花花酱 LeetCode 128. Longest Consecutive Sequence

By zxi on October 2, 2017

Problem:

Given an unsorted array of integers, find the length of the longest consecutive elements sequence.

For example,
Given [100, 4, 200, 1, 3, 2],
The longest consecutive elements sequence is [1, 2, 3, 4]. Return its length: 4.

Your algorithm should run in O(n) complexity.

Idea:

Hashtable / Hashset

Time complexity: O(n)

Space complexity: O(n)

Solution 1: C++ / online

class Solution {

public:

int longestConsecutive(vector<int>& nums) {

unordered_map<int, int> h;

int ans = 0;

for (int num : nums) {

if (h.count(num)) continue;

auto it_l = h.find(num - 1);

auto it_r = h.find(num + 1);

int l = it_l != h.end() ? it_l->second : 0;

int r = it_r != h.end() ? it_r->second : 0;

int t = l + r + 1;

h[num] = h[num - l] = h[num + r] = t;

ans = max(ans, t);

}

return ans;

}

};

Solution 2: C++ / offline

class Solution {

public:

int longestConsecutive(vector<int>& nums) {

unordered_set<int> h(nums.begin(), nums.end());

int ans = 0;

for (int num : nums)

if (!h.count(num - 1)) {

int l = 0;

while (h.count(num++)) ++l;

ans = max(ans, l);

}

return ans;

}

};

花花酱 688. Knight Probability in Chessboard

By zxi on October 1, 2017

https://leetcode.com/problems/knight-probability-in-chessboard/description/

Problem:

On an NxN chessboard, a knight starts at the r-th row and c-th column and attempts to make exactly Kmoves. The rows and columns are 0 indexed, so the top-left square is (0, 0), and the bottom-right square is (N-1, N-1).

A chess knight has 8 possible moves it can make, as illustrated below. Each move is two squares in a cardinal direction, then one square in an orthogonal direction.

Each time the knight is to move, it chooses one of eight possible moves uniformly at random (even if the piece would go off the chessboard) and moves there.

The knight continues moving until it has made exactly K moves or has moved off the chessboard. Return the probability that the knight remains on the board after it has stopped moving.

Example:

Input: 3, 2, 0, 0

Output: 0.0625

Explanation: There are two moves (to (1,2), (2,1)) that will keep the knight on the board.

From each of those positions, there are also two moves that will keep the knight on

the board. The total probability the knight stays on the board is 0.0625.

Note:

N will be between 1 and 25.
K will be between 0 and 100.
The knight always initially starts on the board.

Idea:

Dynamic programming

Count the ways to reach (x, y) after k moves from start.

Time Complexity: O(k*n^2)

Space Complexity: O(n^2)

Solution:

// Author: Huahua

// Runtime: 6 ms

class Solution {

public:

double knightProbability(int N, int K, int r, int c) {

vector<vector<double>> dp0(N, vector<double>(N, 0.0));

dp0[r][c] = 1.0;

int dirs[8][2] = {{1, 2}, {-1, -2}, {1, -2}, {-1, 2},

{2, 1}, {-2, -1}, {2, -1}, {-2, 1}};

for (int k = 0; k < K; ++k) {

vector<vector<double>> dp1(N, vector<double>(N, 0.0));

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

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

for (int m = 0; m < 8; ++m) {

int x = j + dirs[m][0];

int y = i + dirs[m][1];

if (x < 0 || y < 0 || x >= N || y >= N) continue;

dp1[i][j] += dp0[y][x];

}

std::swap(dp0, dp1);

}

double total = 0;

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

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

total += dp0[i][j];

return total / pow(8, K);

}

};

Related problems:

花花酱 LeetCode 687. Longest Univalue Path

By zxi on October 1, 2017

题目大意：给你一棵二叉树，返回一条最长的路径，要求路径上所有的节点值都相同。

https://leetcode.com/problems/longest-univalue-path/description/

Problem:

Given a binary tree, find the length of the longest path where each node in the path has the same value. This path may or may not pass through the root.

