Binary Prefix Divisible By 5 LT1018

Posted 2021-11-25 taste-it-own-it-love-it

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

Idea 1. Modular arithmetic

(a*b + c)%d = (a%d) * (b%d) + c%d

Time complexity: O(n)

Space complexity: O(n)

 1 class Solution {
 2     public List<Boolean> prefixesDivBy5(int[] A) {
 3         int val = 0;
 4         List<Boolean> divisible = new ArrayList<>();
 5         
 6         for(int num: A) {
 7             val = ((val * 2)%5 + num)%5;
 8             divisible.add(val == 0? true : false);
 9         }
10         
11         return divisible;
12     }
13 }

Idea 1.b left shift, bitwise operation

 1 class Solution {
 2     public List<Boolean> prefixesDivBy5(int[] A) {
 3         int val = 0;
 4         List<Boolean> divisible = new ArrayList<>();
 5         
 6         for(int num: A) {
 7             val = ((val << 1) | num)%5;
 8             divisible.add(val == 0);
 9         }
10         
11         return divisible;
12     }
13 }