Find how many times numbers appear in array

My program should print how many times each number appears in array.

OUTPUT:

Number 1 appears 2 times

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Number 2 appears 5 times

Number 3 appears 1 times

Number 4 appears 3 times

Most frequent: 2

This is my output:

Number 1 appears 1 times
Number 2 appears 4 times
Number 1 appears 0 times
Number 4 appears 2 times
Number 2 appears 3 times
Number 4 appears 1 times
Number 2 appears 2 times
Number 2 appears 1 times
Number 3 appears 0 times
Number 4 appears 0 times
Number 2 appears 0 times
Most frequent: 2

#include <stdio.h>

void main() {
    int i, j, count = 0, max_count = 0, max, n;
    int arr[] = {1, 2, 1, 4, 2, 4, 2, 2, 3, 4, 2}, counter[1000] = {0};
    n = sizeof(arr) / sizeof(*arr);

    for (i = 0; i < n; i++) {
        count = 0;

        for (j = i + 1; j < n; j++) {
            if (arr[i] == arr[j]) {
                count++;
                counter[i]++;
            }
            if (count > max_count) {
                max = arr[j];
                max_count = count;
            }
        }
    }

    for (i = 0; i < n; i++)
        printf("Number %d appears %d times\n", arr[i], counter[i]);

    printf("Most frequent: %d\n", max);
}

Could you help me fix my code?

>Solution :

Try this (see below). One of your main problems is that you are using counter[i] but that is using a loop index, not the actual number from the array. So you need to use counter[arr[i]]. The other problem is that you are using two nested loops, which will double up your counts.

Note: The quicksort compare function is taken from this answer.

#include <stdio.h>
#include <string.h>
#include <stdlib.h>

int compare( const void* a, const void* b)
{
     int int_a = * ( (int*) a );
     int int_b = * ( (int*) b );

     if ( int_a == int_b ) return 0;
     else if ( int_a < int_b ) return -1;
     else return 1;
}

int main()
{
  int i, j, count = 0, max_count = 0, max, n;
  int arr[] = {1, 2, 1, 4, 2, 4, 2, 2, 3, 4, 2}, counter[1000] = {0};
  n = sizeof(arr) / sizeof(*arr);
  qsort( arr, n, sizeof(int), compare );
  for (i = 0; i < n; i++) {
    ++counter[arr[i]];
  }
  max = 0;
  for (i = 0; i < n; i++) {
    if (counter[arr[i]] > max) {
      max = arr[i];
    }
  }
  
  int seen[1000];
  memset(seen,0,sizeof seen);
  for (i = 0; i < n; i++) {
    if (seen[arr[i]])
      continue;
    seen[arr[i]] = 1;
    printf("Number %d appears %d times\n", arr[i], counter[arr[i]]);
  }
  printf("Most frequent: %d\n", max);
}

Output:

---------- Capture Output ----------
> "c:\windows\system32\cmd.exe" /c C:\Temp\temp.exe
Number 1 appears 2 times
Number 2 appears 5 times
Number 3 appears 1 times
Number 4 appears 3 times
Most frequent: 4

> Terminated with exit code 0.