Problem Description

You are given an integer n. Your task is to:

1. Calculate n² (n squared) and convert it to hexadecimal (base-16)
2. Calculate n³ (n cubed) and convert it to hexatrigesimal (base-36)
3. Concatenate these two string representations and return the result

Number System Details:

• Hexadecimal (base-16): Uses digits 0-9 and uppercase letters A-F to represent values 0 to 15

  • For example: 10 in decimal = A in hexadecimal, 15 in decimal = F in hexadecimal

• Hexatrigesimal (base-36): Uses digits 0-9 and uppercase letters A-Z to represent values 0 to 35

  • For example: 10 in decimal = A in base-36, 35 in decimal = Z in base-36

Example walkthrough:

If n = 4:

• n² = 16 → hexadecimal representation is "10"
• n³ = 64 → base-36 representation is "1S" (since 64 = 1×36 + 28, and 28 is represented as 'S')
• Result: "10" + "1S" = "101S"

The solution implements a helper function f(x, k) that converts any integer x to its representation in base k. It works by:

1. Repeatedly finding the remainder when dividing by the base
2. Converting each remainder to its appropriate character (digit or letter)
3. Building the result string from least significant to most significant digit
4. Reversing the result to get the correct order

The main function then applies this conversion to n² with base 16 and n³ with base 36, concatenating the results.

Intuition

The core challenge here is converting numbers to different base systems. When we think about how number systems work, any number in base k can be broken down into powers of k.

For example, the number 234 in base 10 means: 2×10² + 3×10¹ + 4×10⁰

To convert a decimal number to any base, we can use the division method:

• Divide the number by the base and keep track of the remainder
• The remainder gives us the rightmost digit in the new base
• Continue with the quotient until it becomes 0
• The remainders, read in reverse order, give us the number in the new base

Let's trace through converting 16 to hexadecimal (base 16):

• 16 ÷ 16 = 1 remainder 0
• 1 ÷ 16 = 0 remainder 1
• Reading remainders in reverse: "10"

The tricky part is handling bases larger than 10, where we need letters. When the remainder is 10 or more:

• In hexadecimal: 10→A, 11→B, ..., 15→F
• In base-36: 10→A, 11→B, ..., 35→Z

This mapping can be achieved using ASCII values: if remainder v > 9, we use chr(ord('A') + v - 10) to get the corresponding letter.

Since both conversions follow the same pattern (just with different bases), we can create a single function f(x, k) that handles any base conversion. This avoids code duplication and makes the solution cleaner.

The final step is straightforward: compute n² and n³, convert them using our function with the appropriate bases (16 and 36), and concatenate the results.

Learn more about Math patterns.

Solution Approach

The solution implements a simulation approach with a helper function for base conversion.

Helper Function f(x, k) - Base Conversion Algorithm:

The function converts an integer x to its string representation in base k:

1. Initialize an empty list res to store the digits/characters in reverse order
2. Extract digits using modulo operation:
   • While x > 0:
     • Calculate remainder: v = x % k
     • If v <= 9: append the digit as a string str(v)
     • If v > 9: convert to letter using chr(ord("A") + v - 10)
       • This maps 10→'A', 11→'B', and so on
     • Update x by integer division: x //= k
3. Reverse and join the result: "".join(res[::-1])
   • We reverse because we built the number from least to most significant digit

Main Function Flow:

1. Calculate the powers:

   • x = n ** 2 (n squared)
   • y = n ** 3 (n cubed)

2. Convert to respective bases:

   • Convert x to hexadecimal: f(x, 16)
   • Convert y to base-36: f(y, 36)

3. Concatenate and return: Simply concatenate the two strings

Example Trace for n = 4:

• x = 16, y = 64
• Converting 16 to hex (base 16):
  • 16 % 16 = 0 → append '0'
  • 16 // 16 = 1
  • 1 % 16 = 1 → append '1'
  • 1 // 16 = 0 → stop
  • Reverse: "10"
• Converting 64 to base-36:
  • 64 % 36 = 28 → 28 > 9, so append chr(65 + 28 - 10) = chr(83) = 'S'
  • 64 // 36 = 1
  • 1 % 36 = 1 → append '1'
  • 1 // 36 = 0 → stop
  • Reverse: "1S"
• Result: "10" + "1S" = "101S"

The time complexity is O(log x + log y) where x = n² and y = n³, as we process each digit once. The space complexity is also O(log x + log y) for storing the result strings.

