Formatted question description: https://leetcode.ca/all/1018.html

1018. Binary Prefix Divisible By 5 (Easy)

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

Related Topics:
Array

Solution 1.

// OJ: https://leetcode.com/problems/binary-prefix-divisible-by-5/

// Time: O(N)
// Space: O(1)
class Solution {
public:
vector<bool> prefixesDivBy5(vector<int>& A) {
vector<bool> ans;
int val = 0;
for (int n : A) {
val = ((val << 1) + n) % 5;
ans.push_back(val == 0);
}
return ans;
}
};


Java

class Solution {
public List<Boolean> prefixesDivBy5(int[] A) {
List<Boolean> list = new ArrayList<Boolean>();
int length = A.length;
int num = 0;
for (int i = 0; i < length; i++) {
num = (num * 2 + A[i]) % 5;