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

https://leetcode.com/problems/find-first-and-last-position-of-element-in-sorted-array/

Problem

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

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

Follow up: Could you 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]

Constraints:

• 0 <= nums.length <= 105

• -109 <= nums[i] <= 109

• nums is a non-decreasing array.

• -109 <= target <= 109

Solution

class Solution {
public:
    vector<int> searchRange(vector<int>& nums, int target) {
        vector<int> null{-1, -1};
        if (nums.empty()) return null;
        int lower = lowerBound(nums, target);
        if (lower == -1) return null;
        return vector<int>{lower, upperBound(nums, target)};
    }
    inline int lowerBound(vector<int>& nums, int target) {
        int begin = 0, end = nums.size() - 1;
        while (begin < end) {
            int mid = begin + (end - begin) / 2;
            if (nums[mid] < target) begin = mid + 1;
            else end = mid;
        }
        return nums[begin] == target ? begin : -1;
    }
    
    inline int upperBound(vector<int>& nums, int target) {
        int begin = 0, end = nums.size() - 1;
        while (begin < end) {
            int mid = begin + (end - begin + 1) / 2;
            if (nums[mid] > target) end = mid - 1;
            else begin = mid;
        }
        return nums[begin] == target ? begin : -1;
    }
};

• #binarysearch