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[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

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
        }
    }
}

468. 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

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"
}