Can I use non-const references in C++ during interviews (onsite or not) ?
Haozhe Sun
154

According to Google C++ Style Guide and Elements of programming interviews, in C++, non-const references are never allowed as input arguments to functions.

Is this convention widely accepted by other companies (Facebook, Linkedin, or even banks)?

I wonder if I should obey this convention during all my programming interviews (onsite or not) ?

In some programming problems, it is convenient (even necessary) to modify variables inside a function. If non-const reference arguments are forbidden, then I have to use pointers instead, which would lead to inconvenient and ugly code like (*pval)++.

For example, if I am asked to solve Leetcode 46. Permutations during an interview, can I write the following code which uses non-const reference arguments?

class Solution {
public:
    vector<vector<int>> permute(vector<int>& nums) {
        vector<vector<int>> res;
        vector<int> current;
        vector<bool> used(nums.size(), false);
        backtrack(res, nums, current, used);
        return res;
    }
    
    void backtrack(vector<vector<int>>& res, vector<int>& nums, vector<int>& current, vector<bool>& used){
        if(current.size() == nums.size()){
            res.push_back(current);
            return;
        }
        for(int i = 0; i < nums.size(); i++){
            if(!used[i]){
                current.push_back(nums[i]);
                used[i] = true;
                backtrack(res, nums, current, used);
                current.pop_back();
                used[i] = false;
            }
        }
    }
};

Or is it better for me to write the following code which substitutes non-const reference arguments for pointers?

class Solution {
public:
    vector<vector<int>> permute(vector<int>& nums) {
        vector<vector<int>> res;
        vector<int> current;
        vector<bool> used(nums.size(), false);
        backtrack(&res, nums, &current, &used);
        return res;
    }
    
    void backtrack(vector<vector<int>>* res, const vector<int>& nums, vector<int>* current, vector<bool>* used){
        if((*current).size() == nums.size()){
            (*res).push_back(*current);
            return;
        }
        for(int i = 0; i < nums.size(); i++){
            if(!(*used)[i]){
                (*current).push_back(nums[i]);
                (*used)[i] = true;
                backtrack(res, nums, current, used);
                (*current).pop_back();
                (*used)[i] = false;
            }
        }
    }
};
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