# 2022-05-29

## [153. Find Minimum in Rotated Sorted Array](https://leetcode.com/problems/find-minimum-in-rotated-sorted-array/)

### Description

Suppose an array of length `n` sorted in ascending order is **rotated** between `1` and `n` times. For example, the array `nums = [0,1,2,4,5,6,7]` might become:

* `[4,5,6,7,0,1,2]` if it was rotated `4` times.
* `[0,1,2,4,5,6,7]` if it was rotated `7` times.

Notice that **rotating** an array `[a[0], a[1], a[2], ..., a[n-1]]` 1 time results in the array `[a[n-1], a[0], a[1], a[2], ..., a[n-2]]`.

Given the sorted rotated array `nums` of **unique** elements, return *the minimum element of this array*.

You must write an algorithm that runs in `O(log n) time.`

**Example 1:**

```
Input: nums = [3,4,5,1,2]
Output: 1
Explanation: The original array was [1,2,3,4,5] rotated 3 times.
```

**Example 2:**

```
Input: nums = [4,5,6,7,0,1,2]
Output: 0
Explanation: The original array was [0,1,2,4,5,6,7] and it was rotated 4 times.
```

**Example 3:**

```
Input: nums = [11,13,15,17]
Output: 11
Explanation: The original array was [11,13,15,17] and it was rotated 4 times. 
```

**Constraints:**

* `n == nums.length`
* `1 <= n <= 5000`
* `-5000 <= nums[i] <= 5000`
* All the integers of `nums` are **unique**.
* `nums` is sorted and rotated between `1` and `n` times.

### Solution

#### Approach #0

```go
func findMin(nums []int) int {
    n := len(nums)
    i, j := 0, n-1
    for i < j {
        mid := (j-i)/2 + i
        if nums[mid] < nums[n-1] {
            j = mid
        } else {
            i = mid + 1
        }
    }
    return nums[i]
}
```

## [162. Find Peak Element](https://leetcode.com/problems/find-peak-element/)

### Description

A peak element is an element that is strictly greater than its neighbors.

Given an integer array `nums`, find a peak element, and return its index. If the array contains multiple peaks, return the index to **any of the peaks**.

You may imagine that `nums[-1] = nums[n] = -∞`.

You must write an algorithm that runs in `O(log n)` time.

**Example 1:**

```
Input: nums = [1,2,3,1]
Output: 2
Explanation: 3 is a peak element and your function should return the index number 2.
```

**Example 2:**

```
Input: nums = [1,2,1,3,5,6,4]
Output: 5
Explanation: Your function can return either index number 1 where the peak element is 2, or index number 5 where the peak element is 6.
```

**Constraints:**

* `1 <= nums.length <= 1000`
* `-2^31 <= nums[i] <= 2^31 - 1`
* `nums[i] != nums[i + 1]` for all valid `i`.

### Solution

#### Approach #0

```go
func findPeakElement(nums []int) int {
    res := 0
    for i, num := range nums {
        if num > nums[res] {
            res = i
        }
    }
    return res
}
```

#### Approach #1

```go
func findPeakElement(nums []int) int {
    n := len(nums)
    get := func(i int) int {
        if i < 0 || i >= n {
            return math.MinInt64
        }
        return nums[i]
    }

    i, j := 0, n-1
    for {
        mid := (j-i)/2 + i
        if get(mid-1) < get(mid) && get(mid) > get(mid+1) {
            return mid
        }
        if get(mid) < get(mid+1) {
            i = mid + 1
        } else {
            j = mid - 1
        }
    }
}
```

## [468. Validate IP Address](https://leetcode.com/problems/validate-ip-address/)

### Description

Given a string `queryIP`, return `"IPv4"` if IP is a valid IPv4 address, `"IPv6"` if IP is a valid IPv6 address or `"Neither"` if IP is not a correct IP of any type.

**A valid IPv4** address is an IP in the form `"x1.x2.x3.x4"` where `0 <= xi <= 255` and `xi` **cannot contain** leading zeros. For example, `"192.168.1.1"` and `"192.168.1.0"` are valid IPv4 addresses while `"192.168.01.1"`, `"192.168.1.00"`, and `"192.168@1.1"` are invalid IPv4 addresses.

**A valid IPv6** address is an IP in the form `"x1:x2:x3:x4:x5:x6:x7:x8"` where:

* `1 <= xi.length <= 4`
* `xi` is a **hexadecimal string** which may contain digits, lowercase English letter (`'a'` to `'f'`) and upper-case English letters (`'A'` to `'F'`).
* Leading zeros are allowed in `xi`.

For example, "`2001:0db8:85a3:0000:0000:8a2e:0370:7334"` and "`2001:db8:85a3:0:0:8A2E:0370:7334"` are valid IPv6 addresses, while "`2001:0db8:85a3::8A2E:037j:7334"` and "`02001:0db8:85a3:0000:0000:8a2e:0370:7334"` are invalid IPv6 addresses.

**Example 1:**

```
Input: queryIP = "172.16.254.1"
Output: "IPv4"
Explanation: This is a valid IPv4 address, return "IPv4".
```

**Example 2:**

```
Input: queryIP = "2001:0db8:85a3:0:0:8A2E:0370:7334"
Output: "IPv6"
Explanation: This is a valid IPv6 address, return "IPv6".
```

**Example 3:**

```
Input: queryIP = "256.256.256.256"
Output: "Neither"
Explanation: This is neither a IPv4 address nor a IPv6 address.
```

**Constraints:**

* `queryIP` consists only of English letters, digits and the characters `'.'` and `':'`.

### Solution

#### Approach #0

```go
func validIPAddress(queryIP string) string {
    if parts := strings.Split(queryIP, "."); len(parts) == 4 {
        for _, part := range parts {
            if len(part) > 1 && part[0] == '0' {
                return "Neither"
            }
            if v, err := strconv.Atoi(part); err != nil || v > 255 {
                return "Neither"
            }
        }
        return "IPv4"
    }
    if parts := strings.Split(queryIP, ":"); len(parts) == 8 {
        for _, part := range parts {
            if len(part) > 4 {
                return "Neither"
            }
            if _, err := strconv.ParseUint(part, 16, 64); err != nil {
                return "Neither"
            }
        }
        return "IPv6"
    }
    return "Neither"
}
```