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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;
            list.add(num == 0);
        }
        return list;
    }
}

  • // 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;
    }
};

  • class Solution(object):
    def prefixesDivBy5(self, A):
        """
        :type A: List[int]
        :rtype: List[bool]
        """
        res = []
        prefix = 0
        for a in A:
            prefix = (prefix * 2 + a) % 5
            res.append(prefix == 0)
        return res

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