Binary Prefix Divisible By 5 LeetCode solution in C++

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class Solution {
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 public:
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  /*
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  1. We notice that 1 = 1; 10 = 1*2 + 0 = 2; 101 = 2*2 + 1 = 5; .Therefore, the
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  current_decimal form of number = previous sum * 2 + nums[i];
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  2. It means: current_number = current_number * 2 + nums[i] (i = current index)
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  3. If we did that, the current_number will exceed integer limit => Therefore,
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  we modulo the current_number by 5. We check if the remainder equals to 0 or
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  not. By doing that, we can keep our current_number from 0 to 4, which will not
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  exceed integer range.
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  */
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  vector<bool> prefixesDivBy5(vector<int> &nums) {
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    vector<bool> answer;
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    int current_number = 0;
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    for (int i = 0; i < nums.size(); i++) {
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      current_binary = (current_number * 2 + nums[i]) % 5;
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      if (current_number == 0)
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        answer.push_back(true);
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      else
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        answer.push_back(false);
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    }
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    return answer;
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  }
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};