This is "big number" calculation.
The logic is the add k to A starting from the last element of array
so the traverse start from index: num.length-1
sum = currentA+ currentK + carry
carry = sum // 10
current = sum % 10
k \=10
i--
you can choose to add the end of array and then do reverse to generate answer which is faster.
Or, you can use array.add(0,current) to always add to the head of the array, which will take O(N) to finish
if using python, you dont need to consider overflow of big number, so this problem can be done like this: transfer array --> num, add with k, transfer back to array for answer
- time complexity: O(N)
- space complexity: O(N) since we need to create an array to store answer
- extra space complexity: O(1) as we only need to create sum, carry to store the value each time.