Problem Description

This problem asks you to convert a given integer into its base 7 representation and return it as a string.

In base 7 numbering system, we only use digits 0 through 6 to represent numbers. Each position represents a power of 7, similar to how base 10 uses powers of 10.

For example:

• The number 100 in base 10 would be 202 in base 7 because: 100 = 2×7² + 0×7¹ + 2×7⁰ = 2×49 + 0×7 + 2×1
• The number 7 in base 10 would be 10 in base 7
• The number -7 in base 10 would be -10 in base 7

The solution handles three cases:

1. If num is 0, it directly returns '0'
2. If num is negative, it recursively converts the positive value and prepends a minus sign
3. For positive numbers, it repeatedly divides by 7, collecting remainders which form the digits of the base 7 representation (in reverse order), then reverses them to get the final result

The algorithm works by:

• Taking the remainder when dividing by 7 (this gives the rightmost digit in base 7)
• Integer dividing by 7 to shift to the next position
• Repeating until the number becomes 0
• Reversing the collected digits since they were gathered from right to left

Intuition

The key insight comes from understanding how number systems work. In any base system, a number can be broken down into digits by repeatedly dividing by the base and keeping track of remainders.

Think about how we normally extract digits from a decimal number. To get the last digit of 123, we can do 123 % 10 = 3. To get the second-to-last digit, we first remove the last digit by doing 123 // 10 = 12, then take 12 % 10 = 2. This same principle applies to any base conversion.

When converting to base 7, each remainder when dividing by 7 tells us what digit should appear at that position. The reason we use modulo (%) operation is that it gives us exactly the value that "doesn't fit" into groups of 7, which becomes our digit for that position.

For example, if we have 100:

• 100 % 7 = 2 (this is our rightmost digit)
• 100 // 7 = 14, then 14 % 7 = 0 (this is our middle digit)
• 14 // 7 = 2, then 2 % 7 = 2 (this is our leftmost digit)
• 2 // 7 = 0 (we stop here)

Reading the remainders in reverse gives us 202 in base 7.

The handling of negative numbers is straightforward - we can convert the positive version and just add a negative sign, since the digit representation itself follows the same pattern regardless of sign. The special case for zero is needed because the while loop wouldn't execute at all for zero, so we handle it explicitly.

Learn more about Math patterns.

Solution Approach

The implementation uses a recursive approach combined with iterative digit extraction:

1. Base Cases Handling:

   • If num == 0, immediately return '0' since zero is represented as '0' in any base
   • If num < 0, recursively call the function with the positive value -num and prepend a minus sign to the result

2. Digit Extraction Process:

   • Initialize an empty list ans to collect the base 7 digits
   • Use a while loop that continues as long as num > 0
   • In each iteration:
     • Calculate num % 7 to get the current rightmost digit in base 7
     • Convert this digit to string and append to ans
     • Update num to num // 7 to remove the processed digit and shift to the next position

3. Result Construction:

   • The digits collected in ans are in reverse order (rightmost digit first)
   • Use ans[::-1] to reverse the list
   • Join all digits with ''.join() to form the final string

Example Walkthrough with num = 100:

• First iteration: 100 % 7 = 2, append '2', num = 100 // 7 = 14
• Second iteration: 14 % 7 = 0, append '0', num = 14 // 7 = 2
• Third iteration: 2 % 7 = 2, append '2', num = 2 // 7 = 0
• Loop ends, ans = ['2', '0', '2']
• Reverse and join: '202'

The algorithm has O(log₇n) time complexity since we divide by 7 in each iteration, and O(log₇n) space complexity for storing the digits.

Example Walkthrough

Let's walk through converting the number 50 from base 10 to base 7.

