34. Find First and Last Position of Element in Sorted Array

2020, Nov 06 2 mins read

1. Description

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.

2. Example

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]

3. Constraints

0 <= nums.length <= 10$^{5}$
-10$^{9}$ <= nums[i] <= 10$^{9}$
nums is a non-decreasing array.
-10$^{9}$ <= target <= 10$^{9}$

4. Solutions

Binary Search

n = nums.size()
Time complexity: O(logn)
Space complexity: O(1)

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
class Solution {
public:
    vector<int> searchRange(vector<int> &nums, int target) {
        const int n = nums.size();

        int left = 0, right = n;
        while (left < right) {
            int mid = left + (right - left) / 2;
            if (nums[mid] < target) {
                left = mid + 1;
            } else {
                right = mid;
            }
        }

        if (left == n || nums[left] != target) {
            return {-1, -1};
        }

        int left_border = left;

        left = 0;
        right = n;
        while (left < right) {
            int mid = left + (right - left) / 2;
            if (nums[mid] <= target) {
                left = mid + 1;
            } else {
                right = mid;
            }
        }

        return {left_border, left - 1};
    }
};