Binary Prefix Divisible By 5

Problem

You are given a binary array nums (0-indexed).

We define xi as the number whose binary representation is the subarray nums[0..i] (from most-significant-bit to least-significant-bit).

• For example, if nums = [1,0,1], then x0 = 1, x1 = 2, and x2 = 5.

Return an array of booleans answer where answer[i] is true if xi is divisible by 5.

Constraints

• 1 <= nums.length <= 105
• nums[i] is either 0 or 1.

Solution

The problem Binary Prefix Divisible By 5 can be solved by adding each digit into a number and checking each step if the number is divisble by 5. However, as 2100000 - 1, the largest value xi can have, is well above the range of integer in C++, we need to find a way to check if xi is divisble by 5 without directly calculating xi. xi can be represented as xi = 5α + β where α and β are non-negative integers and 0 <= β < 5. Then, xi+1 = 2 * (5α + β) + φ = 10α + 2 * β + φ where φ = 0, 1. In this case, it is obvious that subtracting 5α from xi does not have any affect on the modulo operation as it does not affect the remainder of xi or xi+1. Therefore, by performing modulo 5 operation each step on the sum of digits, it is possible to check if the sum is divisible by 5 while not overflowing the variable.

Implementation

static const int fast_io = []()
{
    std::ios::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    return 0;
}();

class Solution
{
  public:
    vector<bool> prefixesDivBy5(vector<int> &nums)
    {
        vector<bool> ret;

        int sum = 0;
        for (int num : nums)
        {
            sum = ((sum << 1) + num) % 5;
            ret.push_back(!sum);
        }

        return ret;
    }
};