# 2022-05-29

## ​153. 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, a, a, ..., a[n-1]]` 1 time results in the array `[a[n-1], a, a, a, ..., 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

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​

### 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

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

#### Approach #1

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
}
}
}

### 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 `"[email protected]"` 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"
Constraints:
• `queryIP` consists only of English letters, digits and the characters `'.'` and `':'`.

### Solution

#### Approach #0

if parts := strings.Split(queryIP, "."); len(parts) == 4 {
for _, part := range parts {
if len(part) > 1 && part == '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"
}