66 Plus One

Sep 2022

You are given a large integer represented as an integer array digits, where each digits[i] is the i^th digit of the integer. The digits are ordered from most significant to least significant in left-to-right order. The large integer does not contain any leading 0's.

Increment the large integer by one and return the resulting array of digits.

Example 1:

Input: digits = [1,2,3]
Output: [1,2,4]
Explanation: The array represents the integer 123.
Incrementing by one gives 123 + 1 = 124.
Thus, the result should be [1,2,4].

Example 2:

Input: digits = [4,3,2,1]
Output: [4,3,2,2]
Explanation: The array represents the integer 4321.
Incrementing by one gives 4321 + 1 = 4322.
Thus, the result should be [4,3,2,2].

Example 3:

Input: digits = [9]
Output: [1,0]
Explanation: The array represents the integer 9.
Incrementing by one gives 9 + 1 = 10.
Thus, the result should be [1,0].

Constraints:

1 <= digits.length <= 100
0 <= digits[i] <= 9
digits does not contain any leading 0's.

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class Solution:
    def plusOne(self, digits: List[int]) -> List[int]:
        digits[-1] += 1
        res = []
        carry = 0
        for i in range(len(digits)-1, -1, -1):
            res.append((digits[i] + carry) % 10)
            carry = (digits[i] + carry) // 10
        if carry > 0:
            res.append(1)
        return reversed(res)


'''
Early termination
'''
class Solution:
    def plusOne(self, digits: List[int]) -> List[int]:
        n = len(digits)

        # move along the input array starting from the end
        for i in range(n):
            idx = n - 1 - i
            # set all the nines at the end of array to zeros
            if digits[idx] == 9:
                digits[idx] = 0
            # here we have the rightmost not-nine
            else:
                # increase this rightmost not-nine by 1
                digits[idx] += 1
                # and the job is done
                return digits

        # we're here because all the digits are nines
        return [1] + digits