Can someone give me an algorithm to count distinct elements of an array of integers in one pass.
for example i can try to traverse through the array using a for loop
I will store the first element in another array.And the subsequent elements will be compared with those in the second array and if it is distinct then i will store it in that array and increment counter.
can someone give me a better algorithm than this.
Using c and c++
Supposing that your elements are integers and their values are between 0 and MAXVAL-1.
#include <stdio.h>
#include <string.h>
#define MAXVAL 50
unsigned int CountDistinctsElements(unsigned int* iArray, unsigned int iNbElem) {
unsigned int ret = 0;
//this array will contains the count of each value
//for example, c[3] will contain the count of the value 3 in your original array
unsigned int c[MAXVAL];
memset(c, 0, MAXVAL*sizeof(unsigned int));
for (unsigned int i=0; i<iNbElem; i++) {
unsigned int elem = iArray[i];
if (elem < MAXVAL && c[elem] == 0) {
ret++;
}
c[elem]++;
}
return ret;
}
int main() {
unsigned int myElements[10] = {0, 25, 42, 42, 1, 2, 42, 0, 24, 24};
printf("Distincts elements : %d\n", CountDistinctsElements(myElements, 10));
return 0;
}
Output : (Ideone link)
Distincts elements : 6
memset initialize all values of c to zero.Maintain a array of structures. structure should have a value and a counter of that value. As soon as you pass an new element in an array being tested create a structure with value and increment the counter by 1.if you pass an existing element in the array then simply access the related structure and increment its counter by 1. Finally after you do a one complete pass of the array, you will have the required result in the array of structures.
Edit: I wasn't aware you wanted just to count the elements. Updated code below.
int countUnique()
{
uniqueArray[numElements];
myArray[numElements];
int counter = 0;
int uniqueElements = 0;
for(int i = 0; i < numElements; i++)
{
element tempElem = myArray[i];
if(!doesUniqueContain(tempElem, counter, uniqueArray)//If it doesn't contain it
{
uniqueArray[counter] = tempElem;
uniqueElements++;
}
}
return uniqueElements;
}
bool doesUniqueContain(element oneElement, int counter, array *uniqueArray)
{
if(counter == 0)
{
return false; //No elements, so it doesn't contain this element.
}
for(int i = 0; i < counter; i++)
{
if(uniqueArray[i] == oneElement)
return true;
}
return false;
}
This is only so you can see the logic
How about using a hash table (in the Java HashMap or C# Dictionary sense) to count the elements? Basically you create an empty hash table with the array element type as the key type and the count as values. Then you iterate over your list. If the element is not yet in the hash table, you add it with count 1, otherwise you increment the count for that element.
