Posts tagged as “graph”

花花酱 LeetCode 863. All Nodes Distance K in Binary Tree

By zxi on June 30, 2018

Problem

题目大意：给你一棵二叉树（根结点root）和一个target节点。返回所有到target的距离为K的节点。

We are given a binary tree (with root node root), a target node, and an integer value K.

Return a list of the values of all nodes that have a distance K from the target node. The answer can be returned in any order.

Example 1:

Input: root = [3,5,1,6,2,0,8,null,null,7,4], target = 5, K = 2
Output: [7,4,1]
Explanation: 
The nodes that are a distance 2 from the target node (with value 5)
have values 7, 4, and 1.

Note that the inputs "root" and "target" are actually TreeNodes.
The descriptions of the inputs above are just serializations of these objects.

Note:

The given tree is non-empty.
Each node in the tree has unique values 0 <= node.val <= 500.
The target node is a node in the tree.
0 <= K <= 1000.

Solution1: DFS + BFS

Use DFS to build the graph, and use BFS to find all the nodes that are exact K steps from target.

Time complexity: O(n)

Space complexity: O(n)

C++

// Author: Huahua

// Running time: 6 ms

class Solution {

public:

vector<int> distanceK(TreeNode* root, TreeNode* target, int K) {

buildGraph(nullptr, root);

vector<int> ans;

unordered_set<TreeNode*> seen;

queue<TreeNode*> q;

seen.insert(target);

q.push(target);

int k = 0;

while (!q.empty() && k <= K) {

int s = q.size();

while (s--) {

TreeNode* node = q.front(); q.pop();

if (k == K) ans.push_back(node->val);

for (TreeNode* child : g[node]) {

if (seen.count(child)) continue;

q.push(child);

seen.insert(child);

}

++k;

}

return ans;

}

private:

unordered_map<TreeNode*, vector<TreeNode*>> g;

void buildGraph(TreeNode* parent, TreeNode* child) {

if (parent) {

g[parent].push_back(child);

g[child].push_back(parent);

}

if (child->left) buildGraph(child, child->left);

if (child->right) buildGraph(child, child->right);

}

};

Array version

// Author: Huahua

// Running time: 6 ms

class Solution {

public:

vector<int> distanceK(TreeNode* root, TreeNode* target, int K) {

constexpr int kMaxN = 500 + 1;

g = vector<vector<int>>(kMaxN);

buildGraph(nullptr, root);

vector<int> ans;

vector<int> seen(kMaxN);

queue<int> q;

seen[target->val] = 1;

q.push(target->val);

int k = 0;

while (!q.empty() && k <= K) {

int s = q.size();

while (s--) {

int node = q.front(); q.pop();

if (k == K) ans.push_back(node);

for (int child : g[node]) {

if (seen[child]) continue;

q.push(child);

seen[child] = 1;

}

++k;

}

return ans;

}

private:

vector<vector<int>> g;

void buildGraph(TreeNode* parent, TreeNode* child) {

if (parent) {

g[parent->val].push_back(child->val);

g[child->val].push_back(parent->val);

}

if (child->left) buildGraph(child, child->left);

if (child->right) buildGraph(child, child->right);

}

};

Solution 2: Recursion

Recursively compute the distance from root to target, and collect nodes accordingly.

Time complexity: O(n)

Space complexity: O(n)

// Author: Huahua

// Running time: 5 ms

class Solution {

public:

vector<int> distanceK(TreeNode* root, TreeNode* target, int K) {

ans.clear();

dis(root, target, K);

return ans;

}

private:

vector<int> ans;

// Returns the distance from root to target.

// Returns -1 if target does not in the tree.

int dis(TreeNode* root, TreeNode* target, int K) {

if (root == nullptr) return -1;

if (root == target) {

collect(target, K);

return 0;

}

int l = dis(root->left, target, K);

int r = dis(root->right, target, K);

// Target in the left subtree

if (l >= 0) {

if (l == K - 1) ans.push_back(root->val);

// Collect nodes in right subtree with depth K - l - 2

collect(root->right, K - l - 2);

return l + 1;

}

// Target in the right subtree

if (r >= 0) {

if (r == K - 1) ans.push_back(root->val);

// Collect nodes in left subtree with depth K - r - 2

collect(root->left, K - r - 2);

return r + 1;

}

return -1;

}

// Collect nodes that are d steps from root.

void collect(TreeNode* root, int d) {

if (root == nullptr || d < 0) return;

if (d == 0) ans.push_back(root->val);

collect(root->left, d - 1);

collect(root->right, d - 1);

}

};

花花酱 LeetCode 847. Shortest Path Visiting All Nodes

By zxi on June 3, 2018

Problem

题目大意：求顶点覆盖的最短路径。

https://leetcode.com/problems/shortest-path-visiting-all-nodes/description/

An undirected, connected graph of N nodes (labeled 0, 1, 2, ..., N-1) is given as graph.

graph.length = N, and j != i is in the list graph[i] exactly once, if and only if nodes i and j are connected.

