How to count the frequency of the number x

Question

I want to print the first number that first counts to 4, for example, I have this random function, I want to see the first number that reaches 4 times. so it's the first number that prints himself 4 times. For example:

    int n;
    int count1 = 0;
    int count2 = 0;
    int count3 = 0;
    while (true) {
        cout << "Enter a number btween 1-3" << endl;
        cin >> n;
        if (n == 1) {
            count1++;
        }
        if (n == 2) {
            count2++;
        }
        if (n == 3) {
            count3++;
        }
        if (count1 == 4) {
            cout << "You printed the number 1 4 times!";
            break;
        }
        if (count2 == 4) {
            cout << "You printed the number 2 4 times!";
            break;
        }
        if (count3 == 4) {
            cout << "You printed the number 3 4 times!";
            break;
        }

But what would I do if it was 1-1000 numbers not just 1-3 what would I do then?

I want to do that but on a random function - the first number that the count of that number is 4 times print the number -

int fun() {
    srand(time(NULL));
    return rand() % 3;
}

Then I want to do in main that first number that reaches for example 4 times print this number.

I tried doing something like this:

for (int i = 0; i < 31; i++) {
        arr[fun()]++;
        cout << arr[fun()];
        if (arr[fun()] == 4) {
            cout << arr[fun()];
        }
    }

Answer 1

You would use an collection (such as a vector) for that, rather than a thousand separate variables :-)

For a start, if you want random numbers in the range 1..3, you would use (rand() % 3) + 1. However, you can use the range 0..n-1 rather than 1..n and just adjust the value after the loop.

First step is to create and initialise the counts of each number to zero:

const int SZ = 1000;
std::vector<int> count(SZ, 0);

Then your loop just generates random numbers and adjusts the relevant count, until one of them reaches the target value:

int num;
for (;;) { // infinite loop
    num = rand() % SZ;
    ++count[num];
    if (count[num] == 4)
        break; // exit infinite loop when one of them reaches four.
}

Then you just simply output the one that reached four first. Note that, since we're doing 0..999, we map that to 1.1000:

std::cout << ++num << " reached a count of four first\n";

---

A complete program showing this can be seen below:

#include <iostream>
#include <vector>
#include <cstdlib>
#include <ctime>

int main() {
    srand(time(nullptr));

    const int SZ = 1000;
    std::vector<int> count(SZ, 0);

    int num;
    for (;;) { // infinite loop
        num = rand() % SZ;
        ++count[num];
        if (count[num] == 4)
            break; // exit loop when one of them reaches four.
    }

    std::cout << ++num << " reached a count of four first\n";
}

A sample run of that (ensuring a delay so random number generator gets different seed):

>> for i in {1..10} ; do sleep 1 ; ./testprog ; done )
296 reached a count of four first
520 reached a count of four first
205 reached a count of four first
239 reached a count of four first
822 reached a count of four first
260 reached a count of four first
421 reached a count of four first
444 reached a count of four first
21 reached a count of four first
92 reached a count of four first

Answer 2

The last answer was out of range. My answer is using only an array

while(1) {
        int f = rand() % 3;
        count[f]++;
        cout << f << endl;
        if (count[f] == 4) {
            cout <<"4 times" <<f;
            break;
        }
    }