花花酱 LeetCode 495. Teemo Attacking

By zxi on March 8, 2018

题目大意：给你攻击的时间序列以及中毒的时长，求总共的中毒时间。

Problem:

https://leetcode.com/problems/teemo-attacking/description/

In LOL world, there is a hero called Teemo and his attacking can make his enemy Ashe be in poisoned condition. Now, given the Teemo’s attacking ascending time series towards Ashe and the poisoning time duration per Teemo’s attacking, you need to output the total time that Ashe is in poisoned condition.

You may assume that Teemo attacks at the very beginning of a specific time point, and makes Ashe be in poisoned condition immediately.

Example 1:

Input: [1,4], 2
Output: 4
Explanation: At time point 1, Teemo starts attacking Ashe and makes Ashe be poisoned immediately. 
This poisoned status will last 2 seconds until the end of time point 2. 
And at time point 4, Teemo attacks Ashe again, and causes Ashe to be in poisoned status for another 2 seconds. 
So you finally need to output 4.

Example 2:

Input: [1,2], 2
Output: 3
Explanation: At time point 1, Teemo starts attacking Ashe and makes Ashe be poisoned. 
This poisoned status will last 2 seconds until the end of time point 2. 
However, at the beginning of time point 2, Teemo attacks Ashe again who is already in poisoned status. 
Since the poisoned status won't add up together, though the second poisoning attack will still work at time point 2, it will stop at the end of time point 3. 
So you finally need to output 3.

Note:

You may assume the length of given time series array won’t exceed 10000.
You may assume the numbers in the Teemo’s attacking time series and his poisoning time duration per attacking are non-negative integers, which won’t exceed 10,000,000.

Idea: Running Process

Compare the current attack time with the last one, if span is more than duration, add duration to total, otherwise add (curr – last).

C++

// Author: Huahua

// Running time: 66 ms

class Solution {

public:

int findPoisonedDuration(vector<int>& t, int duration) {

if (t.empty()) return 0;

int total = 0;

for (int i = 1; i < t.size(); ++ i)

total += (t[i] > t[i - 1] + duration ? duration : t[i] - t[i - 1]);

return total + duration;

}

};

Java

// Author: Huahua

// Running time: 9 ms

class Solution {

public int findPoisonedDuration(int[] t, int duration) {

if (t.length == 0) return 0;

int total = 0;

for (int i = 1; i < t.length; ++ i)

total += (t[i] > t[i - 1] + duration ? duration : t[i] - t[i - 1]);

return total + duration;

}

Python3

"""

Author: Huahua

Running time: 64 ms (beats 100%)

"""

class Solution:

def findPoisonedDuration(self, timeSeries, duration):

if not timeSeries: return 0

l = timeSeries[0]

total = 0

for t in timeSeries:

total += duration if t - l > duration else t - l

l = t

return total + duration

import numpy as np

class Solution:

def findPoisonedDuration(self, timeSeries, duration):

if not timeSeries: return 0

d = np.diff(timeSeries)

d = np.clip(d, 0, duration)

return int(np.sum(d) + duration)

class Solution:

def findPoisonedDuration(self, t, duration):

return sum([min(c - l, duration) for l, c in zip(t, t[1:] + [1e9])])

花花酱 LeetCode 521. Longest Uncommon Subsequence I

By zxi on March 7, 2018

题目大意：求2个字符串最长的不同子序列的长度。

Problem:

https://leetcode.com/problems/longest-uncommon-subsequence-i/description/

Given a group of two strings, you need to find the longest uncommon subsequence of this group of two strings. The longest uncommon subsequence is defined as the longest subsequence of one of these strings and this subsequence should not be any subsequence of the other strings.

A subsequence is a sequence that can be derived from one sequence by deleting some characters without changing the order of the remaining elements. Trivially, any string is a subsequence of itself and an empty string is a subsequence of any string.

The input will be two strings, and the output needs to be the length of the longest uncommon subsequence. If the longest uncommon subsequence doesn’t exist, return -1.

Example 1:

Input: "aba", "cdc"

Output: 3

Explanation: The longest uncommon subsequence is "aba" (or "cdc"),

because "aba" is a subsequence of "aba",

but not a subsequence of any other strings in the group of two strings.

Note:

Both strings’ lengths will not exceed 100.
Only letters from a ~ z will appear in input strings.

Idea:

If two strings are the same, then the longest uncommon sequence does not exist, return -1.

e.g. aaa vs aaa, return -1

Otherwise, the longer string is always a uncommon sequence of the shorter one.

e.g. aaab vs aaa, return 4

Solution 1:

Time complexity: O(n)

Space complexity: O(1)

C++

// Author: Huahua

// Running time: 3 ms

class Solution {

public:

int findLUSlength(string a, string b) {

return a == b ? -1 : max(a.length(), b.length());

}

};

Java

// Author: Huahua

// Running time: 3 ms

class Solution {

public int findLUSlength(String a, String b) {

return a.equals(b) ? -1 : Math.max(a.length(), b.length());

}

Python3

"""

Author: Huahua

Running time: 40 ms (beats 100%)

Website: http://zxi.mytechroad.com/blog/string/leetcode-521-longest-uncommon-subsequence-i/

"""

class Solution:

def findLUSlength(self, a, b):

la, lb = len(a), len(b)

if la != lb: return max(la, lb)

return -1 if a == b else la

花花酱 LeetCode 95. Unique Binary Search Trees II

By zxi on March 7, 2018

Problem

https://leetcode.com/problems/unique-binary-search-trees-ii/description/

Given an integer n, generate all structurally unique BST’s (binary search trees) that store values 1…n.

