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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 0
s and 1
s, 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 <= A.length <= 30000
A[i]
is0
or1
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; list.add(num == 0); } 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; ans.add(x == 0); } 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
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func prefixesDivBy5(nums []int) (ans []bool) { x := 0 for _, v := range nums { x = (x<<1 | v) % 5 ans = append(ans, x == 0) } return }
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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; }