Sum of Digits in the Minimum Number
Here is a simple solution for the problem. The steps are as follows:
 Find the minimum number from the list
nums
.  Convert the minimum number to a string to separate the digits.
 Calculate the sum of these digits.
 If the sum is even, return 1. Otherwise, return 0.
Here is the Python code:


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


 It finds the minimum value in the list A.
 It initializes a variable
ans
to 0.  In a loop, it adds the last digit of
min_val
toans
and then removes that digit frommin_val
.  Finally, it returns 1 if
ans
is even, and 0 ifans
is odd.
This code involves the use of several highlevel coding constructs:
List Minima:
min(A)
is used to find the smallest value in the list. Themin
function is a builtin Python function that returns the smallest item in an iterable (like a list) or the smallest of two or more arguments.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 thewhile
loop executes as long asmin_val > 0
.Arithmetic Operations: Several arithmetic operations are used in this solution:
 The modulus operation (
min_val % 10
) is used to get the last digit ofmin_val
.  The integer division operation (
min_val //= 10
) is used to remove the last digit frommin_val
.  The subtraction operation (
1  ans % 2
) is used to return 1 ifans
is even and 0 ifans
is odd.
 The modulus operation (
Assignment Operation: The
+=
operator is a shorthand forans = ans + min_val % 10
, which adds the value ofmin_val % 10
to the current value ofans
and assigns the result back toans
.Conditional Expressions: The expression
ans % 2
is a condition that checks ifans
is even or odd. Ifans
is even,ans % 2
equals 0, and ifans
is odd,ans % 2
equals 1. Therefore,1  ans % 2
equals 1 ifans
is even and 0 ifans
is odd.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. Thewhile
loop keeps running as long as the minimum number is greater than 0, each time reducing the number by removing its last digit usingmin //= 10
.Transformation: The value of
min
is transformed on each iteration of the loop. This transformation involves digit extraction, which is performed using%
and//
operations.Accumulation: Inside the loop, we accumulate the sum of the digits of the minimum number in
ans
using an addition operation.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 takingans
modulo 2 to find out if it’s even or odd, and then subtracting the result from 1.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 ofmin
by 10. Given thatmin
is an integer,min % 10
will be the last digit ofmin
.This digit is then added to
ans
, accumulating the sum of the digits ofmin
.After the digit is added to
ans
,min
is divided by 10 using integer divisionmin //= 10
, which removes the last digit frommin
. This updatedmin
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, thenans % 2
will equal 0 (since even numbers are divisible by 2 with no remainder). So,1  ans % 2
will be1  0
, which equals 1.If
ans
is odd, thenans % 2
will equal 1 (since odd numbers leave a remainder of 1 when divided by 2). In this case,1  ans % 2
will be1  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.