Note: The length of path between two nodes is represented by the number of edges between them.

Example 1:

Input:

Output:

Example 2:

Input:

Output:

Note: The given binary tree has not more than 10000 nodes. The height of the tree is not more than 1000.

Idea:

DFS

Solution: Recursion

Time complexity: O(n)

Space complexity: O(n)

C++

// Author: Huahua

// Runtime: 69 ms

class Solution {

public:

int longestUnivaluePath(TreeNode* root) {

if (root == nullptr) return 0;

int ans = 0;

univaluePath(root, &ans);

return ans;

}

private:

int univaluePath(TreeNode* root, int* ans) {

if (root == nullptr) return 0;

int l = univaluePath(root->left, ans);

int r = univaluePath(root->right, ans);

int pl = 0;

int pr = 0;

if (root->left && root->val == root->left->val) pl = l + 1;

if (root->right && root->val == root->right->val) pr = r + 1;

*ans = max(*ans, pl + pr);

return max(pl, pr);

}

};

Java

// Author: Huahua

// Runtime: 12 ms

class Solution {

private int ans;

public int longestUnivaluePath(TreeNode root) {

if (root == null) return 0;

this.ans = 0;

univaluePath(root);

return this.ans;

}

private int univaluePath(TreeNode root) {

if (root == null) return 0;

int l = univaluePath(root.left);

int r = univaluePath(root.right);

int pl = 0;

int pr = 0;

if (root.left != null && root.val == root.left.val) pl = l + 1;

if (root.right != null && root.val == root.right.val) pr = r + 1;

this.ans = Math.max(this.ans, pl + pr);

return Math.max(pl, pr);

}

Python3

"""

Author: Huahua

Runtime: 899 ms

"""

class Solution:

def longestUnivaluePath(self, root):

self.ans = 0

self._univaluePath(root)

return self.ans

def _univaluePath(self, root):

if root is None: return 0

l = self._univaluePath(root.left) if root.left is not None else -1

r = self._univaluePath(root.right) if root.right is not None else -1

pl = l + 1 if l >= 0 and root.val == root.left.val else 0

pr = r + 1 if r >= 0 and root.val == root.right.val else 0

self.ans = max(self.ans, pl + pr)

return max(pl, pr)

Related Problems

花花酱 LeetCode 124. Binary Tree Maximum Path Sum

花花酱 LeetCode 678. Valid Parenthesis String

By zxi on September 30, 2017

Problem:

Given a string containing only three types of characters: ‘(‘, ‘)’ and ‘*’, write a function to check whether this string is valid. We define the validity of a string by these rules:

Any left parenthesis '(' must have a corresponding right parenthesis ')'.
Any right parenthesis ')' must have a corresponding left parenthesis '('.
Left parenthesis '(' must go before the corresponding right parenthesis ')'.
'*' could be treated as a single right parenthesis ')' or a single left parenthesis '(' or an empty string.
An empty string is also valid.

Example 1:

1 2	Input: "()" Output: True

Example 2:

1 2	Input: "(*)" Output: True

Example 3:

1 2	Input: "(*))" Output: True

Note:

The string size will be in the range [1, 100].

Idea:

Dynamic Programming / Counting

Solution 1:

C++ / DP / Top-down O(n^3)

// Author: Huahua

// Runtime: 12 ms

class Solution {

public:

bool checkValidString(const string& s) {

int l = s.length();

m_ = vector<vector<int>>(l, vector<int>(l, -1));

return isValid(s, 0, l - 1);

}

private:

vector<vector<int>> m_;

bool isValid(const string& s, int i, int j) {

if (i > j) return true;

if (m_[i][j] >= 0) return m_[i][j];

if (i == j) return m_[i][j] = (s[i] == '*');

if ((s[i] == '(' || s[i] == '*')

