First Position of Target

For a given sorted array (ascending order) and a target number, find the first index of this number in O(log n) time complexity.

If the target number does not exist in the array, return -1.

Example:

If the array is [1, 2, 3, 3, 4, 5, 10], for given target 3, return 2.

Solution:

O(logn) and sorted array indicate this is a binary search problem.

To find the first duplicate, when nums[mid] == target, assign mid to end. This will drop all elements which after mid. The first element is kept either in nums[start] _or nums[mid]_, when there are only two elements left.

Code:

public int binarySearch(int[] nums, int target) {
    if (nums == null || nums.length == 0) {
        return -1;
    }
    int start = 0;
    int end = nums.length - 1;
    while (start + 1 < end) {
        int mid = start + (end - start) / 2;
        if (nums[mid] < target) {
            start = mid;
        } else {
            end = mid;
        }
    }
    if (nums[start] == target) {
        return start;
    }
    if (nums[end] == target) {
        return end;
    }
    return -1;
}