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;

}

Problem:

https://leetcode.com/problems/factorial-trailing-zeroes/description/

Given an integer n, return the number of trailing zeroes in n!.

Note: Your solution should be in logarithmic time complexity.

Idea:

All trailing zeros are come from even_num x 5, we have more even_num than 5, so only count factor 5.

4! = 1x2x3x4 = 24, we haven’t encountered any 5 yet, so we don’t have any trailing zero.

5! = 1x2x3x4x5 = 120, we have one trailing zero. either 2×5, or 4×5 can contribute to that zero.

9! = 362880, we only encountered 5 once, so 1 trailing zero as expected.

10! = 3628800, 2 trailing zeros, since we have two numbers that have factor 5, one is 5 and the other is 10 (2×5)

What about 100! then?

100/5 = 20, we have 20 numbers have factor 5: 5, 10, 15, 20, 25, …, 95, 100.

Is the number of trailing zero 20? No, it’s 24, why?

Within that 20 numbers, we have 4 of them: 25 (5×5), 50 (2x5x5), 75 (3x5x5), 100 (4x5x5) that have an extra factor of 5.

So, for a given number n, we are looking how many numbers <=n have factor 5, 5×5, 5x5x5, …

Summing those numbers up we got the answer.

e.g. 1000! has 249 trailing zeros:

1000/5 = 200

1000/25 = 40

1000/125 = 8

1000/625 = 1

200 + 40 + 8 + 1 = 249

alternatively, we can do the following

1000/5 = 200

200/5 = 40

40/5 = 8

8/5 = 1

1/5 = 0

200 + 40 + 8 + 1 + 0 = 249

Solution 1: Recursion

Time complexity: O(log5(n))

Space complexity: O(log5(n))

C++

// Author: Huahua

// Running time: 4 ms

class Solution {

public:

int trailingZeroes(int n) {

return n < 5 ? 0 : n /5 + trailingZeroes(n / 5);

}

};

Java

// Author: Huahua

// Running time: 1 ms

class Solution {

public int trailingZeroes(int n) {

return n < 5 ? 0 : n /5 + trailingZeroes(n / 5);

}

Python3

"""

Author: Huahua

Running time: 68 ms

"""

class Solution:

def trailingZeroes(self, n):

return 0 if n < 5 else n // 5 + self.trailingZeroes(n // 5)

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

花花酱 LeetCode 172. Factorial Trailing Zeroes

Problem:

Idea:

Solution 1: Recursion

C++

Java

Python3

Related Problems:

Huahua's Tech Road

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

花花酱 LeetCode 172. Factorial Trailing Zeroes

Problem:

Idea:

Solution 1: Recursion

C++

Java

Python3

Related Problems: