/*Classical Binary Search

Find any position of a target number in a sorted array. Return -1 if target does not exist.

Example

Given [1, 2, 2, 4, 5, 5].

For target = 2, return 1 or 2.

For target = 5, return 4 or 5.

For target = 6, return -1.

Challenge O(logn) time

Tags Binary Search

*/

public class ClassicalBinarySearch {

/**

* @param A an integer array sorted in ascending order

* @param target an integer

* @return an integer

*/

public static int findPosition(int[] A, int target) {

// Write your code here

if (A == null || A.length == 0) {

return -1;

}

int start = 0;

int end = A.length - 1;

while (start + 1 < end) {

int mid = start + (end - start) / 2;

if (A[mid] == target) {

return mid;

} else if (A[mid] < target) {

start = mid;

} else {

end = mid;

}

}

if (A[start] == target ) {

return start;

} else if (A[end] == target ) {

return end;

}

return -1;

}

public static void main (String[] args) {

int[] test = {0,1,4,6,6,7,8,14,16,18};

int target = 6;

int result = findPosition(test, target);

System.out.println("Test result should be 4");

System.out.println(result);

}

}