Step-by-step process:

Starting with num = 50 and an empty list ans = []:

Iteration 1:

• Calculate remainder: 50 % 7 = 1
• This gives us the rightmost digit in base 7
• Append '1' to ans: ans = ['1']
• Update num: 50 // 7 = 7

Iteration 2:

• Calculate remainder: 7 % 7 = 0
• This gives us the next digit (moving left)
• Append '0' to ans: ans = ['1', '0']
• Update num: 7 // 7 = 1

Iteration 3:

• Calculate remainder: 1 % 7 = 1
• This gives us the leftmost digit
• Append '1' to ans: ans = ['1', '0', '1']
• Update num: 1 // 7 = 0

Final Step:

• Since num = 0, the loop ends
• We have ans = ['1', '0', '1'] (collected from right to left)
• Reverse the list: ['1', '0', '1']
• Join into string: '101'

Verification: 50 = 1×7² + 0×7¹ + 1×7⁰ = 1×49 + 0×7 + 1×1 = 49 + 0 + 1 = 50 ✓

Therefore, 50 in base 10 equals 101 in base 7.

Solution Implementation

Time and Space Complexity

Time Complexity: O(log₇(n)) where n is the absolute value of the input number.

The algorithm repeatedly divides the number by 7 until it becomes 0. The number of iterations required is equal to the number of digits in the base-7 representation, which is ⌊log₇(n)⌋ + 1. Each iteration performs constant time operations (modulo, integer division, and appending to a list), so the overall time complexity is O(log₇(n)). The final reversal operation ans[::-1] takes O(log₇(n)) time as well, but this doesn't change the overall complexity.

Space Complexity: O(log₇(n)) where n is the absolute value of the input number.

The space is primarily used by:

1. The ans list which stores the digits of the base-7 representation. The maximum number of elements in this list is ⌊log₇(n)⌋ + 1.
2. The recursive call stack in the case of negative numbers adds at most one additional frame.
3. The string created by ''.join(ans[::-1]) also takes O(log₇(n)) space.

Therefore, the overall space complexity is O(log₇(n)).

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

Common Pitfalls

1. Forgetting to Handle the Zero Case

One of the most common mistakes is not properly handling when num = 0. Without the special case check, the while loop while num > 0 would never execute, resulting in an empty string being returned instead of '0'.

Incorrect Implementation:

def convertToBase7(self, num: int) -> str:
    if num < 0:
        return '-' + self.convertToBase7(-num)
  
    digits = []
    while num > 0:  # This loop never runs when num = 0
        digits.append(str(num % 7))
        num //= 7
  
    return ''.join(reversed(digits))  # Returns '' for num = 0

Solution: Always include the zero check at the beginning of the function before any other logic.

2. Forgetting to Reverse the Digits

During the extraction process, digits are collected from least significant to most significant (right to left). A common error is forgetting to reverse them before joining.

Incorrect Implementation:

def convertToBase7(self, num: int) -> str:
    if num == 0:
        return '0'
    if num < 0:
        return '-' + self.convertToBase7(-num)
  
    digits = []
    while num > 0:
        digits.append(str(num % 7))
        num //= 7
  
    return ''.join(digits)  # Wrong! This gives reversed result
    # For num = 100, this returns '202' instead of '202' (happens to be same)
    # For num = 10, this returns '31' instead of '13'

Solution: Always reverse the collected digits using reversed(digits) or digits[::-1] before joining them.

3. Integer Overflow with Minimum Integer Value

In languages with fixed integer sizes (like Java or C++), handling Integer.MIN_VALUE can cause overflow when negating it, since the positive equivalent doesn't fit in the integer range. While Python handles arbitrary precision integers, this is still worth noting for implementations in other languages.

Problematic Pattern in Other Languages:

// In Java, this could overflow
if (num < 0) {
    return "-" + convertToBase7(-num); // -Integer.MIN_VALUE causes overflow
}

Solution: For languages with fixed integer sizes, handle the conversion using unsigned operations or convert to a larger data type first. In Python, this isn't an issue due to arbitrary precision integers.

4. Using String Concatenation in Loop

While not incorrect, using string concatenation in the loop instead of collecting in a list can be inefficient.

Inefficient Implementation:

def convertToBase7(self, num: int) -> str:
    if num == 0:
        return '0'
    if num < 0:
        return '-' + self.convertToBase7(-num)
  
    result = ''
    while num > 0:
        result = str(num % 7) + result  # String concatenation is O(n) each time
        num //= 7
  
    return result

Solution: Use a list to collect digits and join them at the end for O(n) total time complexity instead of O(n²) from repeated string concatenation.