34. Search for a Range

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].

Solution:

Apply two binary searches.

The first search is to find the first occurrence of target.

If we find a match, we include it and search to the left.

The second search is to find the last occurrence of target.

If we find a match, we include it and search to the right.

The time complexity is O(logn) and the space complexity is O(n).



Code:

public class Solution {
    public int[] searchRange(int[] nums, int target) {
        if (nums == null || nums.length == 0) {
            return new int[] {-1, -1};
        }
        int start = 0;
        int end = nums.length - 1;
        while (start + 1 < end) {
            int mid = start + (end - start) / 2;
            if (nums[mid] >= target) {
                end = mid;
            }
            else {
                start = mid;
            }
        }
        int first = -1;
        int last = -1;
        if (nums[start] == target) {
            first = start;
        }
        else if (nums[end] == target) {
            first = end;
        }
        else {
            return new int[] {first, last};
        }
        
        start = 0;
        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[end] == target) {
            last = end;
        }
        else if (nums[start] == target) {
            last = start;
        }
        return new int[] {first, last};
    }
}