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

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`

34. Find First and Last Position of Element in Sorted Array
``````struct Solution;

impl Solution {
fn search_range(nums: Vec<i32>, target: i32) -> Vec<i32> {
let n = nums.len();
match nums.binary_search(&target) {
Ok(i) => {
let mut l = i;
let mut r = i;
while l > 0 && nums[l - 1] == target {
l -= 1;
}
while r + 1 < n && nums[r + 1] == target {
r += 1;
}
vec![l as i32, r as i32]
}
Err(_) => vec![-1, -1],
}
}
}

#[test]
fn test() {
let nums = vec![5, 7, 7, 8, 8, 10];
let target = 8;
let res = vec![3, 4];
assert_eq!(Solution::search_range(nums, target), res);
}
``````