Majority element

/********
A majority element in an array A, of size N is an element that appears more than N/2 times (and hence there is atmost
one such element)

Write a function which takes an array and emits the majority element (if it exists), otherwise prints NONE as follows:

I/P : 3 3 4 2 4 4 2 4 4
O/P : 4

I/P : 3 3 4 2 4 4 2 4
O/P : NONE

Answer:

As we sweep through the sequence, we maintain a pair consisting of a current candidate and a counter. Initially, the current candidate is unknown and the counter is 0. When we move the pointer forward over an element e:
If the counter is 0, we set the current candidate to e and we set the counter to 1.
If the counter is not 0, we increment or decrement the counter according to whether e is the current candidate. When we are done, the current candidate is the majority element, if there is a majority.
If you must confirm that the chosen element is indeed the majority element, you must take one more linear pass through the data to do it.
******/
int find_majority(int *a, int n)
{
int count = 1;
currentIndex = 0;
for (i = 1 ; i lessThan n; i ++)
{
if (a[i] == a[currentIndex])
count++;
else
count--;
if (count == 0)
{
currentIndex = i;
count = 1;
}

}
return a[currentIndex];
}