/*
34. Find First and Last Position of Element in Sorted Array
Medium - 42.9%

Given an array of integers nums sorted in non-decreasing order, find the starting and ending 
position of a given target value.

If target is not found in the array, return [-1, -1].

You must write an algorithm with O(log n) runtime complexity.

Example 1:
Input: nums = [5,7,7,8,8,10], target = 8
Output: [3,4]

Example 2:
Input: nums = [5,7,7,8,8,10], target = 6
Output: [-1,-1]

Example 3:
Input: nums = [], target = 0
Output: [-1,-1]
*/

class Solution {
    public int[] searchRange(int[] nums, int target) {
        int[] result = {-1, -1};
        
        if (nums.length == 0) return result;
        
        // Find first position
        int first = findFirst(nums, target);
        if (first == -1) return result;
        
        // Find last position
        int last = findLast(nums, target);
        
        return new int[]{first, last};
    }
    
    private int findFirst(int[] nums, int target) {
        int left = 0, right = nums.length - 1;
        int first = -1;
        
        while (left <= right) {
            int mid = left + (right - left) / 2;
            
            if (nums[mid] == target) {
                first = mid;
                right = mid - 1; // Continue searching left for first occurrence
            } else if (nums[mid] < target) {
                left = mid + 1;
            } else {
                right = mid - 1;
            }
        }
        
        return first;
    }
    
    private int findLast(int[] nums, int target) {
        int left = 0, right = nums.length - 1;
        int last = -1;
        
        while (left <= right) {
            int mid = left + (right - left) / 2;
            
            if (nums[mid] == target) {
                last = mid;
                left = mid + 1; // Continue searching right for last occurrence
            } else if (nums[mid] < target) {
                left = mid + 1;
            } else {
                right = mid - 1;
            }
        }
        
        return last;
    }
}