Searching and sorting arrays - c++

Question

I am taking an introductory level C++ class. I have to write a boolean function that checks for duplicates along a vectors and returns true and false if no duplicates
#include<iostream>
#include<vector>
using namespace std;
bool has_duplicates(const vector <int> &v);
int main() {
vector<int> Vec(8);
Vec = { 20, 30, 40, 50, 10, 20, 5, 6 };
has_duplicates(Vec);
return 0;
}
bool has_duplicates(const vector<int>& v) {
bool duplicatefound = false;
for (int i = 0; i < 8; i++) { // Check each other number in the array
for (int j = i; j < 8; j++) { // Check the rest of the numbers
while (j != i) {// Makes sure don't check number against itself
if (v[i] == v[j]) {
cout << "duplicate found" << endl;
duplicatefound = true;
break;
}
}
}
}
return duplicatefound; // Reset the boolean after each number entered has been checked
}

Answer 1

There is a very simple solution to your problem:
Iterate through your array, using a loop:
for(int i=0; i < array_length; i++)
Within this loop, make use of another loop, to check whether any value prior to the one, to which i is pointing, is equal to your_array[i]. If this is the case you have found a duplicate and can return true, else the loops will just run until the end of the array is reached and then you can return false.
Thus your entire function would be something like:
bool contains_duplicates(int array[], int len) {
for(int i=0; i < len; i++) {
for(int j=0; j < i; j++) {
if(array[j]==array[i]) {
return true;
}
}
}
return false;
}
Hope I could help, cheers!
lindebear

Answer 2

You can use templates to extend the type contained on vector like this. This solution reduces the algorithm complexity to O(n log n) which is better than O(n^2) from your nested loop, it doesn't mean it will be faster always but it make a difference on large vectors.
template< typename T>
bool hasDuplicates(std::vector<T> vect) {
std::sort(vect.begin(), vect.end());
T last;
typename vector<T>::iterator it;
for(it = vect.begin(); it < vect.end(); it++) {
if (vect.begin() != it) {
if(last == *it) {
return true;
}
}
last = *it;
}
return false;
}