1018. Binary Prefix Divisible By 5 #

Problem #

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

Problem Summary #

Given an array A consisting of several 0s and 1s. We define N_i: the i-th subarray from A[0] to A[i] is interpreted as a binary number (from the most significant bit to the least significant bit). Return a list of booleans answer, where answer[i] is true only when N_i is divisible by 5; otherwise it is false.

Solution Approach #

• Easy problem. For each number scanned in the array, cumulatively convert it into a binary number modulo 5. If the remainder is 0, store true; otherwise store false.

Code #

package leetcode

func prefixesDivBy5(a []int) []bool {
	res, num := make([]bool, len(a)), 0
	for i, v := range a {
		num = (num<<1 | v) % 5
		res[i] = num == 0
	}
	return res
}