&&(s[j] == ')' || s[j] == '*')

&& isValid(s, i + 1, j - 1))

return m_[i][j] = 1;

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

if (isValid(s, i, p) && isValid(s, p + 1, j))

return m_[i][j] = 1;

return m_[i][j] = 0;

}

};

C++ / DP/ Bottom-up O(n^3)

// Author: Huahua

// Runtime: 12 ms

class Solution {

public:

bool checkValidString(const string& s) {

int l = s.length();

if (l == 0) return true;

vector<vector<int>> dp(l, vector<int>(l, 0));

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

if (s[i] == '*') dp[i][i] = 1;

for (int len = 2; len <= l; ++len)

for (int i = 0; i <= l - len; ++i) {

int j = i + len - 1;

// (...), *...), (...*, *...*

if ((s[i] == '(' || s[i] == '*')

&&(s[j] == ')' || s[j] == '*'))

if (len == 2 || dp[i + 1][j - 1]) {

dp[i][j] = 1;

continue;

}

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

if (dp[i][k] && dp[k + 1][j]) {

dp[i][j] = 1;

break;

}

return dp[0][l - 1];

}

};

C++ / Counting O(n)

// Author: Huahua

// Runtime: 3 ms

class Solution {

public:

bool checkValidString(string s) {

int min_op = 0;

int max_op = 0;

for (char c : s) {

if (c == '(') ++min_op; else --min_op;

if (c != ')') ++max_op; else --max_op;

if (max_op < 0) return false;

min_op = max(0, min_op);

}

return min_op == 0;

}

};

Java / DP / Bottom-up O(n^3)

class Solution {

public boolean checkValidString(String s) {

int l = s.length();

if (l == 0) return true;

boolean[][] dp = new boolean[l][l];

char[] ss = s.toCharArray();

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

if (ss[i] == '*') dp[i][i] = true;

for (int len = 2; len <= l; ++len)

for (int i = 0; i + len <= l; ++i) {

int j = i + len - 1;

if ((ss[i] == '*' || ss[i] == '(')

&&(ss[j] == '*' || ss[j] == ')')) {

if (len == 2 || dp[i + 1][j - 1]) {

dp[i][j] = true;

continue;

}

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

if (dp[i][k] && dp[k + 1][j]) {

dp[i][j] = true;

break;

}

return dp[0][l - 1];

}

花花酱 LeetCode 683. K Empty Slots

By zxi on September 29, 2017

题目大意：有n个花盆，第i天，第flowers[i]个花盆的花会开。问是否存在一天，两朵花之间有k个空花盆。

Problem:

There is a garden with N slots. In each slot, there is a flower. The N flowers will bloom one by one in N days. In each day, there will be exactly one flower blooming and it will be in the status of blooming since then.

Given an array flowers consists of number from 1 to N. Each number in the array represents the place where the flower will open in that day.

For example, flowers[i] = x means that the unique flower that blooms at day i will be at position x, where i and x will be in the range from 1 to N.

Also given an integer k, you need to output in which day there exists two flowers in the status of blooming, and also the number of flowers between them is k and these flowers are not blooming.

If there isn’t such day, output -1.

Example 1:

Input: 
flowers: [1,3,2]
k: 1
Output: 2
Explanation: In the second day, the first and the third flower have become blooming.

Example 2:

Input: 
flowers: [1,2,3]
k: 1
Output: -1

Note:

The given array will be in the range [1, 20000].

