66. Plus One

Description of 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
digits does not contain any leading 0's.

Solution

Code (Java)

class Solution {
    public int[] plusOne(int[] digits) {
        int carry = 1;
        for(int i = digits.length - 1; i >= 0; i--){
            int sum = digits[i] + carry;
            carry = sum >= 10 ? 1 : 0;
            digits[i] = sum % 10;
        }

        if (carry > 0){
            int [] new_digits = new int[digits.length + 1];
            new_digits[0] = carry;
            /*
            for(int i = 0; i < digits.length; i++){
                new_digits[i+1] = digits[i];
            }
            */
            return new_digits;
        }else{
            return digits; 
        }
    }
}

Complexity

n is length of digits

Time Complexity

\(T(n)=O(n)\)

Auxiliary Space

\(S(n)=O(1)\)

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