers into single-digit components and optimize for the smallest result.

Problem Statement

• Input:
• num: A positive integer (1 ≤ num ≤ 10⁹).
• Output: An integer—the smallest number whose digits multiply to num, or 0 if impossible within 32-bit bounds.
• Rules:
• Use digits 2-9 (1 doesn’t change product).
• Result must be ≤ 2³¹ - 1.
• Return 0 if no valid factorization exists.

Constraints

• 1 ≤ num ≤ 10⁹.

Examples

• Input: num = 48
• Output: 68 (6 * 8 = 48, smallest valid)
• Input: num = 15
• Output: 35 (3 * 5 = 15)
• Input: num = 100
• Output: 0 (No valid factorization within 32-bit bounds)

These examples set the stage—let’s solve it!

Understanding the Problem: Factoring into Digits

To solve LeetCode 625: Minimum Factorization in Python, we need to factor a number into digits from 2-9 that multiply to it, arrange them to form the smallest possible integer, and ensure it fits within 32-bit bounds. A naive approach might try all combinations, but we can optimize by factoring greedily or systematically. We’ll use:

• Best Solution (Greedy Factorization with Digit Optimization): O(log n) time, O(1) space—fast and intuitive.
• Alternative Solution (Dynamic Programming): O(n * log n) time, O(n) space—thorough but complex.

Let’s start with the greedy solution, breaking it down for beginners.

Best Solution: Greedy Factorization with Digit Optimization

Why This Is the Best Solution

The greedy factorization approach is the top choice for LeetCode 625 because it’s efficient—O(log n) time with O(1) space—and leverages a key insight: to minimize the resulting integer, use the largest possible digits (9 down to 2) in descending order, as this reduces the number of digits needed, which minimizes the number when sorted ascendingly. It fits constraints (num ≤ 10⁹) and is simple to implement. It’s like picking the biggest building blocks to make the smallest stack!

How It Works

Think of this as breaking a number into digit blocks:

• Step 1: Handle Edge Cases:
• If num < 10, return num (already a digit).
• Step 2: Factor Greedily:
• Start with largest digit (9), divide num by it if possible.
• Add digit to list, update num to quotient.
• Repeat down to 2 until num = 1 or no factors work.
• Step 3: Check Validity:
• If num > 1, no valid factorization (remaining factor not 1).
• If digits exceed 32-bit bound when combined, return 0.
• Step 4: Build Smallest Number:
• Sort digits ascending, join into integer.
• Step 5: Return Result:
• Return integer or 0 if invalid.

It’s like a digit sculptor—carve and assemble!

Step-by-Step Example

Example: num = 48

• Step 1: 48 ≥ 10, proceed.
• Step 2: Factor:
• Try 9: 48 % 9 = 3, no.
• Try 8: 48 / 8 = 6, yes, digits = [8], num = 6.
• Try 7: 6 % 7, no.
• Try 6: 6 / 6 = 1, yes, digits = [8, 6], num = 1.
• Stop (num = 1).
• Step 3: Valid (num = 1), digits = [8, 6].
• Step 4: Sort: [6, 8], join: 68.
• Check: 68 < 2³¹ - 1.
• Output: 68.

Example: num = 100

• Step 2: Factor:
• Try 9: 100 % 9, no.
• Try 5: 100 / 5 = 20, digits = [5], num = 20.
• Try 5: 20 / 5 = 4, digits = [5, 5], num = 4.
• Try 4: 4 / 4 = 1, digits = [5, 5, 4], num = 1.
• Step 3: Valid, digits = [5, 5, 4].
• Step 4: Sort: [4, 5, 5], join: 455 < 2³¹ - 1.
• Output: 455.

