• Approach
The approach uses the same idea as adding numbers when learning addition in grade school. We start from the right-most index on both numbers and add digits as we move to the left of both strings until there are no more digits in either number.

For example, if we have the numbers "54" and "5", we start at indices `1` and `0`, respectively. We add both of `4` and `5` to get `9` with a `carry bit` of `0`. This number is appended to added to the `result` array at a separately tracked index (in this case, it's `2` because the sum of the lengths of both string versions of `54` and `5` is `3` but `3-1=2`). Then, we decrement both indices. On the next iteration, we see that we have reached index `-1` for the number `5`, so now we only use `5` from index `0` and the `carry bit` of `0`.

The `result` array is then joined to get `59`.

Code

``````class Solution:
num1 = str(num1)
num2 = str(num2)

l1, l2 = len(num1), len(num2)
idx1, idx2 = len(num1)-1, len(num2)-1
result, resultIdx = [""] * (l1+l2), len(l1+l2)-1=

carry = 0

while idx1 >= 0 or idx2 >= 0 or carry:
v1 = v2 = 0

if idx1 >= 0.:
v1 = num1[idx1]
idx1 -= 1

if idx2 >= 0:
v2 = num2[idx2]
idx2 -= 1

totalSum = int(v1) + int(v2) + carry
carry = int(totalSum / 10)
netSum = totalSum % 10

result[resultIdx] = str(netSum)
resultIdx -= 1

return "".join(result)
``````

Complexity
Let `N = max(len(str(num1), len(str(num2))`. The complexity of this approach is then `O(N)`.

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