Sum of Digits in the Minimum Number

Here is a simple solution for the problem. The steps are as follows:

  1. Find the minimum number from the list nums.
  2. Convert the minimum number to a string to separate the digits.
  3. Calculate the sum of these digits.
  4. If the sum is even, return 1. Otherwise, return 0.

Here is the Python code:

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class Solution:
    def sumOfDigits(self, nums: List[int]) -> int:
        min_num = min(nums)  # step 1
        digits = [int(digit) for digit in str(min_num)]  # step 2
        digit_sum = sum(digits)  # step 3
        return 1 if digit_sum % 2 == 0 else 0  # step 4

This solution works because it follows the problem’s instructions literally. It first identifies the smallest number in the array, then breaks this number into its constituent digits. By summing these digits and checking if the result is even or odd, it can return the correct value.

Alternate Solution

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from typing import List

class Solution:
    def sumOfDigits(self, A: List[int]) -> int:
        min_val = min(A)
        ans = 0
        while min_val > 0:
            ans += min_val % 10
            min_val //= 10
        return ans % 2 == 0
  1. It finds the minimum value in the list A.
  2. It initializes a variable ans to 0.
  3. In a loop, it adds the last digit of min_val to ans and then removes that digit from min_val.
  4. Finally, it returns 1 if ans is even, and 0 if ans is odd.

This code involves the use of several high-level coding constructs:

  1. List Minima: min(A) is used to find the smallest value in the list. The min function is a built-in Python function that returns the smallest item in an iterable (like a list) or the smallest of two or more arguments.

  2. Loops: The while loop is used here to repeatedly execute a block of code as long as a certain condition is met. In this case, the block of code inside the while loop executes as long as min_val > 0.

  3. Arithmetic Operations: Several arithmetic operations are used in this solution:

    • The modulus operation (min_val % 10) is used to get the last digit of min_val.
    • The integer division operation (min_val //= 10) is used to remove the last digit from min_val.
    • The subtraction operation (1 - ans % 2) is used to return 1 if ans is even and 0 if ans is odd.

  4. Assignment Operation: The += operator is a shorthand for ans = ans + min_val % 10, which adds the value of min_val % 10 to the current value of ans and assigns the result back to ans.

  5. Conditional Expressions: The expression ans % 2 is a condition that checks if ans is even or odd. If ans is even, ans % 2 equals 0, and if ans is odd, ans % 2 equals 1. Therefore, 1 - ans % 2 equals 1 if ans is even and 0 if ans is odd.

  6. Reduction: The problem initially reduces the given array to find the minimum number using min(A). After finding the minimum number, a further reduction process takes place where the problem is narrowed down to working with the digits of this minimum number. The while loop keeps running as long as the minimum number is greater than 0, each time reducing the number by removing its last digit using min //= 10.

  7. Transformation: The value of min is transformed on each iteration of the loop. This transformation involves digit extraction, which is performed using % and // operations.

  8. Accumulation: Inside the loop, we accumulate the sum of the digits of the minimum number in ans using an addition operation.

  9. Mapping: The sum of the digits (ans) is then mapped to the problem’s required output format (0 if the sum is odd, 1 if the sum is even). This mapping is achieved by taking ans modulo 2 to find out if it’s even or odd, and then subtracting the result from 1.

  10. Output: The result of the mapping operation is the final output, representing whether the sum of the digits of the minimum number is even or odd.

The extraction of digits that need to be added is performed in the while loop within the function.

In particular, the line ans += min % 10 is responsible for this extraction. Here’s how it works:

  • The modulus operation % returns the remainder of the division of min by 10. Given that min is an integer, min % 10 will be the last digit of min.

  • This digit is then added to ans, accumulating the sum of the digits of min.

  • After the digit is added to ans, min is divided by 10 using integer division min //= 10, which removes the last digit from min. This updated min then goes through the next iteration of the while loop.

This process continues until all digits of min have been extracted and added to ans.

The expression 1 - ans % 2 determines whether the sum of the digits (ans) is odd or even.

  • If ans is even, then ans % 2 will equal 0 (since even numbers are divisible by 2 with no remainder). So, 1 - ans % 2 will be 1 - 0, which equals 1.

  • If ans is odd, then ans % 2 will equal 1 (since odd numbers leave a remainder of 1 when divided by 2). In this case, 1 - ans % 2 will be 1 - 1, which equals 0.

So, if the sum of the digits is even, the function returns 1. If the sum of the digits is odd, the function returns 0.