Solution Explanation for Sum of Digits in the Minimum Number

The problem asks to find the sum of digits of the minimum number in an integer array and determine if the sum is odd or even. The solution returns 0 if the sum is odd, and 1 if the sum is even.

Approach:

1. Find the Minimum: First, we find the minimum element in the input array nums. This can be done efficiently using built-in functions like min() in Python or min_element() in C++.

2. Calculate the Sum of Digits: Next, we calculate the sum of the digits of the minimum number. This can be done using a loop, repeatedly extracting the last digit using the modulo operator (%) and integer division (// or /).

3. Check for Odd or Even: Finally, we determine if the sum of digits is odd or even. A bitwise AND operation (& 1) efficiently checks for oddness (result is 1 if odd, 0 if even). We use the XOR operator (^ 1) to invert the result, giving 1 for even and 0 for odd, as per the problem's requirements.

Code Implementation and Explanation (Python):

class Solution:
    def sumOfDigits(self, nums: List[int]) -> int:
        x = min(nums)  # Find the minimum number
        s = 0          # Initialize the sum of digits
        while x:        # Loop until x becomes 0
            s += x % 10  # Add the last digit to the sum
            x //= 10     # Remove the last digit
        return s & 1 ^ 1 # Return 1 if even, 0 if odd

Time and Space Complexity Analysis:

• Time Complexity: O(log(m)), where 'm' is the minimum number in the input array. The time complexity is dominated by the loop that calculates the sum of digits. The number of iterations in this loop is proportional to the number of digits in 'm', which is logarithmic with respect to 'm'. Finding the minimum using min() is O(n), where n is the length of the array, but this is overshadowed by the logarithmic time complexity of digit summation if m is relatively large.

• Space Complexity: O(1). The algorithm uses a constant amount of extra space regardless of the input size.

Code Implementation and Explanation (Java):

class Solution {
    public int sumOfDigits(int[] nums) {
        int x = 100; // Initialize with a large value to ensure correct minimum finding
        for (int v : nums) {
            x = Math.min(x, v);
        }
        int s = 0;
        for (; x > 0; x /= 10) {
            s += x % 10;
        }
        return s & 1 ^ 1;
    }
}

The Java code follows the same logic as the Python code. Note that Math.min() is used to find the minimum element, and the loop for summing digits uses integer division (/=) and the modulo operator (%).

Code Implementation and Explanation (C++):

class Solution {
public:
    int sumOfDigits(vector<int>& nums) {
        int x = *min_element(nums.begin(), nums.end()); //Efficiently finds minimum using STL
        int s = 0;
        for (; x > 0; x /= 10) {
            s += x % 10;
        }
        return s & 1 ^ 1;
    }
};

The C++ code utilizes min_element() from the Standard Template Library (STL) for efficient minimum finding. The rest of the logic remains the same.

Code Implementation and Explanation (Go):

func sumOfDigits(nums []int) int {
	s := 0
	for x := slices.Min(nums); x > 0; x /= 10 {
		s += x % 10
	}
	return s&1 ^ 1
}
 

Go's implementation leverages the slices.Min() function from the golang.org/x/exp/slices package to efficiently find the minimum value. The rest of the logic remains consistent with the other languages. Remember to import the necessary package: import "golang.org/x/exp/slices"

The time and space complexity analysis remains the same across all languages. The primary difference lies in the syntax and specific functions used for finding the minimum element.