Code with Detailed Line-by-Line Explanation

Here’s the Python code, explained clearly:

Copy codeclass Solution:
    def smallestFactorization(self, num: int) -> int:
        # Step 1: Handle edge case
        if num < 10:
            return num

        # Step 2: Greedily factorize into digits
        digits = []
        temp = num
        for d in range(9, 1, -1):  # From 9 to 2
            while temp % d == 0:   # While divisible
                digits.append(d)
                temp //= d

        # Step 3: Check if factorization is complete
        if temp != 1:  # Remaining factor not 1
            return 0

        # Step 4: Build smallest number
        digits.sort()  # Ascending for smallest
        result = 0
        for digit in digits:
            result = result * 10 + digit
            if result > 2**31 - 1:  # 32-bit bound
                return 0

        # Step 5: Return result
        return result

• Lines 4-5: Edge case: num < 10 is itself.
• Lines 8-13: Factor:
• Try digits 9 to 2, divide and collect.
• Lines 16-17: Check: Must fully factor (temp = 1).
• Lines 20-25: Build:
• Sort digits, construct number, check bounds.
• Time Complexity: O(log n)—divisions reduce num logarithmically.
• Space Complexity: O(1)—digits list is small (max ~10).

This is like a digit optimizer—fast and neat!

Alternative Solution: Dynamic Programming Approach

Why an Alternative Approach?

The dynamic programming (DP) approach builds all possible factorizations—O(n * log n) time, O(n) space. It’s thorough, finding the smallest number systematically, but overkill for this problem due to complexity and space. It’s like exploring every digit combination to find the best!

How It Works

Picture this as a digit explorer:

• Step 1: DP array for smallest number up to num.
• Step 2: For each value, try digits 2-9.
• Step 3: Build smallest number with valid factors.
• Step 4: Check bounds and return.

It’s like a number builder—exhaustive but complex!

Step-by-Step Example

Example: num = 15

• Step 1: DP[0] = "" (empty), others infinity.
• Step 2: Fill DP:
• DP[3] = "3" (3).
• DP[5] = "5" (5).
• DP[15] = min("35", "53") = "35" (3*5).
• Step 3: DP[15] = "35".
• Step 4: 35 < 2³¹ - 1, return 35.
• Output: 35.

Code for DP Approach

Copy codeclass Solution:
    def smallestFactorization(self, num: int) -> int:
        # Step 1: Handle edge case
        if num < 10:
            return num

        # Step 2: DP to find smallest factorization
        dp = [float('inf')] * (num + 1)
        dp[1] = 1  # Base case

        for i in range(2, num + 1):
            for d in range(2, 10):
                if i % d == 0:
                    prev = dp[i // d]
                    if prev != float('inf'):
                        curr = int(str(prev) + str(d))
                        dp[i] = min(dp[i], curr)

        # Step 3: Get result
        result = dp[num]
        if result == float('inf') or result > 2**31 - 1:
            return 0

        # Step 4: Return result
        return result

• Lines 8-11: Init DP with infinity, DP[1] = 1.
• Lines 13-18: Fill DP:
• Try digits, build numbers, take min.
• Lines 21-23: Check validity and bounds.
• Time Complexity: O(n * log n)—n values, log n digits.
• Space Complexity: O(n)—DP array.

It’s a number explorer—thorough but heavy!

Comparing the Two Solutions

• Greedy (Best):
• Pros: O(log n) time, O(1) space, fast and simple.
• Cons: Assumes greedy works (proven correct here).
• DP (Alternative):
• Pros: O(n * log n) time, O(n) space, exhaustive.
• Cons: Slower, more space.

Greedy wins for efficiency.

Additional Examples and Edge Cases

• Input: num = 1
• Output: 1.
• Input: num = 13
• Output: 0 (prime, no digit factorization).

Both handle these well.

Complexity Breakdown

• Greedy: Time O(log n), Space O(1).
• DP: Time O(n * log n), Space O(n).

Greedy is optimal.

Key Takeaways

• Greedy: Largest digits—smart!
• DP: Full exploration—clear!
• Numbers: Factorization is fun.
• Python Tip: Loops rock—see Python Basics.

Final Thoughts: Factor That Number

LeetCode 625: Minimum Factorization in Python is a cool number-crunching challenge. Greedy factorization offers speed and elegance, while DP provides a comprehensive view. Want more? Try LeetCode 263: Ugly Number or LeetCode 313: Super Ugly Number. Ready to crack it? Head to Solve LeetCode 625 on LeetCode and find that smallest factorization today!