tags: `leetcode` `easy` `array` `math`

66. Plus One

Description

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.

Examples

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.

Code

/**
 * Note: The returned array must be malloced, assume caller calls free().
 */
int* plusOne(int* digits, int digitsSize, int* returnSize) {
    int carry = 1;
    // Sequentially checking each digit is carry or not
    for (int i = digitsSize - 1; i >= 0; i--) {
        digits[i] += carry;
        if (digits[i] == 10) {
            digits[i] = 0;
            carry = 1;
            continue;
        }
        carry = 0;
        break;
    }
    // Using carry bit to assign correst return size
    *returnSize = digitsSize + carry;
    int* ans = malloc(*returnSize * sizeof(int));
    // Assign number from digits with shifting of digits by carry
    for (int i = *returnSize - 1; i > 0; i--)
        ans[i] = digits[i - carry];
    ans[0] = carry ? 1 : digits[0];
    return ans;
}

Complexity

Space	Time
$O(n)$	$O(n)$

Result

Runtime: 2 ms, 76.18%
Memory Usage: 5.8 MB, 30.18%

tags: leetcode easy array math