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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.

• Java
• C++
• Python
• Go
• TypeScript

• 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
  

• 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;
  }
  
  

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