Return the length of the shortest path that visits every node. You may start and stop at any node, you may revisit nodes multiple times, and you may reuse edges.

Example 1:

Input: [[1,2,3],[0],[0],[0]]
Output: 4
Explanation: One possible path is [1,0,2,0,3]

Example 2:

Input: [[1],[0,2,4],[1,3,4],[2],[1,2]]
Output: 4
Explanation: One possible path is [0,1,4,2,3]

Solution: BFS

Time complexity: O(n*2^n)

Space complexity: O(n*2^n)

C++

// Author: Huahua

// Running time: 34 ms

class Solution {

public:

int shortestPathLength(vector<vector<int>>& graph) {

const int kAns = (1 << (graph.size())) - 1;

queue<pair<int, int>> q;

unordered_set<int> visited; // (cur_node << 16) | state

for (int i = 0; i < graph.size(); ++i)

q.push({i, 1 << i});

int steps = 0;

while (!q.empty()) {

int s = q.size();

while (s--) {

auto p = q.front();

q.pop();

int n = p.first;

int state = p.second;

if (state == kAns) return steps;

int key = (n << 16) | state;

if (visited.count(key)) continue;

visited.insert(key);

for (int next : graph[n])

q.push({next, state | (1 << next)});

}

++steps;

}

return -1;

}

};

C++ / vector

// Author: Huahua

// Running time: 10 ms

class Solution {

public:

int shortestPathLength(vector<vector<int>>& graph) {

const int n = graph.size();

const int kAns = (1 << n) - 1;

queue<pair<int, int>> q;

vector<vector<int>> visited(n, vector<int>(1 << n));

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

q.push({i, 1 << i});

int steps = 0;

while (!q.empty()) {

int s = q.size();

while (s--) {

auto p = q.front();

q.pop();

int node = p.first;

int state = p.second;

if (state == kAns) return steps;

if (visited[node][state]) continue;

visited[node][state] = 1;

for (int next : graph[node])

q.push({next, state | (1 << next)});

}

++steps;

}

return -1;

}

};

Problem

In a 2D grid of 0s and 1s, we change at most one 0 to a 1.

After, what is the size of the largest island? (An island is a 4-directionally connected group of 1s).

Example 1:

Input: [[1, 0], [0, 1]]
Output: 3
Explanation: Change one 0 to 1 and connect two 1s, then we get an island with area = 3.

Example 2:

Input: [[1, 1], [1, 0]]
Output: 4
Explanation: Change the 0 to 1 and make the island bigger, only one island with area = 1.

Example 3:

Input: [[1, 1], [1, 1]]
Output: 4
Explanation: Can't change any 0 to 1, only one island with area = 1.

Notes:

1 <= grid.length = grid[0].length <= 50.
0 <= grid[i][j] <= 1.

Solution

Step 1: give each connected component a unique id and count its ara.

Step 2: for each 0 zero, check its 4 neighbours, sum areas up by unique ids.

Time complexity: O(n*m)

Space complexity: O(n*m)

C++

// Author: Huahua

// Running time: 14 ms

class Solution {

public:

int largestIsland(vector<vector<int>>& grid) {

color_ = 1;

g_ = &grid;

m_ = grid.size();

n_ = grid[0].size();

unordered_map<int, int> areas; // color -> area

int ans = 0;

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

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

if (grid[i][j] == 1) {

++color_;

areas[color_] = getArea(j, i);

ans = max(ans, areas[color_]);

}

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

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

if (grid[i][j] == 0) {

int area = 1;

for (int color : set<int>{getColor(j, i - 1), getColor(j, i + 1),

getColor(j - 1, i), getColor(j + 1, i)})

area += areas[color];

ans = max(ans, area);

}

return ans;

}

private:

int m_;

int n_;

int color_;

vector<vector<int>>* g_; // does not own the object.

int getColor(int x, int y) {

return (x < 0 || x >= n_ || y < 0 || y >= m_) ? 0 : (*g_)[y][x];

}

// Return the area of a connected component, set all elements to color_.

int getArea(int x, int y) {

if (x < 0 || x >= n_ || y < 0 || y >= m_ || (*g_)[y][x] != 1) return 0;

(*g_)[y][x] = color_;

return 1 + getArea(x - 1, y) + getArea(x + 1, y)

+ getArea(x, y - 1) + getArea(x, y + 1);

}

};

花花酱 LeetCode 818. Race Car

By zxi on April 15, 2018

Problem

题目大意：初始位置0速度+1，每次你可以加速（速度*2）或者倒车（速度变成-1*dir）。问最少需要执行多少步操作能够到达T。

https://leetcode.com/problems/race-car/description/

Your car starts at position 0 and speed +1 on an infinite number line. (Your car can go into negative positions.)

Your car drives automatically according to a sequence of instructions A (accelerate) and R (reverse).

When you get an instruction “A”, your car does the following: position += speed, speed *= 2.

When you get an instruction “R”, your car does the following: if your speed is positive then speed = -1 , otherwise speed = 1. (Your position stays the same.)

