# 1018. Binary Prefix Divisible By 5

## Description

You are given a binary array nums (0-indexed).

We define xi as the number whose binary representation is the subarray nums[0..i] (from most-significant-bit to least-significant-bit).

• For example, if nums = [1,0,1], then x0 = 1, x1 = 2, and x2 = 5.

Return an array of booleans answer where answer[i] is true if xi is divisible by 5.

Example 1:

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


Constraints:

• 1 <= nums.length <= 105
• nums[i] is either 0 or 1.

## Solutions

• 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;
}
}

• class Solution {
public:
vector<bool> prefixesDivBy5(vector<int>& nums) {
vector<bool> ans;
int x = 0;
for (int v : nums) {
x = (x << 1 | v) % 5;
ans.push_back(x == 0);
}
return ans;
}
};

• class Solution:
def prefixesDivBy5(self, nums: List[int]) -> List[bool]:
ans = []
x = 0
for v in nums:
x = (x << 1 | v) % 5
ans.append(x == 0)
return ans


• 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;
}