# 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; 

}