1073. Adding Two Negabinary Numbers

Problem

Given two numbers arr1 and arr2 in base -2, return the result of adding them together.

Each number is given in array format:  as an array of 0s and 1s, from most significant bit to least significant bit.  For example, arr = [1,1,0,1] represents the number (-2)^3 + (-2)^2 + (-2)^0 = -3.  A number arr in array, format is also guaranteed to have no leading zeros: either arr == [0] or arr[0] == 1.

Return the result of adding arr1 and arr2 in the same format: as an array of 0s and 1s with no leading zeros.

  Example 1:

Input: arr1 = [1,1,1,1,1], arr2 = [1,0,1]
Output: [1,0,0,0,0]
Explanation: arr1 represents 11, arr2 represents 5, the output represents 16.

Example 2:

Input: arr1 = [0], arr2 = [0]
Output: [0]

Example 3:

Input: arr1 = [0], arr2 = [1]
Output: [1]

  Constraints:

• 1 <= arr1.length,&nbsp;arr2.length <= 1000

• arr1[i] and arr2[i] are 0 or 1

• arr1 and arr2 have no leading zeros

Solution (Java)

class Solution {
    public int[] addNegabinary(int[] arr1, int[] arr2) {
        int len1 = arr1.length;
        int len2 = arr2.length;
        int[] reverseArr1 = new int[len1];
        for (int i = len1 - 1; i >= 0; i--) {
            reverseArr1[len1 - i - 1] = arr1[i];
        }
        int[] reverseArr2 = new int[len2];
        for (int i = len2 - 1; i >= 0; i--) {
            reverseArr2[len2 - i - 1] = arr2[i];
        }
        int[] sumArray = new int[Math.max(len1, len2) + 2];
        System.arraycopy(reverseArr1, 0, sumArray, 0, len1);
        for (int i = 0; i < sumArray.length; i++) {
            if (i < len2) {
                sumArray[i] += reverseArr2[i];
            }
            if (sumArray[i] > 1) {
                sumArray[i] -= 2;
                sumArray[i + 1]--;
            } else if (sumArray[i] == -1) {
                sumArray[i] = 1;
                sumArray[i + 1]++;
            }
        }
        int resultLen = sumArray.length;
        for (int i = sumArray.length - 1; i >= 0; i--) {
            if (sumArray[i] == 0) {
                resultLen--;
            } else {
                break;
            }
        }
        if (resultLen == 0) {
            return new int[] {0};
        }
        int[] result = new int[resultLen];
        for (int i = resultLen - 1; i >= 0; i--) {
            result[resultLen - i - 1] = sumArray[i];
        }
        return result;
    }
}

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).