How to check how many times a number has been entered - c++

Question

I'm trying to solve one of the questions on a task sheet I've been received, to help me further in my understanding of C++ code from my class.
The question is (and I quote):
Write a program that:
Asks the user to enter 10 numbers between 1 and 5 into an array and displays the array on screen
Creates a second array of size 5 and fills it with zeros
Counts how many 1s, 2s, , … 5s have been entered into the first array and stores this number in the second array.
Displays the second array as shown in the example below.
The problem is how to go about checking how many times a number was entered. I was thinking of a for loop, but the way I wrote it is fundamentally incorrect, so I find myself struggling to see the mistake I am having. Perhaps I am missing something simple? Any help would be great.
Here is my (terrible) for loop attempt, so you can see my error.
#include <iostream>
#include <windows.h>
using namespace std;
int main()
{
int input[10];
const int MAX_NO = 5;
int COUNT[5] = { 0,0,0,0,0 };
int count = 10;
for (int i = 0; i < count; i++)
{
cout << "Please enter a number for value " << i + 1 << " :";
cin >> input[i];
while (input[i] < 1 || input[i] > 5)
{
cout << "Error: Enter another number between 1 and 5: ";
cin >> input[i];
}
}
cout << endl << "You entered ";
for (int i = 0; i < count; i++)
{
cout << input[i] << " ";
}
cout << "\n";
// show how many times 1 number appears
for (int i = 1; i <= 5; i++)
{
if (input[i] == i)
{
COUNT[i]++;
}
}
for (int i = 0; i < MAX_NO; i++)
{
cout << i + 1 << " appears " << COUNT[i]
<< " times in the input" << endl;
}
cout << endl;
system("pause");
return 0;
}

Answer 1

Put
COUNT[ input[i]-1 ]++;
in your first loop (after validation). Once you do that, you don't need a second loop to tally up the results.
This works from the inside out by first getting what input[i] is, then using it to modify the (input[i]-1)'th location in the COUNT array. If the user enters 4 on the first run of the loop, then i == 0 and input[i] == 4. Since arrays are 0-based, it will increment COUNT[input[i]-1] which in this case is COUNT[4-1] == COUNT[3].
After your initial loop runs the number of 1's will be in COUNT[0], the number of 2's will be in COUNT[1] and so on.

Answer 2

#include <iostream>
#include <windows.h>
using namespace std;
int main()
{
//declare a constant values
int input[10];
int count = 10; //all constant MUST be in capital letters
//second array filled with zeros
const int MAX_NO = 5;
int COUNT[5] = { 0, 0, 0, 0, 0 };
//ask user for 10 input values
for (int i = 0; i < count; i++)
{
cout << "Please enter a number for value " << i + 1 << " :";
cin >> input[i];
//check if input numbers are between 1 and 5 inclusive
while (input[i] < 1 || input[i] > 5)
{
cout << "Error: Enter another number between 1 and 5: ";
cin >> input[i];
}
/* show how many times 1 number appears.
this section should be in the main loop which would enable the program to check how many times a
number is entered so that it is stored in the second array. changed i to secondCount because this is the counting index of the second array not the first which you've called i (one of the reason you'd all zero as output when u ran your code)*/
for (int secondCount = 1; secondCount <= MAX_NO; secondCount++)
{
if (input[i] == secondCount)
{
COUNT[secondCount-1]+= 1; //use minus 1 from i and increment. += 1 is the same as COUNT++
}
}
}
//display number entered in the first array
cout << endl << "You entered ";
for (int i = 0; i < count; i++)
{
cout << input[i] << " ";
}
cout << "\n";
//display how many times a number is entered.
for (int secondCount = 0; secondCount < MAX_NO; secondCount++)
{
cout << secondCount + 1 << " appears " << COUNT[secondCount]
<< " times in the input" << endl;
}
cout << endl;
system("pause");
return 0;
}
OUTPUT:
Please enter a number for value 1 = 1
Please enter a number for value 2 = 1
Please enter a number for value 3 = 1
Please enter a number for value 4 = 2
Please enter a number for value 5 = 3
Please enter a number for value 6 = 2
Please enter a number for value 7 = 4
Please enter a number for value 8 = 4
Please enter a number for value 9 = 3
Please enter a number for value 10 = 2
You entered: 1 1 1 2 3 2 4 4 3 2
1 appears 3 times in the input
2 appears 3 times in the input
3 appears 2 times in the input
4 appears 2 times in the input
5 appears 0 times in the input

Answer 3

for (int i = 1; i <= 5; i++)
{
if (input[i] == i)
{
COUNT[i]++;
}
}
Let's examine what this is doing. It starts by checking input[1]. (This should be input[0] as array indices start at 0). Then it checks if input[1] is equal to 1. If it is, then it increments COUNT[1].
Next, it checks input[2]. Then it checks if input[2] is equal to 2. If it is, it increments COUNT[2]. And so on until it has gone through input[5].
Do you see the problem with this? You're only checking the first 5 inputs and only checking them against a single value.