For example,
Given n = 3, your program should return all 5 unique BST’s shown below.

1 3 3 2 1

\ / / / \ \

3 2 1 1 3 2

/ / \ \

2 1 2 3

Idea: Recursion

for i in 1..n: pick i as root,
left subtrees can be generated in the same way for n_l = 1 … i – 1,
right subtrees can be generated in the same way for n_r = i + 1, …, n
def gen(s, e):
return [tree(i, l, r) for l in gen(s, i – 1) for r in gen(i + 1, e) for i in range(s, e+1)

# of trees:

n = 0: 1
n = 1: 1
n = 2: 2
n = 3: 5
n = 4: 14
n = 5: 42
n = 6: 132
…
Trees(n) = Trees(0)*Trees(n-1) + Trees(1)*Trees(n-2) + … + Tress(n-1)*Trees(0)

Time complexity: O(3^n)

Space complexity: O(3^n)

C++

// Author: Huahua

// Running time: 16 ms

class Solution {

public:

vector<TreeNode*> generateTrees(int n) {

if (n == 0) return {};

const auto& ans = generateTrees(1, n);

cout << ans.size() << endl;

return ans;

}

private:

vector<TreeNode*> generateTrees(int l, int r) {

if (l > r) return { nullptr };

vector<TreeNode*> ans;

for (int i = l; i <= r; ++i)

for (TreeNode* left : generateTrees(l, i - 1))

for (TreeNode* right : generateTrees(i + 1, r)) {

ans.push_back(new TreeNode(i));

ans.back()->left = left;

ans.back()->right = right;

}

return ans;

}

};

Java

// Author: Huahua 4 ms

class Solution {

public List<TreeNode> generateTrees(int n) {

if (n == 0) return new ArrayList<TreeNode>();

return generateTrees(1, n);

}

private List<TreeNode> generateTrees(int l, int r) {

List<TreeNode> ans = new ArrayList<>();

if (l > r) {

ans.add(null);

return ans;

}

for (int i = l; i <= r; ++i)

for (TreeNode left : generateTrees(l, i - 1))

for (TreeNode right: generateTrees(i + 1, r)) {

TreeNode root = new TreeNode(i);

root.left = left;

root.right = right;

ans.add(root);

}

return ans;

}

Python 3

"""

Author: Huahua

Running time: 84 ms

"""

class Solution:

def generateTrees(self, n):

def newTree(x, l, r):

t = TreeNode(x)

t.left = l

t.right = r

return t

def gen(s, e):

return [newTree(i, l, r) for i in range(s, e + 1) for l in gen(s, i - 1) for r in gen(i + 1, e)] or [None]

return gen(1, n) if n > 0 else []

Solution 2: DP

Java

// Author: Huahua, 3 ms

class Solution {

public List<TreeNode> generateTrees(int n) {

if (n == 0) return new ArrayList<TreeNode>();

List<TreeNode>[] dp = new List[n + 1];

dp[0] = new ArrayList<TreeNode>();

dp[0].add(null);

for (int i = 1; i <= n; ++i) {

dp[i] = new ArrayList<TreeNode>();

for (int j = 0; j < i; ++j)

for (TreeNode left : dp[j])

for (TreeNode right: dp[i - j - 1]) {

TreeNode root = new TreeNode(j + 1);

root.left = left;

root.right = clone(right, j + 1);

dp[i].add(root);

}

return dp[n];

}

private TreeNode clone(TreeNode root, int delta) {

if (root == null) return root;

TreeNode node = new TreeNode(root.val + delta);

node.left = clone(root.left, delta);

node.right = clone(root.right, delta);

return node;

}

花花酱 LeetCode 495. Teemo Attacking

花花酱 LeetCode 521. Longest Uncommon Subsequence I

花花酱 LeetCode 95. Unique Binary Search Trees II

Problem

Idea: Recursion

C++

Java

Python 3

Solution 2: DP

Java

Related Problems

花花酱 LeetCode 620. Not Boring Movies

花花酱 LeetCode 15. 3Sum

Huahua's Tech Road

花花酱 LeetCode 495. Teemo Attacking

花花酱 LeetCode 521. Longest Uncommon Subsequence I

花花酱 LeetCode 95. Unique Binary Search Trees II

Problem

Idea: Recursion

C++

Java

Python 3

Solution 2: DP

Java

Related Problems

花花酱 LeetCode 620. Not Boring Movies

花花酱 LeetCode 15. 3Sum