Filling an 1D array in C++ - c++

Question

I have an integer array:
int listint[10] = {1,2,2,2,4,4,5,5,7,7,};
What I want to do is to create another array in terms of the multiplicity. So I define another array by:
int multi[7]={0};
the first index of the multi array multi[0] will tell us the number of multiplicity of the array listint that has zero. We can easily see that, there is no zero in the array listint, therefore the first member would be 0. Second would be 1 spice there are only 1 member in the array. Similarly multi[2] position is the multiplicity of 2 in the listint, which would be 3, since there are three 2 in the listint.
I want to use an for loop to do this thing.
#include <iostream>
#include <stdio.h>
using namespace std;
int main()
{
unsigned int count;
int j;
int listint[10] = { 1,2,2,2,4,4,5,5,7,7, };
int multi[7] = { 0 };
for (int i = 0; i < 9; i++)
{
if (i == listint[i])
count++;
j = count;
multi[j] = 1;
}
cout << "multi hit \n" << multi[1] << endl;
return 0;
}
After running this code, I thought that I would want the multiplicity of the each element of the array of listint. So i tried to work with 2D array.
#include <iostream>
#include <stdio.h>
using namespace std;
int main()
{
unsigned int count;
int i, j;
int listint[10] = { 1,2,2,2,4,4,5,5,7,7, };
int multi[7][10] = { 0 };
for (int i = 0; i < 9; i++)
{
if (i == listint[i])
count++;
j = count;
for (j = 0; j < count; j++) {
multi[j][i] = 1;
}
}
cout << "multi hit \n" << multi[4][i] << endl;
return 0;
}
The first code block is something that I wanted to print out the multiplicity. But later I found that, I want in a array that multiplicity of each elements. SO isn't the 2D array would be good idea?
I was not successful running the code using 2D array.
Another question. When I assign j = count, I mean that that's the multiplicity. so if the value of count is 2; I would think that is a multiplicity of two of any element in the array listint.

Answer 1

A 2d array is unnecessary if you're just trying to get the count of each element in a list.
#include <iostream>
int main() {
int listint[10] = { 1,2,2,2,4,4,5,5,7,7, };
int multi[8] = { 0 };
for (int i : listint)
++multi[i];
for (int i = 0; i < 8; ++i)
std::cout << i << ": " << multi[i] << '\n';
return 0;
}
There's also a simpler and better way of doing so using the standard collection std::map. Notably, this doesn't require you to know what the largest element in the array is beforehand:
#include <map>
#include <iostream>
int main() {
int listint[10] = {1,2,2,2,4,4,5,5,7,7,};
std::map<int, int> multi;
for (int i : listint)
multi[i]++;
for (auto [k,v] : multi)
std::cout << k << ": " << v << '\n';
}

Answer 2

Try this incase maps won't work for you since you're a beginner, simple:
#include <iostream>
#include <stdio.h>
using namespace std;
int main()
{
unsigned int count;
int j;
int listint[10] = {1,2,2,2,4,4,5,5,7,7};
int multi[8]={0};
for(int i=0; i<10; i++)
{
multi[listint[i]]++; // using listint arrays elements as index of multi to increase count.
}
for( int i=1; i<8; i++)
{
cout << "multi hit of "<<i<<" : "<< multi[i]<<endl;
}
return 0;
}
OR if numbers could get large and are unknown but sorted
#include <iostream>:
#include <stdio.h>
using namespace std;
int main()
{
unsigned int count = 0;
int index = 0; // used to fill elements in below arrays
int Numbers[10] = {0}; // storing unique numbers like 1,2,4,5,7...
int Count[10] = {0}; // storing their counts like 1,3,2,2,2...
int listint[10] = {1, 2, 2, 2, 4, 4, 5, 5, 7, 7};
for(int i = 0; i < sizeof(listint) / sizeof(listint[0]); i++)
{
count++;
if (listint[i] != listint[i+1]) {
Numbers[index] = listint[i];
Count[index] = count;
count=0;
index++;
}
}
for(int i=0; i<index; i++)
{
cout << "multi hit of "<<Numbers[i]<<" is " << Count[i]<<endl;
}
return 0;
}