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)

class Solution {
public:
    vector<int> searchRange(const vector<int> &nums, int target) {
        if (nums.empty()) {
            return {-1, -1};
        }

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

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

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

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