More elegant way to check for duplicates in C++ ar

Question

I wrote this code in C++ as part of a uni task where I need to ensure that there are no duplicates within an array:

// Check for duplicate numbers in user inputted data
    int i; // Need to declare i here so that it can be accessed by the 'inner' loop that starts on line 21
    for(i = 0;i < 6; i++) { // Check each other number in the array
        for(int j = i; j < 6; j++) { // Check the rest of the numbers
            if(j != i) { // Makes sure don't check number against itself
                if(userNumbers[i] == userNumbers[j]) {
                    b = true;
                }
            }
            if(b == true) { // If there is a duplicate, change that particular number
                cout << "Please re-enter number " << i + 1 << ". Duplicate numbers are not allowed:" << endl;
                cin >> userNumbers[i];
            }
        } // Comparison loop
        b = false; // Reset the boolean after each number entered has been checked
    } // Main check loop

It works perfectly, but I'd like to know if there is a more elegant or efficient way to check.

Answer 1

The following solution is based on sorting the numbers and then removing the duplicates:

#include <algorithm>

int main()
{
    int userNumbers[6];

    // ...

    int* end = userNumbers + 6;
    std::sort(userNumbers, end);
    bool containsDuplicates = (std::unique(userNumbers, end) != end);
}

Answer 2

//std::unique(_copy) requires a sorted container.
std::sort(cont.begin(), cont.end());

//testing if cont has duplicates
std::unique(cont.begin(), cont.end()) != cont.end();

//getting a new container with no duplicates
std::unique_copy(cont.begin(), cont.end(), std::back_inserter(cont2));

Answer 3

I'm not sure why this hasn't been suggested but here is a way in base 10 to find duplicates in O(n).. The problem I see with the already suggested O(n) solution is that it requires that the digits be sorted first.. This method is O(n) and does not require the set to be sorted. The cool thing is that checking if a specific digit has duplicates is O(1). I know this thread is probably dead but maybe it will help somebody! :)

/*
============================
Foo
============================
* 
   Takes in a read only unsigned int. A table is created to store counters 
   for each digit. If any digit's counter is flipped higher than 1, function
   returns. For example, with 48778584:
    0   1   2   3   4   5   6   7   8   9
   [0] [0] [0] [0] [2] [1] [0] [2] [2] [0]

   When we iterate over this array, we find that 4 is duplicated and immediately
   return false.

*/
bool Foo( unsigned const int &number)
{
    int temp = number;
    int digitTable[10]={0};

    while(temp > 0)
    {
        digitTable[temp % 10]++; // Last digit's respective index.
        temp /= 10; // Move to next digit
    }

    for (int i=0; i < 10; i++)
    {
        if (digitTable [i] > 1)
        {
            return false;
        }
    }
    return true;
}

Answer 4

You can add all elements in a set and check when adding if it is already present or not. That would be more elegant and efficient.

Answer 5

You could sort the array in O(nlog(n)), then simply look until the next number. That is substantially faster than your O(n^2) existing algorithm. The code is also a lot cleaner. Your code also doesn't ensure no duplicates were inserted when they were re-entered. You need to prevent duplicates from existing in the first place.

std::sort(userNumbers.begin(), userNumbers.end());
for(int i = 0; i < userNumbers.size() - 1; i++) {
    if (userNumbers[i] == userNumbers[i + 1]) {
        userNumbers.erase(userNumbers.begin() + i);
        i--;
    }
}

I also second the reccomendation to use a std::set - no duplicates there.

Answer 6

#include<iostream>
#include<algorithm>

int main(){

    int arr[] = {3, 2, 3, 4, 1, 5, 5, 5};
    int len = sizeof(arr) / sizeof(*arr); // Finding length of array

    std::sort(arr, arr+len);

    int unique_elements = std::unique(arr, arr+len) - arr;

    if(unique_elements == len) std::cout << "Duplicate number is not present here\n";
    else std::cout << "Duplicate number present in this array\n";

    return 0;
}