Difficult(maybe it is simple) question from another website: Simple Array Sum

Question

#include <bits/stdc++.h>

using namespace std;

vector<string> split_string(string);

/*
* Complete the simpleArraySum function below.
*/
int simpleArraySum(vector<int> ar) {
/*
* Write your code here.

*/int sum = 0;

cout << sizeof(ar);
for(int n =0;n<(sizeof(ar)/sizeof(ar[0]));n++){
cout << sizeof(ar[0]);
sum += ar[n];

}
return sum;
}

int main()
{
ofstream fout(getenv("OUTPUT_PATH"));

int ar_count;
cin >> ar_count;
cin.ignore(numeric_limits<streamsize>::max(), '\n');

string ar_temp_temp;
getline(cin, ar_temp_temp);

vector<string> ar_temp = split_string(ar_temp_temp);

vector<int> ar(ar_count);

for (int ar_itr = 0; ar_itr < ar_count; ar_itr++) {
int ar_item = stoi(ar_temp[ar_itr]);

ar[ar_itr] = ar_item;
}

int result = simpleArraySum(ar);

fout << result << "\n";

fout.close();

return 0;
}

vector<string> split_string(string input_string) {
string::iterator new_end = unique(input_string.begin(), input_string.end(), [] (const char &x, const char &y) {
return x == y and x == ' ';
});

input_string.erase(new_end, input_string.end());

while (input_string[input_string.length() - 1] == ' ') {
input_string.pop_back();
}

vector<string> splits;
char delimiter = ' ';

size_t i = 0;
size_t pos = input_string.find(delimiter);

while (pos != string::npos) {
splits.push_back(input_string.substr(i, pos - i));

i = pos + 1;
pos = input_string.find(delimiter, i);
}

splits.push_back(input_string.substr(i, min(pos, input_string.length()) - i + 1));

return splits;
}

It requires me to input some codes there. But what should I input there? I am confused...I originally thought that It was fine but it didn't past the test cases of that website. Please help..

Answer 1

That is a big code, you should instead save it and attach the code link. Reading code as raw text in Description is acceptable for short codes, but not for big one.

So save that code, edit your question, remove code text (keep the problem description) and paste the code link in place of it. Hopefully with a code for review your question may gain more responses.

Good luck! 👍