Problem Description

You are given a positive integer n. The task is to return the maximum product of any two digits in n. Note that you may use the same digit twice if it appears more than once in n.

Intuition

To solve this problem, we need to determine which two digits from the given integer n yield the highest product. The most straightforward approach is to identify the two largest digits in the number and calculate their product. This makes sense because the multiplication of the highest numbers results in the largest product. We will iterate through each digit of n, maintaining two variables: a and b. a holds the largest digit encountered thus far, and b holds the second largest. As we examine each digit, we compare it with a. If it surpasses a, we update b to take on the old value of a, and then set a to the new larger digit. If the digit does not exceed a but is greater than b, we simply assign it to b. Finally, the maximum product is a * b.

Solution Approach

In the solution, we are using a simple iterative approach to determine the two largest digits in the integer n. Here's how the implementation works:

1. Initialize Two Variables: We start by initializing two variables, a and b, to zero. These will store the largest and second largest digits, respectively.

2. Iterate Over Digits: We use a while loop to process every digit of n. In each iteration, we use the divmod function, which returns the quotient and remainder when n is divided by 10. Here, the quotient becomes the updated value of n, and the remainder, x, is the current rightmost digit.

3. Updating Largest and Second Largest:

   • If the current digit x is greater than a, then b is updated to the value of a, and a is updated to x.
   • If x is not greater than a but is greater than b, then b is updated to x.

4. Calculate and Return the Product: Once the loop completes, we have the two largest digits stored in a and b. We simply return their product a * b.

This approach efficiently finds the largest and second largest digits in O(d) time complexity, where d is the number of digits in n, and calculates the desired maximum product.

Example Walkthrough

Let's walk through an example with n = 29458.

1. Initialization: Start with a = 0 and b = 0.

2. Iteration Begins:

   • Step 1:

     • Current n = 29458. Using divmod(29458, 10), we get quotient 2945 and remainder 8.
     • Compare 8 with a (0). Since 8 > 0, update b to a's value (0), and set a = 8.

   • Step 2:

     • Current n = 2945. Using divmod(2945, 10), we get quotient 294 and remainder 5.
     • Compare 5 with a (8) and b (0). 5 isn't greater than a, but greater than b, so update b = 5.

   • Step 3:

     • Current n = 294. Using divmod(294, 10), we get quotient 29 and remainder 4.
     • Compare 4 with a (8) and b (5). 4 isn't greater than a or b, hence no updates.

   • Step 4:

     • Current n = 29. Using divmod(29, 10), we get quotient 2 and remainder 9.
     • Compare 9 with a (8). Since 9 > 8, update b = 8, and set a = 9.

   • Step 5:

     • Current n = 2. Using divmod(2, 10), we get quotient 0 and remainder 2.
     • Compare 2 with a (9) and b (8). 2 isn't greater than a or b, hence no updates.

3. Result: After processing all digits, largest a = 9 and second largest b = 8. The maximum product is 9 * 8 = 72.

Thus, the maximum product of any two digits in 29458 is 72.

Solution Implementation

Time and Space Complexity

The time complexity of the given code is O(\log n), where n is the input number. This is because in each iteration of the while loop, the number n is reduced by a factor of 10 (using divmod(n, 10)), and the loop continues until n becomes 0. Therefore, the number of iterations is proportional to the number of digits in n, which is O(\log n).

The space complexity of the code is O(1). This is because the algorithm uses a constant amount of space regardless of the size of the input. The space is used for storing the variables a, b, and x, and no additional data structures that grow with input size are used.