Example Walkthrough

Let's walk through the solution with n = 12:

Step 1: Calculate the powers

• n² = 12² = 144
• n³ = 12³ = 1728

Step 2: Convert 144 to hexadecimal (base-16)

Using the division method with our helper function f(144, 16):

• 144 ÷ 16 = 9 remainder 0 → append '0'
• 9 ÷ 16 = 0 remainder 9 → append '9'
• Stop (quotient is 0)
• Reverse the digits: "90"

Step 3: Convert 1728 to base-36

Using our helper function f(1728, 36):

• 1728 ÷ 36 = 48 remainder 0 → append '0'
• 48 ÷ 36 = 1 remainder 12 → Since 12 > 9, convert to letter: chr(65 + 12 - 10) = chr(67) = 'C' → append 'C'
• 1 ÷ 36 = 0 remainder 1 → append '1'
• Stop (quotient is 0)
• Reverse the characters: "1C0"

Step 4: Concatenate the results

• Hexadecimal of 144: "90"
• Base-36 of 1728: "1C0"
• Final result: "90" + "1C0" = "901C0"

The key insight is that both conversions follow the same pattern - we repeatedly divide by the base and collect remainders, converting values above 9 to letters. By implementing this once in the helper function f(x, k), we can handle both base-16 and base-36 conversions efficiently.

Solution Implementation

Time and Space Complexity

The time complexity is O(log n). This is because:

• Computing n² and n³ takes constant time
• The function f(x, k) performs divisions by base k until x becomes 0
• For f(n², 16): The number of iterations is O(log₁₆(n²)) = O(2 * log₁₆(n)) = O(log n)
• For f(n³, 36): The number of iterations is O(log₃₆(n³)) = O(3 * log₃₆(n)) = O(log n)
• Since both conversions are O(log n) and they run sequentially, the overall time complexity is O(log n)

The space complexity is O(log n), not O(1) as stated in the reference answer. This is because:

• The res list in function f stores the digits of the number in the target base
• For n² in base 16, we need O(log₁₆(n²)) = O(log n) digits
• For n³ in base 36, we need O(log₃₆(n³)) = O(log n) digits
• The final string concatenation also requires O(log n) space for the result
• Therefore, the space complexity is O(log n)

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

Common Pitfalls

1. Forgetting to Handle Zero Edge Case

The most common pitfall is not properly handling when n = 0. Without a special check, the base conversion function would return an empty string since the while loop condition number > 0 would never execute.

Problem Code:

def convert_to_base(number: int, base: int) -> str:
    result = []
    while number > 0:  # This loop never runs when number = 0
        remainder = number % base
        # ... conversion logic
        number //= base
    return ''.join(result[::-1])  # Returns "" for number = 0

Solution: Add an explicit check at the beginning of the conversion function:

def convert_to_base(number: int, base: int) -> str:
    if number == 0:
        return "0"
    # ... rest of the conversion logic

2. Using Lowercase Letters Instead of Uppercase

The problem specifically requires uppercase letters (A-Z) for digits above 9. A common mistake is using lowercase letters or forgetting to specify the case.

Problem Code:

# Wrong: produces lowercase letters
result.append(chr(ord('a') + remainder - 10))

Solution: Always use uppercase 'A' as the starting point:

result.append(chr(ord('A') + remainder - 10))

3. Building the Result in Wrong Order

When converting to a different base, digits are extracted from least significant to most significant. Forgetting to reverse the result leads to backwards numbers.

Problem Code:

def convert_to_base(number: int, base: int) -> str:
    result = []
    while number > 0:
        # ... append digits to result
        number //= base
    return ''.join(result)  # Wrong: not reversed!

Solution: Remember to reverse the result before joining:

return ''.join(result[::-1])

4. Off-by-One Error in Letter Mapping

A subtle but critical error occurs when mapping numbers to letters. The mapping should be: 10→'A', 11→'B', ..., 35→'Z'.

Problem Code:

# Wrong: off-by-one error
if remainder >= 10:
    result.append(chr(ord('A') + remainder - 9))  # Should be -10, not -9

Solution: Use the correct offset of -10:

if remainder > 9:
    result.append(chr(ord('A') + remainder - 10))

5. Integer Overflow Concerns (Language-Specific)

While Python handles arbitrary precision integers automatically, in other languages like C++ or Java, calculating n³ for large values of n could cause integer overflow.

Solution for Other Languages:

• Use appropriate data types (long long in C++, BigInteger in Java)
• Add input validation if needed
• Consider the maximum value of n based on problem constraints

6. Incorrect Base Validation

Not validating that the base is within the supported range (2-36) could lead to incorrect results or runtime errors.

Solution: Add base validation:

def convert_to_base(number: int, base: int) -> str:
    if base < 2 or base > 36:
        raise ValueError(f"Base must be between 2 and 36, got {base}")
    # ... rest of the function