I had a question in c++

**it-girl** · 17-03-2011

hi .
i have two questions in my book .
that i can't understand them

The mode of an array of numbers is the number m in the array that is repeated most frequently. If more than one number is repeated with equal maximal frequencies, there is no mode. Write a program that accepts an array of numbers and returns the mode or an indication that the mode does not exist

who can write this code .
my idea is i should be their two function the first is to sort the array aompare the elements .
and other function is to chick if there mode or not
if any one can explain ..
because i did't study the sorting until now .

2
the median of an array of numbers is the element m of the array such that half the remaining numbers in the array are greater than or equal to m and half are less than or equal to m . if the numbers of elements in the array is odd . if the number of elements is even .the median is the average of the two elements m1 and m2 , such that half the remaining elements are greater than or equal to m1,m2 and half the elements are less than or equal to m1,m2 . write a c function that accepts an array of numbers and return the median of the numbers in the array

what the mean of median ?

....
also
plz if you have any goog refrence to learn programming
that can help me .

thanks alot

...

**Janos™** · 18-03-2011

Why cant you just create a 2-dimensional array in function mode(), that contains the value in the original array and the count of the number of times it appears in the original array.

Code:

int mode (int arr[MAX])
{
int totals[2][MAX] = {0};
int i, nItemsAdded = 0;
for(i = 0; i < MAX; i++)
{
   // search totals to see if the value is already there.  If not, then
   // add it.  If its already in totals array then just increment
   // the counter for it.
   int j;
   int found = 0;
   for(j = 0; found == 0 && j < nItemsAdded; ++j)
   {
        if( totals[0][j] == arr[j])
        {
           found = 1; // set flag indicating value found
           ++totals[1][j]; // bump count
         }
    }
    if( found == 0)
    {
       totals[0][nItemsAdded] = arr[i];
       totals[1][nItemsAdded] = 1;
       ++nItemsAdded;
     }
}