C++: Removing a duplicate of the user input in an array - c++

Question

I have been trying to solve this problem since yesterday. The program should accept N numbers from the user and will place them in an array. I have done this. However, I don't seem to know how to "warn" the user that the input is a duplicate and lets the user enter again.
Here is my code:
# include <iostream>
using namespace std;
int main () {
int N, pos = -1, arr [N], i, j;
int startLoop = 1;
// the 'startLoop' variable is used so that the first user input can have the break function
bool found = false;
while (startLoop != 0) {
cout << "Enter the number of integers to be entered: ";
cin >> N;
if (N <= 4) {
cout << "Invalid. \n";
}
if (N > 4) {
cout << "\n\nEnter " << N << " unique numbers: " ;
for (i = 0; i < N; i++) {
cin >> arr [i];
for (j = 0; j < N && !found; j++) {
if (arr [j] == arr [i]) {
found = true;
cout << "Enter again: ";
}
}
break;
}
}
}

Answer 1

The following is the copy paste of your code:
By adding break you will jump out of the outer for loop and you cannot read n inputs.
the inner loop range is not correct since i is the total number of values read by now. i.e. if N= 10 and in third iteration you are trying to loop among 10 elements to double check if the latest number is a duplicate, while you have read 3 elements by now.
if (N > 4) {
cout << "\n\nEnter " << N << " unique numbers: " ;
for (i = 0; i < N; i++) {
cin >> arr [i];
for (j = 0; j < N && !found; j++) {
if (arr [j] == arr [i]) {
found = true;
cout << "Enter again: ";
}
}
break;
}
}
How to fix?
One approach would be to read the input in a separate loop and then try to find the duplicates and remove them.
Time Complexity will be O(n) since you need to traverse through the list twice, however you need to implement removing elements from the array efficiently without shifting all the elements
Second approach is like what you are trying to implement but it is not efficient since every time a user is adding a new elements you need to re-evaluate i numbers (in worst case n numbers when you are filling the end of array). this process should be repeated every time user insert a repeated number.
The best approach would be store all input numbers inside a hash (set) and every-time a new number is inserted check whether it is already suggested by the user or not. Time complexity of searching elements through unordered_set will be O(1)

Answer 2

Problems:
Your code has many several problems:
You shouldn't use VLAs (Variable Length Arrays) as they are non-standard.
Using N before initializing it is UB (Undefined Behaviour).
Your approach is not efficient as it checks on average n/2 elements n times and therefore it has a time complexity of O(n^2).
You have many unused variables and variables that should be in a more specific scope.
Solutions:
Use std::vector insted of VLAs.
Resize the vector after initializing N.
Change your approach.
Change the scope of your variables.
Additional information:
using namespace std; is considered a bad practice (More info here).
You should probably add input checking to std::cin to verify that the input is a number.
Full code:
Checking duplicates once at the end of input with std::vector (Time complexity: O(n * log(n)). Space complexity: O(n)):
#include <iostream>
#include <vector>
#include <algorithm>
bool hasDuplicates(std::vector<int> toCheck)
{
std::sort(toCheck.begin(),toCheck.end());
for(unsigned int i = 1; i < toCheck.size(); i++)
if(toCheck[i] == toCheck[i-1])
return true;
return false;
}
int main () {
std::vector<int> arr;
int N;
while (true) {
std::cout << "Enter the number of integers to be entered: ";
std::cin >> N;
if (N <= 4) {
std::cout << "Invalid: The number of integers to be entered must be bigger than 4. \n";
}
else {
arr.resize(N);
std::cout << "\n\nEnter " << N << " unique numbers: " ;
for (int i = 0; i < N; i++)
std::cin >> arr [i];
if(hasDuplicates(arr))
std::cout << "Invalid: The numbers must be unique. \n";
else
break;
}
}
}
Checking duplicates each time a number is entered with std::unordered_map (Time complexity: Amortized O(n). Space complexity: O(n)):
#include <iostream>
#include <vector>
#include <unordered_set>
int main () {
std::vector<int> arr;
std::unordered_set<int> alreadyAppeared;
int N;
std::cout << "Enter the number of integers to be entered: ";
std::cin >> N;
while (N <= 4) {
std::cout << "Invalid: The number of integers to be entered must be bigger than 4. \n";
std::cout << "Enter the number of integers to be entered: ";
std::cin >> N;
}
arr.resize(N);
std::cout << "\n\nEnter " << N << " unique numbers: \n";
for (int i = 0; i < N; i++)
{
int entering;
std::cout << "Enter unique number: ";
std::cin >> entering;
if(alreadyAppeared.find(entering) == alreadyAppeared.end())
{
alreadyAppeared.insert(entering);
arr[i] = entering;
}
else
{
std::cout << "Invalid: The numbers must be unique. \n";
i--;
}
}
}