Binary Prefix Divisible By 5

LeetCode Q 1018 - Binary Prefix Divisible By 5

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 <= A.length <= 30000
• A[i] is 0 or 1

Solution :

Solution 1: Not Correct

We update each number num = num * 2 + A[i], then let res[i] = num % 5.
This method is not correct. It will cause Integer overflow.

Code:

public List<Boolean> prefixesDivBy5(int[] A) {
	List<Boolean> res = new ArrayList<>();
	int num = 0;
	for (int a : A) {
		num = num * 2 + a;
		res.add(num % 5 == 0);
	}

	return res;
}

Solution 2:

Instead of updating each number, we can update remainder.

Code:

public List<Boolean> prefixesDivBy5(int[] A) {
	List<Boolean> res = new ArrayList<>();
	int remainder = 0;
	for (int a : A) {
		remainder = (remainder * 2 + a) % 5;
		res.add(remainder == 0);
	}

	return res;
}