Efficient way to count occurrences of a key in a sorted array

Question

This was asked in on-site Microsoft interview.

Count the number of occurrences of a given key in an array.

I answered linear search because the elements may be s

Answer 1

** Time Complexity = O ( lg N ) where N is the size of array

** Arguments for binarySearchXXXXX:**

1. int[] array is a sorted array of length >= 1
2. int k : key to search

**

package array;

 public class BinarySearchQuestion {

public static int binarySearchFirst(int[] array, int k) {
    int begin = 0;
    int end = array.length-1;
    int mid = -1;
    while (begin <= end) {
        mid = begin + (end - begin) / 2;
        if (array[mid] < k) {
            begin = mid + 1;
        } else {
            end = mid - 1;
        }
    }
    //System.out.println("Begin index :: " + begin + " ,  array[begin] " + array[begin]);
    return (begin <= array.length - 1  && begin >= 0 && array[begin] != k) ? -1 : begin;
    //      return begin;
}

public static int binarySearchLast(int[] array, int k) {
    int begin = 0;
    int end = array.length - 1;
    int mid = -1;
    while (begin <= end) {
        mid = begin + (end - begin) / 2;
        if (array[mid] > k) {
            end = mid - 1;
        } else {
            begin = mid + 1;
        }
    }
    //System.out.println("Last index end :: " + end + " ,  array[mid] " + array[end]);
    return (end <= array.length - 1  && end >= 0 &&  array[end] != k) ? -1 : end;
    //return end;
}

/**
 * @param args
 */
public static void main(String[] args) {
             //     int[] array = { 0, 1,1,1, 2, 3, 4,4,4,5, 5, 5, 5, 5, 5, 5, 5, 5, 5,5,6,6,6,6, 6, 7, 7, 7,
             //             7, 8, 9 };
            //      int[] array = {-1, 0,1, 1,1,2,3};
    int[] array = {1,1,1};

    int low = binarySearchFirst(array, 1);
    int high = binarySearchLast(array, 1);
    int total = (high >= low && low != -1 && high != -1) ? ( high - low + 1 ): 0;
    System.out.println("Total Frequency " + total);
}

   }