Find First and Last Position of Element in Sorted Array

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.

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

Code

Time Complexity: $O(\log n)$, Space Complexity: $O(1)$

upper_bound 回傳的會是嚴格大於的位置，沒找到的話會回傳 end，因此若回傳 begin 代表 begin 的位置上必定是嚴格大於 target 的。

if(last_iter != nums.begin() && *prev(last_iter) == target) 

class Solution {
public:
    vector<int> searchRange(vector<int>& nums, int target) {
        int first, last;
        auto first_iter = lower_bound(nums.begin(), nums.end(), target);
        if(first_iter == nums.end() || *first_iter != target)
            first = -1;
        else
            first = first_iter - nums.begin();
 
        auto last_iter = upper_bound(nums.begin(), nums.end(), target);
        if(last_iter != nums.begin() && *prev(last_iter) == target) {
            last = last_iter - nums.begin() - 1;
        } else {
            last = -1;
        }
 
        return {first, last};
    }
};

Source

• Find First and Last Position of Element in Sorted Array - LeetCode