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