abort called is showing in Hackerrank - c++

Question

I was solving a problem of the array in which I have to do a left rotation of an array. wrote the code and submit successfully some test cases passed in some it showing abort called I don't know whats the problem. I googled it shows it's due to storage becomes full. what if I declare my temp array in heap. would it make any difference? forgive me for my indentation.
#include <bits/stdc++.h>
using namespace std;
string ltrim(const string &);
string rtrim(const string &);
vector<string> split(const string &);
vector<int> rotateLeft(int d, vector<int> arr) {
vector<int> temp(d);
for (int i = 0; i < d; i++) {
temp[i] = arr[i];
}
for (int i = 0; i < arr.size(); i++) {
arr[i] = arr[d + i];
}
for (int i = 0; i < arr.size(); i++) {
arr[arr.size() - d + i] = temp[i];
}
return arr;
}
int main() {
ofstream fout(getenv("OUTPUT_PATH"));
string first_multiple_input_temp;
getline(cin, first_multiple_input_temp);
vector<string> first_multiple_input = split(rtrim(first_multiple_input_temp));
int n = stoi(first_multiple_input[0]);
int d = stoi(first_multiple_input[1]);
string arr_temp_temp;
getline(cin, arr_temp_temp);
vector<string> arr_temp = split(rtrim(arr_temp_temp));
vector<int> arr(n);
for (int i = 0; i < n; i++) {
int arr_item = stoi(arr_temp[i]);
arr[i] = arr_item;
}
vector<int> result = rotateLeft(d, arr);
for (int i = 0; i < result.size(); i++) {
fout << result[i];
if (i != result.size() - 1) {
fout << " ";
}
}
fout << "\n";
fout.close();
return 0;
}
string ltrim(const string & str) {
string s(str);
s.erase(s.begin(), find_if(s.begin(), s.end(), not1(ptr_fun<int, int>(isspace))));
return s;
}
string rtrim(const string & str) {
string s(str);
s.erase(find_if(s.rbegin(), s.rend(), not1(ptr_fun<int, int>(isspace))).base(), s.end());
return s;
}
vector<string> split(const string & str) {
vector<string> tokens;
string::size_type start = 0;
string::size_type end = 0;
while ((end = str.find(" ", start)) != string::npos) {
tokens.push_back(str.substr(start, end - start));
start = end + 1;
}
tokens.push_back(str.substr(start));
return tokens;
}

Answer 1

The problem is in your 3 loops:
for (int i = 0; i < d; i++) {
temp[i] = arr[i];
}
for (int i = 0; i < arr.size(); i++) {
arr[i] = arr[d + i];
}
for (int i = 0; i < arr.size(); i++) {
arr[arr.size() - d + i] = temp[i];
}
first loop will crash if d > n. It can be the case that your task may have d > n sometimes.
Second loop should crash always when d > 0, because arr[d + i] is out of bounds if d + i >= arr.size() which will always happen because loop is until i < arr.size().
Third loop will also always crash because out of bounds. For example if d == 1 and i == arr.size() - 1 then you get arr[arr.size() - 1 + arr.size() - 1] which is out of bounds.
Also you have to watch if d > n in some tests, then you have to make d %= n;.
Also just a notice - inputs of most HackerRank problems can be read using just things like int i = 0; std::cin >> i; and loops. No need for strings operations like stoi/ltrim/rtrim/split. For example to read array of n numbers,
provided as one or several lines of input, you can do:
std::vector<int> nums;
for (size_t i = 0; i < n; ++i) {
int i = 0;
std::cin >> i;
nums.push_back(i);
}
You can use just standard std::rotate to rotate array left.
One interesting way to solve the problem in O(N) time and O(1) extra memory without std::rotate and std::reverse functions and without temporary storage is to reverse order of first d elements, then reverse order of last n - d elements, then reverse whole array. Of cause usually you just use std helpers, but sometimes you want to implement algorithms from scratch. I.e. something like following:
void reverse(int * begin, int * end) {
--end;
while (begin < end)
std::swap(*begin++, *end--);
}
void rotate(int * begin, int * end, int d) {
d %= (end - begin);
reverse(begin, begin + d);
reverse(begin + d, end);
reverse(begin, end);
}
The shortest way to fix your algorithm (it takes O(N) time and O(N) extra memory) to make it work is:
vector<int> rotateLeft(int d, vector<int> arr) {
d %= arr.size();
vector<int> temp(d);
for (int i = 0; i < d; i++) {
temp[i] = arr[i];
}
for (int i = 0; i < arr.size() - d; i++) {
arr[i] = arr[d + i];
}
for (int i = 0; i < d; i++) {
arr[arr.size() - d + i] = temp[i];
}
return arr;
}