Idea:

BST/Buckets

Solution 2: BST

C++

// Author: Huahua

// Runtime: 228 ms

class Solution {

public:

int kEmptySlots(vector<int>& flowers, int k) {

int n = flowers.size();

if (n == 0 || k >= n) return -1;

set<int> xs;

for (int i = 0; i < n; ++i) {

int x = flowers[i];

auto r = xs.insert(x).first;

auto l = r;

if (++r != xs.end() && *r == x + k + 1) return i + 1;

if (l != xs.begin() && *(--l) == x - k - 1) return i + 1;

}

return -1;

}

};

Solution 3: Buckets

C++

// Author: Huahua

// Runtime: 196 ms (better than 94%)

class Solution {

public:

int kEmptySlots(vector<int>& flowers, int k) {

int n = flowers.size();

if (n == 0 || k >= n) return -1;

++k;

int bs = (n + k - 1) / k;

vector<int> lows(bs, INT_MAX);

vector<int> highs(bs, INT_MIN);

for (int i = 0; i < n; ++i) {

int x = flowers[i];

int p = x / k;

if (x < lows[p]) {

lows[p] = x;

if (p > 0 && highs[p - 1] == x - k) return i + 1;

}

if (x > highs[p]) {

highs[p] = x;

if (p < bs - 1 && lows[p + 1] == x + k) return i + 1;

}

return -1;

}

};

Solution 1: TLE with latest test cases

C++

// Author: Huahua

// Runtime: 192 ms (better than 97.85%)

class Solution {

public:

int kEmptySlots(vector<int>& flowers, int k) {

int n = flowers.size();

if (n == 0 || k >= n) return -1;

std::unique_ptr<char[]> f(new char[n+1]);

memset(f.get(), 0, (n + 1)*sizeof(char));

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

if (IsValid(flowers[i], k, n, f.get()))

return i + 1;

return -1;

}

private:

bool IsValid(int x, int k, int n, char* f) {

f[x] = 1;

if (x + k + 1 <= n && f[x + k + 1]) {

bool valid = true;

for (int i = 1; i <= k; ++i)

if (f[x + i]) {

valid = false;

break;

}

if (valid) return true;

}

if (x - k - 1 > 0 && f[x - k - 1]) {

for (int i = 1; i <= k; ++i)

if (f[x - i]) return false;

return true;

}

return false;

}

};

Java

// Author: Huahua

// Runtime: 17 ms

class Solution {

public int kEmptySlots(int[] flowers, int k) {

int n = flowers.length;

if (n == 0 || k >= n) return -1;

int[] f = new int[n + 1];

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

if (IsValid(flowers[i], k, n, f))

return i + 1;

return -1;

}

private boolean IsValid(int x, int k, int n, int[] f) {

f[x] = 1;

if (x + k + 1 <= n && f[x + k + 1] == 1) {

boolean valid = true;

for (int i = 1; i <= k; ++i)

if (f[x + i] == 1) {

valid = false;

break;

}

if (valid) return true;

}

if (x - k - 1 > 0 && f[x - k - 1] == 1) {

for (int i = 1; i <= k; ++i)

if (f[x - i] == 1) return false;

return true;

}

return false;

}

Python

"""

Author: Huahua

Runtime: 398 ms

"""

class Solution:

def kEmptySlots(self, flowers, k):

n = len(flowers)

f = [0] * (n + 1)

i = 0

def isValid(x, k, n, f):

f[x] = 1

if x + k + 1 <= n and f[x + k + 1] == 1:

valid = True

for i in range(k):

if f[x + i + 1] == 1:

valid = False

break

if valid: return True

if x - k - 1 > 0 and f[x - k - 1] == 1:

for i in range(k):

if f[x - i - 1] == 1:

return False

return True

return False

for x in flowers:

i += 1

if isValid(x, k, n, f): return i

return -1

Huahua's Tech Road

花花酱 LeetCode 128. Longest Consecutive Sequence

花花酱 688. Knight Probability in Chessboard

花花酱 LeetCode 687. Longest Univalue Path

Solution: Recursion

C++

Java

Python3

Related Problems

花花酱 LeetCode 678. Valid Parenthesis String

花花酱 LeetCode 683. K Empty Slots

Solution 2: BST

C++

Solution 3: Buckets

C++

Solution 1: TLE with latest test cases

C++

Java

Python