How to count duplicated numbers in a sorted array - c++

Question

I am trying to finish a problem where I read a file into the program and output a file with the average, min, max, and the count of how many times that number occurred in the program. However, I cannot figure out how to create an array for duplicated number of the "counts".
If the file that I was trying to read in had the values 19 5 26 5 5 19 16 8 1,
I need the outputted file to read 5---3 times; 8---1 time; 16---1 time; 19--2 times; 26--1 times.
I first sorted my array to read 5 5 5 8 16 19 19 26.
Below is my code with explanations of what I was trying to do:
#include <iostream>
#include <fstream>
#include <cstdlib>
#include <string>
using namespace std;
double averageArray(int a[], int length); // user defined function to get average
int maxval(int a[], int length); //user defined function to get max val
int minval(int a[], int length); //user defined function to get min val
void bsort(int arr[], int length);// udf to sort array from min to max
int countArray(int a[], int length); //attempt to create a function to get the number of occurrences of a number that is duplicated
int main()
{
char infilename[16];
int nums[50];//newly created array to read in the numbers from the file
int length(0);//array has a length defined by "length"
ifstream fin;
ofstream fout;
cout << "Please enter an input file name: ";
cin >> infilename;
cout << endl;
fin.open(infilename);
if (fin.fail())
{
cerr << "The file " << infilename << " can't be open!"<<endl;
return 1;
}
cout<<"The output to the file statistics.txt should be as follows: "<<endl;
fout.open("statistics.txt");
fout<<"N"<<"\t"<<"Count"<<endl;
cout<<"N"<<"\t"<<"Count"<<endl;
while (fin >> nums[length])
length++;
bsort(nums, length);
for (int i=0; i<length; i++) {
if (nums[i]==nums[i-1]) {
continue;
}
cout<<nums[i]<<"\t"<<countArray(nums,length)<<endl;
fout<<nums[i]<<"\t"<<endl;
}
cout << "\nAverage: " << averageArray(nums,length) << endl;
cout << "Max: "<< maxval(nums,length)<<endl;
cout << "Min: "<< minval(nums,length)<<endl;
fin.close();
return 0;
}
double averageArray (int a[], int length)
{
double result(0);
for (int i = 0; i < length ; i++)
result += a[i];
return result/length;
}
int maxval(int a[], int length)
{
int max(0);
for (int i=1; i<length; i++)
{
if (a[i]>max)
max=a[i];
}
return max;
}
int minval(int a[], int length)
{
int min(100);
for (int i=1; i<length; i++)
{
if (a[i]<min)
min=a[i];
}
return min;
}
void bsort(int a[], int length)
{
for (int i=length-1; i>0; i--)
for (int j=0; j<i; j++)
if (a[j]>a[j+1])
{
int temp=a[j+1];
a[j+1]=a[j];
a[j]=temp;
}
}
int countArray(int a[], int length)
{
int counter(0);
for (int i=0; i<length; i++){
if (a[i]==a[i+1]) //loop through array and if the number after is the same as the previous number, then count one
counter++;
}
return counter;
}
Though it compiles, the count only shows "3"s as shown in the picture below:
.

Answer 1

Before I give you the solution, please take a moment to remember, you are programming in C++, not C. As such, you ought to use vectors, istream iterators and std::sort. You also ought to use std::map, which easily accomplishes this purpose:
template <typename It>
std::map<int, int> count_occurrences(It it, It end)
{
std::map<int, int> output;
while (it != end) output[*it++]++;
return output;
}
How to combine this with your existing code is left as an exercise for the reader. I suggest you ought to read about iterators.

Answer 2

Your function int countArray(int a[], int length) has no input for the actual number. It always counts how often there are the same numbers behind each other in your array. That happens two times for fives and once for 19 => 3 times.
Solution:
int countArray(int a[], int length, int num)
{
int counter(0);
for (int i=0; i<length; i++){
if (a[i]==num) //loop through array and if the number is the one you are looking for
counter++;
}
return counter;
}
and call you function: countArray(nums, length, nums[i]);

Answer 3

void countArray(int a[], int length)
{
int counter(1);
bool flag(false);
//you take (i+1) index, it can go out of range
for (int i = 0; i < length - 1; i++){
if (a[i]==a[i+1]){ //loop through array and if the number after is the same as the previous number, then count one
flag = true;
counter++;
}
else {
if (flag){
cout<<a[i] << counter << endl;
}
flag = false;
counter = 1;
}
}
}
I didn't code on C for a long time, but I hope it'll help.
This procedure will print you the answer, and you have to call it just once.

Answer 4

I suggest to use std::map which is the best solution to solve your problem. I will try to explain easily the differents steps to do this:
I consider your variables initialized, for instance:
int length = 9;
int nums[length] = {19, 5, 26, 5, 5, 19, 16, 8, 1};
Create the std::map<int,int>, where the key (first int) will be your number and the value (second int) the number of occurence of this number store in the key.
std::map<int,int> listNumber;
Fill your map
// For all numbers read in your file
for(int i=0; i<length; ++i)
{
// Get number value
int n = nums[i];
// Find number in map
std::map<int, int>::iterator it = listNumber.find(n);
// Doesn't exists in map, add it with number of occurence set to 1...
if(it == listNumber.end())
{
listNumber.insert(std::pair<int,int>(n,1));
}
// ... otherwise add one to the number of occurence of this number
else
{
it->second = it->second+1;
}
}
Read the map
// Read all numbers and display the number of occurence
std::cout << "N" << "\t" << "Count" << std::endl;
for(std::map<int, int>::iterator it = listNumber.begin(); it!=listNumber.end(); ++it)
{
std::cout << it->first << "\t" << it->second << std::endl;
}