625. Minimum Factorization

Problem Description

The problem is to find the smallest positive integer (x) that can be formed such that the product of all its digits equals a given positive integer (num). For instance, if num is 18, x could be 29 (since 2 * 9 = 18). We want to find the smallest such x if it exists. However, there are a couple of constraints: if x does not exist or it is not a 32-bit signed integer (which means x must be less than 2^31), the function should return 0.

Intuition

To find the smallest integer x meeting our criteria, we need to consider a couple of key observations:

1. To minimize x, we should try to use the largest digits possible (except for 0 and 1 since they don't change the product). Hence, we should start checking from 9 down to 2.
2. Digits must multiply to num, so we'll repeatedly divide num by these digits, ensuring divisibility at each step.
3. We build x by appending digits to its right, meaning we first find the highest place-value digit and then move towards the lower ones.

The approach works by iterating from 9 to 2 and checking if num can be evenly divided by these digits. When it can, the digit divides num, and itself is added to what will become x. This process repeats until num is reduced to 1 (if it can't be reduced to 1, x is not possible under the problem constraints). After each division, we scale x up by a factor of 10 (to push previously added digits leftward) before adding the new digit. In the end, x must be within the 32-bit signed integer range, or else we return 0.

Learn more about Greedy and Math patterns.

Solution Approach

The implementation follows the intuition closely and uses a straightforward iterative method, with primary focus on the following steps:

1. Early return for numbers less than 2: Given that our smallest possible positive integer that is not 1 must consist of multiple digits, the cases where num is 0 or 1 are special. The function returns the num itself since no multiplication is needed.

   1if num < 2:
   2    return num

2. Initializing variables: ans is initialized to 0 - it will hold the answer. mul is set to 1 and is our multiplying factor which helps in building the integer x from its least significant digit to the most significant digit.

   1ans, mul = 0, 1

3. Iterating from 9 down to 2: The loop runs backwards from 9 to 2, checking at each step if num can be divided by i without remainder (using the modulus operator %).

   1for i in range(9, 1, -1):

4. Dividing num by the digit if possible: When num is divisible by i, we divide num by i using integer division //= and update ans. The new digit is placed in its correct position by the current value of mul. mul is then increased by a factor of 10 to make space for the next digit.

   1while num % i == 0:
   2    num //= i
   3    ans = mul * i + ans
   4    mul *= 10

5. Checking the final conditions: Once we break out of the loop, we check if num has been reduced to 1. If it's not, it means we could not fully factorize num using digits 2-9, so x cannot exist under our constraints. Moreover, we verify that the answer fits into a 32-bit signed integer by comparing it to 2**31 - 1. If either condition is not met, we return 0.

   1return ans if num < 2 and ans <= 2**31 - 1 else 0

The algorithm does not use any complex data structures, but it effectively leverages arithmetic operations and a simple for-loop to achieve the goal. This approach is efficient because it processes each digit in num at most once and avoids unnecessary computations or storage.

Example Walkthrough

Let's walk through a small example to illustrate the solution approach using the number num = 26. We are tasked with finding the smallest positive integer x such that the product of its digits equals 26.

Firstly, the function would check if num is less than 2 and would return num itself if that's the case. Since 26 is greater than 1, this early return doesn't get triggered.

1if num < 2:
2    return num

Then, the variables ans and mul are initialized:

1ans, mul = 0, 1

Here, ans will be built up to form our final number x, and mul is the multiplying factor which helps place the found factors into their correct positions within x.

The loop starts at 9 and checks if num is divisible by any number from 9 down to 2. It proceeds as follows:

• 9 does not divide 26, so it moves to the next digit.
• 8 does not divide 26, the algorithm continues to iterate.
• ...

Until we reach the digit 2:

1for i in range(9, 1, -1):

• 2 does divide 26, leaving us with a quotient of 13 (26 // 2 = 13).

The current state of our variables is now:

• num becomes 13
• ans is updated to mul * i + ans, which is 1 * 2 + 0 = 2
• mul is increased, multiplying by 10, becoming 10.

Now with num as 13, we continue to iterate. Our range is now exhausted since no digit between 2 to 9 divides 13. The loop terminates.

Finally, we check:

• Is num now 1? No, it's 13, so we can't fully factorize num using digits 2-9.
• Would ans be within a 32-bit signed integer range if num was 1? We can't check this since the first condition has already failed.

Since we couldn't reduce num to 1 by dividing by digits from 2 to 9, x does not exist within the problem constraints. Hence the function should return 0.

1return ans if num < 2 and ans <= 2**31 - 1 else 0

This example showed that the number 26 cannot be factorized into a product of digits between 2 and 9, which means there is no such x that would satisfy the problem's conditions.

Solution Implementation

Time and Space Complexity

The time complexity of the given code is O(log(num)). This is because the while loop that reduces num by a factor of i will run at most O(log(num)) times. This is similar to how division works in terms of complexity while reducing the number digit by digit. Since the outer for loop is constant and only runs 8 times (from 9 to 2), it does not affect the time complexity significantly.

The space complexity of the code is O(1). No additional space that grows with the input size num is used. We use a fixed number of variables (ans, mul, and i) that do not depend on the size of the input num.

Learn more about how to find time and space complexity quickly using problem constraints.