Problem Description

You are given an integer array nums. Your task is to:

1. Find the minimum integer in the array
2. Calculate the sum of all digits in this minimum integer
3. Return 0 if this digit sum is odd, or 1 if the digit sum is even

For example:

• If nums = [34, 23, 1, 24, 75, 33, 54, 8], the minimum is 1

  • Sum of digits of 1 is 1 (which is odd)
  • Return 0

• If nums = [99, 77, 33, 66, 55], the minimum is 33

  • Sum of digits of 33 is 3 + 3 = 6 (which is even)
  • Return 1

The solution finds the minimum value using min(nums), then iterates through each digit by repeatedly taking the remainder when divided by 10 (x % 10) and integer division by 10 (x //= 10). The final expression s & 1 ^ 1 checks if the sum is even (returns 1) or odd (returns 0) by using bitwise operations - s & 1 gives 0 for even and 1 for odd, then XOR with 1 flips the result.

Intuition

The problem breaks down into two main tasks: finding the minimum value and then checking if its digit sum has a specific parity (odd or even).

For finding the minimum, we can simply use Python's built-in min() function since we need to examine all elements anyway.

For calculating the digit sum, we need to extract each digit from the number. The standard approach is to use the modulo operation - any number modulo 10 gives us its last digit. We can repeatedly extract the last digit with x % 10, add it to our sum, then remove it by doing integer division x //= 10. This process continues until the number becomes 0.

The interesting part is determining whether to return 0 or 1 based on the parity. We need to return 0 for odd sums and 1 for even sums. This is the opposite of what we'd get from a simple parity check. Using s % 2 would give us 1 for odd and 0 for even. To flip this result, we could use 1 - (s % 2) or utilize bitwise operations.

The solution uses s & 1 ^ 1 which is elegant: s & 1 performs the same function as s % 2 (extracting the least significant bit, which tells us if the number is odd or even), and then XOR-ing with 1 flips the bit - turning 0 to 1 and 1 to 0. This gives us exactly what we need: 1 for even digit sums and 0 for odd digit sums.

Learn more about Math patterns.

Solution Approach

The implementation follows a straightforward approach with three main steps:

1. Find the minimum element: Use Python's built-in min() function to find the smallest integer in the array nums. This operation has O(n) time complexity where n is the length of the array.

   x = min(nums)

2. Calculate the sum of digits: Extract and sum all digits from the minimum value using a while loop:

   s = 0
   while x:
       s += x % 10    # Extract the last digit and add to sum
       x //= 10       # Remove the last digit

   • x % 10 extracts the rightmost digit (units place)
   • x //= 10 performs integer division, effectively removing the rightmost digit
   • The loop continues until x becomes 0 (all digits processed)

3. Determine the return value based on parity: Use bitwise operations to check if the digit sum is odd or even:

   return s & 1 ^ 1

   • s & 1 performs a bitwise AND with 1, which extracts the least significant bit
     • Results in 1 if s is odd, 0 if s is even
   • ^ 1 performs XOR with 1, which flips the bit
     • Changes 0 to 1 (even sum returns 1)
     • Changes 1 to 0 (odd sum returns 0)

Time Complexity: O(n + log m) where n is the length of the array and m is the value of the minimum element (for digit extraction)

Space Complexity: O(1) as we only use a constant amount of extra space

Example Walkthrough

Let's walk through a small example with nums = [42, 15, 38].

Step 1: Find the minimum element

• Compare all elements: 42, 15, 38
• The minimum is 15
• x = 15

Step 2: Calculate the sum of digits of 15

• Initialize s = 0
• First iteration:
  • x = 15, extract digit: 15 % 10 = 5
  • Add to sum: s = 0 + 5 = 5
  • Remove digit: x = 15 // 10 = 1
• Second iteration:
  • x = 1, extract digit: 1 % 10 = 1
  • Add to sum: s = 5 + 1 = 6
  • Remove digit: x = 1 // 10 = 0
• Loop ends since x = 0
• Final digit sum: s = 6

Step 3: Determine return value based on parity

• We have s = 6 (which is even)
• Calculate s & 1:
  • Binary of 6 is 110
  • 110 & 001 = 0 (6 is even, so least significant bit is 0)
• Apply XOR with 1:
  • 0 ^ 1 = 1
• Return 1

Since the digit sum (6) is even, we return 1 as expected.

Let's verify with another quick example where nums = [20]:

• Minimum is 20
• Digit sum: 2 + 0 = 2 (even)
• 2 & 1 = 0 (binary 10 & 01 = 0)
• 0 ^ 1 = 1
• Return 1 ✓

Solution Implementation

Time and Space Complexity

Time Complexity: O(n + log(m)) where n is the length of the input array and m is the value of the minimum element.

• Finding the minimum element using min(nums) takes O(n) time as it needs to traverse through all elements.
• The while loop extracts digits from the minimum number, which runs O(log₁₀(m)) times (the number of digits in base 10).
• Each operation inside the while loop (modulo, integer division, addition) takes O(1) time.
• The final bitwise operations s & 1 ^ 1 take O(1) time.
• Overall: O(n) + O(log(m)) + O(1) = O(n + log(m))

Space Complexity: O(1)

• The algorithm uses only a constant amount of extra space.
• Variables x and s are the only additional storage used, regardless of input size.
• No recursive calls or additional data structures are employed.

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

Common Pitfalls

1. Handling Negative Numbers Incorrectly

The most significant pitfall in this solution is the improper handling of negative numbers. When the minimum value is negative, the digit extraction logic fails because:

• Negative numbers in Python maintain their sign through modulo operations
• -45 % 10 returns 5 (correct digit) but -45 // 10 returns -5 (not -4)
• This causes the while loop to potentially run indefinitely or produce incorrect results

Example Problem Case:

nums = [-45, 23, 67]  # minimum is -45
# Expected: sum of digits = 4 + 5 = 9 (odd), return 0
# Actual: The loop may not terminate correctly

Solution: Always work with the absolute value of the minimum number:

def sumOfDigits(self, nums: List[int]) -> int:
    min_value = min(nums)
    # Convert to absolute value for digit extraction
    min_value = abs(min_value)

    digit_sum = 0
    while min_value > 0:
        digit_sum += min_value % 10
        min_value //= 10

    return (digit_sum & 1) ^ 1

2. Integer Overflow in Other Languages

While not an issue in Python (which handles arbitrary precision integers), implementing this in languages like Java or C++ requires consideration of integer overflow when dealing with very large arrays or values.

3. Empty Array Input

If the input array is empty, min(nums) will raise a ValueError. Although this might be outside the problem constraints, defensive programming suggests checking:

def sumOfDigits(self, nums: List[int]) -> int:
    if not nums:
        return 1  # or handle according to requirements

    min_value = abs(min(nums))
    # ... rest of the code

4. Misunderstanding the Bitwise Operations

Developers might incorrectly assume operator precedence or misunderstand the XOR operation:

• Remember that & has higher precedence than ^
• The expression s & 1 ^ 1 is evaluated as (s & 1) ^ 1, not s & (1 ^ 1)
• For clarity, consider using parentheses or a more explicit approach:

# More readable alternative:
return 1 if digit_sum % 2 == 0 else 0