Problem Description

You are given an integer array nums containing positive integers. We define a function encrypt such that encrypt(x) replaces every digit in x with the largest digit in x. For example, encrypt(523) = 555 and encrypt(213) = 333.

Return the sum of encrypted elements.

Intuition

To approach this problem, we need to simulate the encryption process as described. The idea is to implement a function encrypt(x) which processes each integer x and replaces its digits with the largest digit found within that number.

The steps to encrypt a number x are:

1. Extract Each Digit: Break down the number x to its digits. This can be done using modulus and integer division by 10 in a loop until the number is reduced to zero.
2. Identify the Largest Digit: As we extract digits, keep track of the maximum digit found, denoted as mx.
3. Form the Encrypted Number: For each digit in the original number, replace it with mx. Construct this new number by appending mx repeatedly to form 111...1 of the same digit length as the original number. This is done by maintaining a base multiplier p (starting at 1 and forming numbers like 1, 11, 111, etc.).
4. Compute the Final Sum: Once each number in the input array is encrypted, sum all these encrypted values to get the solution.

This approach efficiently transforms each element and keeps track of transformations to provide the correct sum of encrypted numbers.

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Solution Approach

To solve this problem, we implement the function encrypt(x) which simulates the encryption process for each number in the array nums. The approach is as follows:

1. Define the Encrypt Function:

   def encrypt(x: int) -> int:
       mx = p = 0
       while x:
           x, v = divmod(x, 10)
           mx = max(mx, v)
           p = p * 10 + 1
       return mx * p

   • Extract Digits: The loop extracts each digit from x using divmod(x, 10), which divides x by 10 and provides both quotient and remainder. The remainder v is a digit from x.
   • Track Maximum Digit: mx is updated to be the maximum of itself and the current digit v.
   • Form Multiplier: p keeps record of the number of digits processed, forming numbers like 1, 11, 111, etc., which helps to reconstruct the encrypted number by multiplying mx.

2. Calculate the Sum of Encrypted Elements:

   return sum(encrypt(x) for x in nums)

   • Each number in nums is passed through encrypt(x) and the encrypted versions are summed up using the built-in sum function.

This algorithm effectively performs the encryption by transforming each digit as required and then accumulates the results, using basic arithmetic and iteration constructs.

Example Walkthrough

Let's illustrate the solution approach using a small example.

Suppose we have the integer array nums = [523, 213].

Step-by-step Encryption Process:

1. Encrypt the First Number: 523

   • Initialize mx = 0 and p = 0.
   • Extract digits from 523:
     • 523 % 10 = 3, update mx = max(0, 3) = 3, and update p = p * 10 + 1 = 1.
     • 52 % 10 = 2, update mx = max(3, 2) = 3, and update p = p * 10 + 1 = 11.
     • 5 % 10 = 5, update mx = max(3, 5) = 5, and update p = p * 10 + 1 = 111.
   • As 523 is reduced to 0, the largest digit mx is 5, and p is 111. Multiply to get the encrypted number: 5 * 111 = 555.

2. Encrypt the Second Number: 213

   • Initialize mx = 0 and p = 0.
   • Extract digits from 213:
     • 213 % 10 = 3, update mx = max(0, 3) = 3, and update p = p * 10 + 1 = 1.
     • 21 % 10 = 1, update mx = max(3, 1) = 3, and update p = p * 10 + 1 = 11.
     • 2 % 10 = 2, update mx = max(3, 2) = 3, and update p = p * 10 + 1 = 111.
   • As 213 is reduced to 0, the largest digit mx is 3, and p is 111. Multiply to get the encrypted number: 3 * 111 = 333.

3. Calculate the Sum of Encrypted Elements:

   • Encrypted numbers are [555, 333].
   • Sum of encrypted numbers is 555 + 333 = 888.

Thus, the sum of the encrypted elements of the array nums = [523, 213] is 888. This example illustrates the encryption process of each element followed by summation of the results to obtain the final answer.

Solution Implementation

Time and Space Complexity

The time complexity of the given code is O(n \times \log M), where n is the length of the nums list, and M is the maximum integer value in the list. This is because the encrypt function processes each digit of an integer, and the number of digits in the worst case is O(\log M). The outer loop iterates through each number in the list, leading to the overall complexity.

The space complexity is O(1). The function uses a constant amount of additional space, regardless of the size of the input list nums, because no data structures that grow with the input size are used.

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