Problem Description

You are given an integer array nums containing positive integers. You need to apply an encryption function to each element and return the sum of all encrypted values.

The encryption function encrypt(x) works by replacing every digit in the integer x with the largest digit found in x.

For example:

• encrypt(523) = 555 because the largest digit in 523 is 5, so all digits (5, 2, 3) are replaced with 5
• encrypt(213) = 333 because the largest digit in 213 is 3, so all digits (2, 1, 3) are replaced with 3

The task is to:

1. Apply the encryption function to each element in the array
2. Sum all the encrypted values
3. Return the total sum

The solution implements this by:

• Creating an encrypt function that finds the maximum digit in a number while tracking the number of digits
• For each number, it extracts digits using modulo 10 and integer division
• It keeps track of the maximum digit (mx) and builds a placeholder value (p) like 1, 11, 111, etc.
• The encrypted value is calculated as mx * p, which effectively replaces all digits with the maximum digit
• Finally, it returns the sum of all encrypted values using a generator expression

Intuition

The key insight is recognizing that encrypting a number means creating a new number with the same number of digits, where all digits are replaced by the maximum digit from the original number.

To achieve this transformation, we need two pieces of information:

1. The maximum digit in the number
2. How many digits the number has

Once we have these, we can construct the encrypted number. For instance, if we know the maximum digit is 7 and the number has 3 digits, the encrypted result would be 777.

The clever part of the solution is how it builds the result. Instead of converting to a string or counting digits separately, we can extract digits and count them in a single pass through the number. As we process each digit through divmod(x, 10):

• We compare and update the maximum digit
• We build a "pattern" number p that grows as 1, 11, 111, 1111, etc.

This pattern number p serves as a multiplier. When we multiply the maximum digit by this pattern (like 5 * 111 = 555), we get exactly what we want - the same digit repeated for the correct number of positions.

The beauty of this approach is its efficiency: we traverse the digits only once, avoid string conversions, and use simple arithmetic operations to construct the encrypted value directly.

Learn more about Math patterns.

Solution Approach

The solution implements a simulation approach with a helper function encrypt(x) that processes each number in the array.

Implementation walkthrough:

1. Define the encrypt function: The function takes an integer x and returns its encrypted form.

2. Initialize variables:

   • mx = 0: Tracks the maximum digit found in the number
   • p = 0: Builds the pattern number (1, 11, 111, ...)

3. Extract and process digits: Using a while loop that continues until x becomes 0:

   • x, v = divmod(x, 10): This simultaneously gets the last digit v and removes it from x
   • mx = max(mx, v): Update the maximum digit if current digit is larger
   • p = p * 10 + 1: Build the pattern by shifting left and adding 1

4. Calculate encrypted value: Return mx * p

   • This multiplication creates a number where all digits are the maximum digit
   • For example: if mx = 5 and p = 111, then 5 * 111 = 555

5. Sum all encrypted values: Use a generator expression sum(encrypt(x) for x in nums) to:

   • Apply the encrypt function to each element in the array
   • Sum all the encrypted results
   • Return the final sum

Example trace for encrypt(523):

• Initial: x = 523, mx = 0, p = 0
• Iteration 1: v = 3, x = 52, mx = 3, p = 1
• Iteration 2: v = 2, x = 5, mx = 3, p = 11
• Iteration 3: v = 5, x = 0, mx = 5, p = 111
• Return: 5 * 111 = 555

The time complexity is O(n * m) where n is the length of the array and m is the average number of digits per number. The space complexity is O(1) as we only use a constant amount of extra space.

Example Walkthrough

Let's walk through the solution with nums = [523, 213, 82].

