(Easy) Binary Prefix Divisible by 5. LeetCode (to be continued)

Description:

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

Accepted

12,976

Submissions

27,721

 

Solution:

This solution is only partially correct, as it will have overflow issue. 

class Solution {
    public List<Boolean> prefixesDivBy5(int[] A) {
        
        List<Boolean> list = new ArrayList<Boolean>();
        
        for(int i = 0; i<A.length; i++){
            int num = 0;
            int t = i;
            for(int j = 0; j<=i; j++){
                
                num = num+ A[j]*(int)Math.pow(2,t--);
                
            }
            
                     if(num%5==0){
                list.add(true);
            }
            else{
                list.add(false);
            }
        }
             return list;
        }
     
    }

 

Optimized Solution with Explanation: 

https://leetcode.com/problems/binary-prefix-divisible-by-5/discuss/265601/Detailed-Explanation-using-Modular-Arithmetic-O(n)