README.md

1073. Adding Two Negabinary Numbers

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, arr2.length <= 1000
• arr1[i] and arr2[i] are 0 or 1
• arr1 and arr2 have no leading zeros

Solutions (Rust)

1. Solution

impl Solution {
    pub fn add_negabinary(arr1: Vec<i32>, arr2: Vec<i32>) -> Vec<i32> {
        let mut arr1 = arr1;
        let mut arr2 = arr2;
        let mut carry = 0;
        let mut ret = vec![];

        while !arr1.is_empty() || !arr2.is_empty() || carry != 0 {
            let x = arr1.pop().unwrap_or(0) + arr2.pop().unwrap_or(0) + carry;
            carry = 1 - (x + 2) / 2;
            ret.push(x.rem_euclid(2));
        }

        while ret.len() > 1 && *ret.last().unwrap() == 0 {
            ret.pop();
        }

        ret.reverse();

        ret
    }
}