Step 1: Encrypt first number (523)

• Start with x = 523, mx = 0, p = 0
• Extract rightmost digit: 523 ÷ 10 = 52 remainder 3
  • Update: x = 52, v = 3, mx = 3, p = 1
• Extract next digit: 52 ÷ 10 = 5 remainder 2
  • Update: x = 5, v = 2, mx = 3, p = 11
• Extract last digit: 5 ÷ 10 = 0 remainder 5
  • Update: x = 0, v = 5, mx = 5, p = 111
• Calculate encrypted value: 5 × 111 = 555

Step 2: Encrypt second number (213)

• Start with x = 213, mx = 0, p = 0
• Extract rightmost digit: 213 ÷ 10 = 21 remainder 3
  • Update: x = 21, v = 3, mx = 3, p = 1
• Extract next digit: 21 ÷ 10 = 2 remainder 1
  • Update: x = 2, v = 1, mx = 3, p = 11
• Extract last digit: 2 ÷ 10 = 0 remainder 2
  • Update: x = 0, v = 2, mx = 3, p = 111
• Calculate encrypted value: 3 × 111 = 333

Step 3: Encrypt third number (82)

• Start with x = 82, mx = 0, p = 0
• Extract rightmost digit: 82 ÷ 10 = 8 remainder 2
  • Update: x = 8, v = 2, mx = 2, p = 1
• Extract last digit: 8 ÷ 10 = 0 remainder 8
  • Update: x = 0, v = 8, mx = 8, p = 11
• Calculate encrypted value: 8 × 11 = 88

Step 4: Sum all encrypted values

• Total = 555 + 333 + 88 = 976

The key insight is how the pattern number p (1, 11, 111...) acts as a multiplier. When we multiply the maximum digit by this pattern, we get a number with all digits replaced by that maximum digit. The solution efficiently builds this pattern while finding the maximum digit in a single pass through each number.

Solution Implementation

Time and Space Complexity

Time Complexity: O(n × log M)

The algorithm iterates through each number in the nums array (n elements). For each number, the encrypt function processes the digits by repeatedly dividing by 10 until the number becomes 0. The number of iterations in the while loop equals the number of digits in the number, which is O(log₁₀ x) for a number x. In the worst case, x can be the maximum value M in the array, giving us O(log M) operations per number. Therefore, the total time complexity is O(n × log M).

Space Complexity: O(1)

The algorithm uses only a constant amount of extra space. The encrypt function uses a fixed number of variables (mx, p, x, v) regardless of the input size. The sum operation and generator expression don't create additional data structures that scale with input size, maintaining constant space usage throughout the execution.

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

Common Pitfalls

1. Handling Zero as Input

The current implementation doesn't handle the edge case where the input number is 0. When number = 0, the while loop never executes, resulting in place_value_pattern = 0 and returning 0 * 0 = 0. However, the correct encryption of 0 should still be 0 (single digit with value 0).

Fix:

def encrypt(number: int) -> int:
    # Handle zero as a special case
    if number == 0:
        return 0
  
    max_digit = 0
    place_value_pattern = 0
  
    while number > 0:
        number, current_digit = divmod(number, 10)
        max_digit = max(max_digit, current_digit)
        place_value_pattern = place_value_pattern * 10 + 1
  
    return max_digit * place_value_pattern

2. Integer Overflow for Large Numbers

When dealing with very large numbers or arrays with many large elements, the multiplication max_digit * place_value_pattern and the final sum could potentially cause integer overflow in languages with fixed integer sizes. While Python handles arbitrary precision integers automatically, this could be an issue when porting to other languages.

Mitigation Strategy:

• Be aware of integer limits in the target language
• Consider using long/BigInteger types where necessary
• Add validation for maximum input sizes if needed

3. Inefficient String Conversion Alternative

A common alternative approach might be converting the number to a string to find digits:

def encrypt(number: int) -> int:
    str_num = str(number)
    max_digit = max(int(d) for d in str_num)
    return int(str(max_digit) * len(str_num))

While this works and is more readable, it's less efficient due to string conversions and allocations. The mathematical approach in the original solution is more performant.

4. Forgetting to Handle Negative Numbers

The problem states "positive integers", but if the constraints change or input validation is needed, negative numbers would break the current logic. The modulo operation with negative numbers behaves differently, and the encryption concept would need redefinition.

Defensive Programming:

def encrypt(number: int) -> int:
    # Add input validation if needed
    if number < 0:
        raise ValueError("Encryption only supports non-negative integers")
  
    # Rest of the implementation...