66. Plus One

• python3

Problem

You are given a large integer represented as an integer array digits, where each digits[i] is the ith 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

Pseudocode

1. First, I checked all possible conditions.
2. If a digit is 9 or larger, it should be set to zero and the digit above it will be incremented by one.
3. If a digit is smaller than 9, it should simply be incremented by one.
4. After all steps, if the first element of the array is zero, insert 1 at the beginning of the array.

Code

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

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

        return digits

Result

• Time Complexity: O(n) as we traverse the array only once.
• Space Complexity: O(1) as the size of array is almost same.

Review

• Checking all possibility may be the best way.