1018. Binary Prefix Divisible By 5

Given an array A of 0s and 1s, consider N_i: the i-th subarray from A[0] to A[i] interpreted as a binary number (from most-significant-bit to least-significant-bit.)

Return a list of booleans answer, where answer[i] is true if and only if N_i is divisible by 5.

Example 1:

Input: [0,1,1]
Output: [true,false,false]
Explanation: 
The input numbers in binary are 0, 01, 011; which are 0, 1, and 3 in base-10.  Only the first number is divisible by 5, so answer[0] is true.

Example 2:

Input: [1,1,1]
Output: [false,false,false]

Example 3:

Input: [0,1,1,1,1,1]
Output: [true,false,false,false,true,false]

Example 4:

Input: [1,1,1,0,1]
Output: [false,false,false,false,false]

 

Note:

  1. 1 <= A.length <= 30000
  2. A[i] is 0 or 1

1018. Binary Prefix Divisible By 5
struct Solution;

impl Solution {
    fn prefixes_div_by5(a: Vec<i32>) -> Vec<bool> {
        let mut x = 0;
        let n = a.len();
        let mut res: Vec<bool> = vec![false; n];
        for i in 0..n {
            x = (x * 2 + a[i]) % 5;
            res[i] = x == 0;
        }
        res
    }
}

#[test]
fn test() {
    let a = vec![0, 1, 1];
    let res = vec![true, false, false];
    assert_eq!(Solution::prefixes_div_by5(a), res);
    let a = vec![1, 1, 1];
    let res = vec![false, false, false];
    assert_eq!(Solution::prefixes_div_by5(a), res);
    let a = vec![0, 1, 1, 1, 1, 1];
    let res = vec![true, false, false, false, true, false];
    assert_eq!(Solution::prefixes_div_by5(a), res);
    let a = vec![1, 1, 1, 0, 1];
    let res = vec![false, false, false, false, false];
    assert_eq!(Solution::prefixes_div_by5(a), res);
}