Count the number of occurrences of N in the array A
# include <stdio.h>
/* if x is present in arr[] then returns
the index of FIRST occurrence
of x in arr[0..n-1], otherwise returns -1 */
int first(int arr[], int low, int high, int x, int n)
{
if(high >= low)
{
int mid = (low + high)/2; /*low + (high - low)/2;*/
if( ( mid == 0 || x > arr[mid-1]) && arr[mid] == x)
return mid;
else if(x > arr[mid])
return first(arr, (mid + 1), high, x, n);
else
return first(arr, low, (mid -1), x, n);
}
return -1;
}
/* if x is present in arr[] then returns the
index of LAST occurrence of x in arr[0..n-1],
otherwise returns -1 */
int last(int arr[], int low, int high, int x, int n)
{
if (high >= low)
{
int mid = (low + high)/2; /*low + (high - low)/2;*/
if( ( mid == n-1 || x < arr[mid+1]) && arr[mid] == x )
return mid;
else if(x < arr[mid])
return last(arr, low, (mid -1), x, n);
else
return last(arr, (mid + 1), high, x, n);
}
return -1;
}
/* if x is present in arr[] then returns the count
of occurrences of x, otherwise returns -1. */
int count(int arr[], int x, int n)
{
int i; // index of first occurrence of x in arr[0..n-1]
int j; // index of last occurrence of x in arr[0..n-1]
/* get the index of first occurrence of x */
i = first(arr, 0, n-1, x, n);
/* If x doesn't exist in arr[] then return -1 */
if(i == -1)
return i;
/* Else get the index of last occurrence of x.
Note that we are only looking in the subarray
after first occurrence */
j = last(arr, i, n-1, x, n);
/* return count */
return j-i+1;
}
/* driver program to test above functions */
int main()
{
int arr[] = {1, 2, 2, 3, 3, 3, 3};
int x = 3; // Element to be counted in arr[]
int n = sizeof(arr)/sizeof(arr[0]);
int c = count(arr, x, n);
printf(" %d occurs %d times ", x, c);
getchar();
return 0;
}