how to check if specific element not existed in array

Question

What I try to do is count the frequency of input number but now my problem is it will print an already printed element.

Ex.

Input: 1 3 5 2 3 1

It will output:

Expect ouput:

1 2
3 2
5 1

How do I fix this any suggestion?

#include<stdio.h>
int n = 0;
int number[100] = {0};
int frequency[100] = {0};
int count = 0;
int i = 0, j = 0, k = 0;

int main()
{
    scanf("%d",&n);
    for(count=0; count<n; count++)
    {
        scanf("%d",&number[count]);
    }
    ///////////////////////////////////////////////
    for(i = 0; i<n; i++)
    {
        for(j = 0; j<n; j++)
        {
            if(number[i] == number[j])
            {
                frequency[k] = frequency[k]+1;
            }
        }
        k = k+1;
    }
    ////////////////////////////////////////////////
    for(k=0, i=0; k<n; i++, k++)
    {
                printf("%d %d\n",number[i],frequency[k]);
    }
}

Beside other errors in your algorithm; (1) this will only work if your data (`number`) is in range from 0 to 99, inclusive, (2) if 1 holds true, you do not need nested loop, (3) in any case, `k` is redundant in your nested loop. — ssd, Sep 10 '22 at 08:55
Are you sure that your posted input is your actual input? The first number entered should specify how many numbers follow. — Andreas Wenzel, Sep 10 '22 at 12:36

Andreas Wenzel · Answer 1 · 2022-09-11T14:19:07.033

The nested loop with 3 different loop counters is completely unnecessary. Also, the final loop is wrong. For example, it does not make sense to limit that loop to n iterations.

Here is a working solution:

#include<stdio.h>

//numbers must be in the range of 0 to NUM_MAX
#define NUM_MAX 99

//maximum number of numbers that this program can handle
#define MAX_NUMBERS 100

int main()
{
    int n;
    int numbers[MAX_NUMBERS];
    int frequencies[NUM_MAX+1] = {0};

    //prompt user for input
    printf( "How many numbers? " );

    //read amount of numbers from user
    scanf( "%d", &n );

    //read the individual numbers from user
    for( int i = 0; i < n; i++ )
    {
        //prompt user
        printf( "Enter #%d: ", i + 1 );

        //read number
        scanf( "%d", &numbers[i] );
    }

    //count the frequencies of the individual numbers
    for( int i = 0; i < n; i++ )
    {
        frequencies[numbers[i]]++;
    }

    //print the frequencies
    for ( int i = 0; i < NUM_MAX + 1; i++ )
    {
        if ( frequencies[i] != 0 )
            printf( "The number %d occurred %d times.\n", i, frequencies[i] );
    }
}

This program has the following behavior:

How many numbers? 6
Enter #1: 1
Enter #2: 3
Enter #3: 5
Enter #4: 2
Enter #5: 3
Enter #6: 1
The number 1 occurred 2 times.
The number 2 occurred 1 times.
The number 3 occurred 2 times.
The number 5 occurred 1 times.

Note that this program does not perform any input validation at all, so that it will likely misbehave if the input is invalid.

how to check if specific element not existed in array

1 Answers1