Arrays homework question - c++

Question

I have this homework question:
Write and test a program that read in n integers (max value for n is 20), each integer has
a value between 0 and 100 inclusive. You program should then print out the unique values
among the input numbers and the count of these values.
Sample input:
Enter a the number of integers = 8
Enter 8 integers: 5 6 7 6 6 17 17 35
Sample output:
Number 5: 1
Number 6: 3
Number 7: 1
Number 17: 2
Number 35: 1
This is what I did:
#include<iostream>
using namespace std;
int main(){
int a[20], n;
cout<< "Please enter the number of integers= ";
cin>> n;
cout<<"Please enter"<< n<<" integers: ";
for (int i=0; i<n; i++)
cin >> a[i];
for (int k=0; k< n; k++){
int sum=0;
for (int i=0; i< n; i++){
if (a[i]==a[k])
sum= sum+1;
}
cout<< "Number "<< a[k]<<" : "<< sum<< endl;
}
}

Answer 1

Consider that when you iterate through your list, you're checking all values with both i and k. So essentially, if you had a list of 1 1 2 2, then the first one will count itself, and the 1 at a[1]. The second 1 will count the first 1 and itself, giving you your repeated output.
A way to simplify this would be to make use of a hash_map, or some similar structure (I'm not as familiar with C++) that maps a key to a value and doesn't allow repeats. This would allow you to record the unique numbers as keys, and increment them with only one pass through the list. The advantage to using the hashMap is that you make your program linear (although I don't think that's really a concern at this stage).
The simplest way to solve your problem, however would be to use a Bin sort technique. The underlying idea here is that your number range is simply 0 to 100, meaning you could create bins for 0 to 100 and increment each one. Again, this is Java code, and doesn't have any actual input for a.
// Count is the key, it uses indexes from 0 to 100, with null values of
// 0 after initialized. Simply iterate the loop, and use the value of
// a[k] to increment the corresponding count in the count array.
// Finally, print the results
int[] a = new int[20];
int[] count = new int [101];
for (int k = 0; k < a.length; k++){
count[a[k]]++;
for (int i = 0; i < count.length; i++){
if (count[i] > 0)
System.out.println(i + ": " + count[i]);
}

Answer 2

Add another bool b[20] ,initialize it with true. Then every time you detect a[k] is a dupe, you set b[k] = false. Only print a[k] if b[k] == true
for (int k = 0; k < n; k++) {
if (!b[k]) {
continue;
}
int sum = 0;
for (int i = 0; i < n; i++) {
if (a[i] == a[k]) {
sum = sum + 1;
b[i] = false;
}
}
cout << "Number " << a[k] << " : " << sum << endl;
}

Answer 3

You have to keep a running count of the items you processed in a separate array, and before running your inner loop to count the items, check if he item you're trying to count isn't in your second array already.

Answer 4

Before you print the result, check if you already printed it for this number

Answer 5

Here's my new attempt.
A quick fix (while not the most professional) would be to create another loop checking for repeats right before printing out.
I took your current big loop and turned it into an even bigger monster.
I also tested it out and it works for me. =D
for (int k=0; k< n; k++){
int sum=0;
for (int i=0; i< n; i++)
{
if (a[i]==a[k])
sum= sum+1;
}
bool repeat = false;
for(int i = 0; i < k; i++)
{
if(a[k] == a[i])
{
repeat = true;
}
}
if(!repeat)
cout<< "Number "<< a[k]<<" : "<< sum<< endl;
}

Answer 6

An alternate implementation (and more memory-hungry with your current limit of 20 input values) would be to create an array of 100 "count" values. Increment the appropriate item for each input value, then iterate through the count array outputting non-zero values.
Apparently that description wasn't good enough... perhaps some code would help (NOTE:This code is untested, but should be enough for you to understand the concept):
#include<iostream>
using namespace std;
int main(){
int a[101], n, v;
cout<< "Please enter the number of integers= ";
cin>> n;
cout<<"Please enter"<< n<<" integers: ";
for (int i=0; i<n; i++)
{
cin >> v;
a[v] ++;
}
for (int k=0; k< 100; k++){
if (a[k] > 0)
{
cout<< "Number "<< k + 1 <<" : "<< a[k] << endl;
}
}
}
}

Answer 7

getting familiar with Standart Template Library is the key to writing good programs in my humble opinion, since this is a homework, you do the controlling for 0 and 100 ;-))
#include <iostream>
#include <map>
using std::cin;
using std::map;
using std::cout;
using std::endl;
int main()
{
int limit = 20;
int cnt=0;
int n;
map<int, int> counters;
while( cnt++ < limit )
{
cin >> n;
++counters[n];
}
for(map<int, int>::iterator it = counters.begin();
it!=counters.end(); ++ it)
cout << it->first << " " << it->second << endl;
return 0;
}