How to check if 3 elements of array have the same value - c++

Question

I am trying to write a program which checks if 3 (or more) elements of an array are the same.
I have written a code which works almost perfectly, but it gets stuck when there are 3 pairs of equal elements and I'm not sure how to fix it.
#include <iostream>
#include <algorithm>
using namespace std;
int main() {
int n, a[10],skirt=0;
cin >> n;
for(int i = 0; i < n; i++)
{
cin >> a[i];
}
for(int i = 0; i < n; i++)
{
for(int j = i + 1; j < n; j++)
{
if(a[i] == a[j])
{
skirt++;
}
}
}
cout<<skirt<<endl;
if(skirt>=3)
{
cout << "TAIP" << endl;
}
else
{
cout << "NE" << endl;
}
}
When I input
6
3 3 2 2 1 1 i
get "TAIP" but I need to get "NE".

Answer 1

You can use the following algorithm: first sort the array. Then iterate each adjacent pair. If they are equal, then increment counter, if not then reset counter to 1. If counter is 3, return true. If loop does not return true, then return false.

Answer 2

Add the following condition in the outer for loop
for(int i = 0; i < n - 2 && skirt != 3; i++)
^^^^^^^^^^^^^^^^^^^^^^^
{
skirt = 1;
^^^^^^^^^
for(int j = i + 1; j < n; j++)
{
if(a[i] == a[j])
{
skirt++;
}
}
}
Of course before the loop you should check whether n is not less than 3. For example
if ( not ( n < 3 ) )
{
for(int i = 0; i < n - 2 && skirt != 3; i++)
{
skirt = 1;
for(int j = i + 1; j < n; j++)
{
if(a[i] == a[j])
{
skirt++;
}
}
}
}
Here is a demonstrative program
#include <iostream>
using namespace std;
int main()
{
int a[] = { 6, 3, 3, 2, 2, 1, 1 };
int n = 7;
int skirt = 0;
if ( not ( n < 3 ) )
{
for(int i = 0; i < n - 2 && skirt != 3; i++)
{
skirt = 1;
for(int j = i + 1; j < n; j++)
{
if ( a[i] == a[j] )
{
skirt++;
}
}
}
}
cout << skirt << endl;
if ( skirt == 3 )
{
cout << "TAIP" << endl;
}
else
{
cout << "NE" << endl;
}
return 0;
}
Its output is
1
NE
because the array does not have 3 equal elements.

Answer 3

Reset skirt to 0 every time you increase i if it is less than 3, or break out the loop otherwise.

Answer 4

Another way to do this is using a std::map, which keeps a count of the number of times a given value occurs in your array. You would stop looking as soon as you have a number that has three occurrences.
Here's a 'minimalist' code version:
#include <iostream>
#include <map>
using std::cin; // Many folks (especially here on SO) don't like using the all-embracing
using std::cout; // ... statement, "using namespace std;". So, these 3 lines only 'use'
using std::endl; // ... what you actually need to!
int main() {
int n, a[10], skirt = 0;
std::map<int, int> gots;
cin >> n;
for (int i = 0; i < n; i++) {
cin >> a[i];
}
for (int i = 0; i < n && skirt < 3; i++) {
skirt = 1;
if (gots.find(a[i]) != gots.end()) skirt = gots[a[i]] + 1;
gots.insert_or_assign(a[i], skirt);
}
cout << (skirt >= 3 ? "TAIP" : "NE") << endl;
return 0;
}
I'm not saying this is any better (or worse) than the other answers - just another way of approaching the problem, and making use of what the Standard Library has to offer. Also, with this approach, you could easily modify the code to count how many numbers occur three or more times, or any number of time.
Feel free to ask for further clarification and/or explanation.