Search for a Range

01/16/2018 update
Solution with current chosen template. Longer but much safer. Note that when finding upper bound, check r index first.

class Solution {
    public int[] searchRange(int[] nums, int target) {
        if (nums.length == 0) {
            return new int[]{-1, -1};
        }
        int start = -1, end = -1;


        int low = 0, high = nums.length - 1;  
        while (low + 1< high) {  
            int mid = low + (high - low)/2;  
            if (nums[mid] >= target){  
                high = mid;  
            }  else {
                low = mid;
            }
        }
        if (nums[low] == target) start = low;
        else  if (nums[high] == target) start = high;
        else return new int[]{-1, -1};

        low = 0; 
        high = nums.length - 1;  
        while (low + 1< high) {  
            int mid = low + (high - low)/2;  
            if (nums[mid] <= target){  
                low = mid;  
            }  else {
                high = mid;
            }
        }
        if (nums[high] == target) end = high;
        else  if (nums[low] == target) end = low;
        else return new int[]{-1, -1};

         return new int[]{start, end};
    }


}

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Problem:

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

Your algorithm's runtime complexity must be in the order of O(log n).

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

For example,
Given [5, 7, 7, 8, 8, 10] and target value 8,
return [3, 4].

Analysis:

Binary search's capability has been extended by this problem. Use binary search to find lower bound and upper bound separately. 

lower bound: midValue >= target go left (hi = mid)

upper bound: midValue <= target go right (lo = mid + 1)

Solution:

class Solution {
    public int[] searchRange(int[] nums, int target) {
        
        int lo = findBoundry(nums, target, true);
        int hi = findBoundry(nums, target, false) - 1;
        if (lo == nums.length || nums[lo] != target) return new int[]{-1, -1};
        else return new int[]{lo, hi};
    }

    private int findBoundry(int[] nums, int target, boolean left) {
        int low = 0, high = nums.length;  
        while (low < high) {  
            int mid = low + (high - low)/2;  
            if (nums[mid] > target || (left &&  nums[mid] == target)){  
                high = mid;  
            }  else {
                low = mid + 1;
            }
        }
        return low;

    }
}