C++ How to the total number of duplicate elements in the array? - c++

Question

I am trying to finish my homework, but I have encountered some problems in the item of Duplicate Elements. I have tried to find out where the problem is, but I cannot find it. My code works fine when the sequence is small, but it becomes problematic when the sequence is large.
In this sequence, I expect to get the number in the picture, but I always get the number in the second picture. As you can see, there is always a problem in this item in the duplicate element.
#include <iostream>
using namespace std;
int main()
{
int count = 0;
int arrayA[25];
int arrayB[25];
int min,max;
int num;
int i = 0;
int n = 0;
int temp = 0;
int sum = 0;
//Input
cout << "Input the number of elements to store in the array: ";
cin >> n;
cout << n << endl;
cout << "Input "<< n <<" integers:" << endl;
//Store arrayA
if(count < n){
for(i = 0; i < n; i++){
cin >> num;
cout << "integer - " << count;
cout << " : " << num <<endl;
arrayA[count] = num;
count++;
}
}
//store arrayB
for(i = 0;i < count; i++){
arrayB[i] = arrayA[count - i - 1];
}
//Forwards Array
cout << "The values stored into the array are :" << endl;
for(i = 0;i < count; i++){
cout << arrayA[i] << " ";
}
cout << endl;
//Backwards Array
cout << "The values stored into the array in reverse are :" << endl;
for(i = 0; i < count; i++){
cout << arrayB[i] <<" ";
}
cout << endl;
//Sum
for(i = 0; i < count; i++){
sum += arrayA[i];
}
//Max & Min
for(i = 0; i < count;i++){
if(max < arrayA[i]){
max = arrayA[i];
}
}
for(i = 0; i < count; i++){
if(min > arrayA[i]){
min = arrayA[i];
}
if(arrayA[i]== 0){
min = 0;
break;
}
}
//Duplicate elements
count = 0;
for(i = 0; i < n; i++){
for(temp = i + 1; temp < n; temp++){
if(arrayA[i] == arrayA[temp] ){
count++;
break;
}
}
}
cout << "The sum of all elements of the array is ";
cout << " " << sum << endl;
cout << "The total number of duplicate elements in the array is ";
cout << count << endl;
cout << "The maximum and minimum element in the array are ";
cout << min << " , " << max;
return 0;
}

Answer 1

You should try debugging more by printing, to know what is happening. You are not really taking into account that a number can be repeated several times and depending on the spread, it will count them several times. And here is why:
//Duplicate elements
count = 0;
for(i = 0; i < n; i++){
for(temp = i + 1; temp < n; temp++){
//cout << "Comparing "<< arrayA[i] << " to "<< arrayA[temp]<<endl;
if(arrayA[i] == arrayA[temp] ){
cout << "ITEM DUPLICATED at i , "<<i<< " item "<< arrayA[i]<< " and temp "<< temp<< " item "<< arrayA[temp] << endl;
count++;
break;
}
}
}
ITEM DUPLICATED at i , 0 item 4 and temp 7 item 4
ITEM DUPLICATED at i , 2 item -2 and temp 10 item -2
ITEM DUPLICATED at i , 4 item -3 and temp 14 item -3
ITEM DUPLICATED at i , 5 item 3 and temp 11 item 3
ITEM DUPLICATED at i , 7 item 4 and temp 8 item 4
ITEM DUPLICATED at i , 12 item -1 and temp 13 item -1
As you can see, there is a 4 - 4 twice. It can be seen quite easily if you do it in paper (although you shouldn't)
4A -4B -2C 1D -3E 3F 5G 4H 4I 0J -2K 3L -1M -1N -3O
4a -4b -2c 1d -3e 3f 5g 4h 4i 0j -2k 3l -1m -1n -3o
Connections made:
4A 4h
-2C -2k
-3E -3o
3F 3l
4H 4i
So the code technically is doing what you are asking it to do. if you want to count UNIQUELY (if a 4 appears 15 times, count it as only that 4 is repeated), you should be checking if the number has already been checked. Using a vector or map (dictionaries in C++) can allow you to do that! Or lists with lists (the element, and the number of occurences)