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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.

• 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;
}
return list;
}
}

############

class Solution {
public List<Boolean> prefixesDivBy5(int[] nums) {
List<Boolean> ans = new ArrayList<>();
int x = 0;
for (int v : nums) {
x = (x << 1 | v) % 5;
}
return ans;
}
}

• // 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

• func prefixesDivBy5(nums []int) (ans []bool) {
x := 0
for _, v := range nums {
x = (x<<1 | v) % 5
ans = append(ans, x == 0)
}
return
}

• function prefixesDivBy5(nums: number[]): boolean[] {
const ans: boolean[] = [];
let x = 0;
for (const v of nums) {
x = ((x << 1) | v) % 5;
ans.push(x === 0);
}
return ans;
}