Given a non-empty array of digits representing a non-negative integer, plus one to the integer.

The digits are stored such that the most significant digit is at the head of the list, and each element in the array contain a single digit.

You may assume the integer does not contain any leading zero, except the number 0 itself.

Example 1:

Input: [1,2,3]

Output: [1,2,4]

Explanation: The array represents the integer 123.

Example 2:

Input: [4,3,2,1]

Output: [4,3,2,2]

Explanation: The array represents the integer 4321.

Solution

from typing import List

class Solution:
    def plusOne(self, digits: List[int]) -> List[int]:
        digits[len(digits) - 1] += 1

        for i in range(len(digits) - 1, -1, -1):
            if digits[i] > 9:
                digits[i] %= 10
                if (i - 1 < 0):
                    digits.insert(0, 1)
                else:
                    digits[i - 1] += 1

        return digits

Improved Solution

from typing import List

class Solution:
    def plusOne(self, digits: List[int]) -> List[int]:
        for i in range(len(digits) - 1, -1, -1):
            if (digits[i] < 9):
                digits[i] += 1
                return digits
            else:
                digits[i] = 0

        digits[0] = 1
        digits.append(0)

        return digits

Test Cases

test = Solution()
answer = test.plusOne([0])
assert answer == [1]
answer = test.plusOne([9])
assert answer == [1,0]
answer = test.plusOne([9,9])
assert answer == [1,0,0]
answer = test.plusOne([9,9,1])
assert answer == [9,9,2]
answer = test.plusOne([9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9])
assert answer == [1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0]
print('All Passed!')

Big O Analysis

Space Complexity: O(1)

Time Complexity: O(N)