Add to ArrayForm of Integers
This can be solved by using a simple approach of arithmetic and array manipulation.
 Iterate over the array from the end (i.e., least significant digit)
 In each iteration, add the current digit to the remaining value of
k
. It’s crucial to addk
to the least significant digits becausek
could have more digits than thenum
array.  Use the division and modulus operations to update
k
and replace the current digit. The updated digit becomes(current digit + k) % 10
, and the remainingk
becomes(current digit + k) / 10
.  After going through all the digits of
num
, ifk
is not zero, it means we have remaining carry, ork
had more digits thannum
. We need to add each digit ofk
to the beginning of thenum
array.  Finally, return the
num
array as the result.
Python solution:


This solution treats num
and k
as numbers and follows the standard algorithm for adding numbers: add corresponding digits and propagate the carry. This approach ensures that each digit of both numbers is taken into account. This solution will pass all test cases because it correctly implements the arithmetic of adding two numbers.
What are the building blocks of the solution?
The solution to this problem relies on three main building blocks:
Arithmetic operations: The solution uses the modulus and division operations to calculate the new digit and carry over (if any) when the current digit of
num
andk
are added. Specifically, the new digit is(current digit + k) % 10
, and the remainingk
is(current digit + k) // 10
.Iteration: The solution involves iterating over the
num
array from the end (i.e., the least significant digit). This corresponds to how we usually perform addition manually  from right to left, adding corresponding digits of two numbers.Array manipulation: If
k
is still greater than 0 after iterating through all the digits ofnum
, we add each remaining digit ofk
to the beginning of thenum
array. This uses theinsert
operation of an array.
Together, these building blocks allow us to accurately perform the addition operation and transform the resulting number back into an array form.
I recognize Reduction, can you go up on the abstract level and identify more ?
At a higher level of abstraction, there are several important concepts at play in this problem:
DigitbyDigit Calculation: This problem emphasizes the process of adding numbers in a digitbydigit manner, akin to how we perform manual addition. This procedure is crucial in the context of larger arithmetic and numerical problems.
Carry Handling: The mechanism for dealing with the carry from adding two digits is integral to the solution. It not only affects the current digit but also the computation of the subsequent digit, demonstrating the concept of state propagation in computational problems.
Reverse Iteration: This problem involves iterating over the input in reverse, which is a common technique for problems involving digits or sequences where the last element affects the outcome.
Problem Transformation: The initial problem, which seems to be about arrays, is essentially reduced to a problem of numerical addition with carry handling. Recognizing this transformation is a critical part of devising the solution.
By identifying these higherlevel concepts, we can better understand the nature of the problem and design an efficient algorithm to solve it.
The //
operator in Python performs floor division. This means it divides the operand on its left by the operand on its right and rounds down the result to the nearest whole number. For example:
5 // 2 = 2
9 // 4 = 2
In the first example, 5 divided by 2 is 2.5. When we use //
, Python rounds this down to 2. Similarly, in the second example, 9 divided by 4 is 2.25, which is rounded down to 2.
This operation is different from the standard division operator /
, which performs floating point division and does not round down the result. For instance, 5 / 2
would give 2.5
in Python.
The //
operator is used in this solution to help handle the carry in the addition operation.
Let’s take a closer look at this line: carry, curr = divmod(carry + k % 10, 10)
.
Here, divmod()
is a builtin Python function that takes two arguments a
and b
, and returns a pair of numbers (a // b, a % b)
.
carry + k % 10
adds the carry from the previous step and the last digit of k
. The result is then divided by 10. The quotient represents the next carry, and the remainder represents the current digit.
By using carry // 10
, we are extracting the next carry value. This is important because when you’re adding two digits and the result is greater or equal to 10, the “tens place” digit becomes the carry for the addition of the next set of digits in the numbers being added.
In the context of this problem, //
allows us to correctly propagate the carry value from right to left as we perform the addition of the numbers represented by num
and k
.
Coding Drills
Exercise 1: Understanding and using the divmod
function
The divmod
function is crucial to this solution. Understand how it works with different values and how it returns a pair of numbers  the quotient and the remainder of the division.


This means that when 17 is divided by 5, the quotient is 3 and the remainder is 2.
Exercise 2: Reversing a list
The solution uses list reversal to align the digits of num
and k
from right to left. Understand how to reverse a list in Python.


Exercise 3: List comprehension and the enumerate
function
The solution makes use of list comprehension to simplify the addition of corresponding digits from num
and k
. This technique is crucial to understanding how the solution works.


Also, the enumerate
function is used in combination with list comprehension. Understand how enumerate
works and how it can be used in list comprehension.


Exercise 4: Combining it all together
Take the components from the above exercises and try to write a function that performs addition of a list of digits and an integer, similar to the problem description.
In each exercise, focus on the specific concept at hand, and by the end, try to put it all together to solve the original problem. This approach will help in breaking down complex problems into more manageable components.