For example, after commands “AAR”, your car goes to positions 0->1->3->3, and your speed goes to 1->2->4->-1.

Now for some target position, say the length of the shortest sequence of instructions to get there.

Example 1:
Input: 
target = 3
Output: 2
Explanation: 
The shortest instruction sequence is "AA".
Your position goes from 0->1->3.

Example 2:
Input: 
target = 6
Output: 5
Explanation: 
The shortest instruction sequence is "AAARA".
Your position goes from 0->1->3->7->7->6.

Note:

1 <= target <= 10000.

Visualization of the Solution

Solution 1: BFS

C++/Str

// Author: Huahua

// Running time: 221 ms

class Solution {

public:

int racecar(int target) {

queue<pair<int, int>> q;

q.push({0, 1});

unordered_set<string> v;

v.insert("0_1");

v.insert("0_-1");

int steps = 0;

while (!q.empty()) {

int size = q.size();

while (size--) {

auto p = q.front(); q.pop();

int pos = p.first;

int speed = p.second;

{

int pos1 = pos + speed;

int speed1 = speed * 2;

pair<int, int> p1{pos1, speed1};

if (pos1 == target) return steps + 1;

if (p1.first > 0 && p1.first < 2 * target)

q.push(p1);

}

{

int speed2 = speed > 0 ? -1 : 1;

pair<int, int> p2{pos, speed2};

string key2 = to_string(pos) + "_" + to_string(speed2);

if (!v.count(key2)) {

q.push(p2);

v.insert(key2);

}

++steps;

}

return -1;

}

};

C++/Int

// Author: Huahua

// Running time: 16 ms

class Solution {

public:

int racecar(int target) {

queue<pair<int, int>> q;

q.push({0, 1});

unordered_set<int> v;

v.insert({0, 2});

int steps = 0;

while (!q.empty()) {

int size = q.size();

while (size--) {

auto p = q.front(); q.pop();

int pos = p.first;

int speed = p.second;

pair<int, int> p1 = {pos + speed, speed * 2};

if (p1.first == target) return steps + 1;

if (p1.first > 0 && p1.first + speed < 2 * target)

q.push(p1);

int speed2 = speed > 0 ? -1 : 1;

pair<int, int> p2 = {pos, speed2};

int key = (pos << 2) | (speed2 + 1);

if (!v.count(key) && p2.first >= target / 2) {

q.push(p2);

v.insert(key);

}

++steps;

}

return -1;

}

};

Solution 2: DP O(TlogT)

C++

// Author: Huahua

// Running time: 4 ms

class Solution {

public:

int racecar(int target) {

m_ = vector<int>(target + 1, 0);

return dp(target);

}

private:

vector<int> m_;

int dp(int t) {

if (m_[t] > 0) return m_[t];

int n = ceil(log2(t + 1));

// AA...A (nA) best case

if (1 << n == t + 1) return m_[t] = n;

// AA...AR (nA + 1R) + dp(left)

m_[t] = n + 1 + dp((1 << n) - 1 - t);

for (int m = 0; m < n - 1; ++m) {

int cur = (1 << (n - 1)) - (1 << m);

//AA...ARA...AR (n-1A + 1R + mA + 1R) + dp(left)

m_[t] = min(m_[t], n + m + 1 + dp(t - cur));

}

return m_[t];

}

};

Solution 3: DP O(T^2)

m[t][d] : min steps to reach t and facing d (0 = right, 1 = left)

Time Complexity: O(n^2)

Space complexity: O(n)

C++

// Author: Huahua

// Running time: 469 ms

constexpr int kMaxT = 10000;

int m[kMaxT + 1][2]={0};

class Solution {

public:

int racecar(int target) {

if (m[1][0] == 0) {

for (int t = 1; t <= kMaxT; ++t) {

const int n = ceil(log2(t + 1));

if (1 << n == t + 1) {

m[t][0] = n;

m[t][1] = n + 1;

continue;

}

const int l = (1 << n) - 1 - t;

m[t][0] = n + 1 + min(m[l][1], m[l][0] + 1);

m[t][1] = n + 1 + min(m[l][0], m[l][1] + 1);

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

const int mi = min(m[i][0] + 2, m[i][1] + 1);

m[t][0] = min(m[t][0], m[t - i][0] + mi);

m[t][1] = min(m[t][1], m[t - i][1] + mi);

}

return min(m[target][0], m[target][1]);

}

};

C++/opt

// Author: Huahua

// Running time: 361 ms

typedef unsigned char uint8;

constexpr int kMaxT = 10000;

uint8 m[kMaxT + 1][2]={0};

inline uint8 MIN(uint8 a, uint8 b) {

return a < b ? a : b;

}

class Solution {

public:

int racecar(int target) {

if (m[1][0] == 0) {

for (int t = 1; t <= kMaxT; ++t) {

const int n = ceil(log2(t + 1));

if (1 << n == t + 1) {

m[t][0] = n;

m[t][1] = n + 1;

continue;

}

const int l = (1 << n) - 1 - t;

m[t][0] = n + 1 + MIN(m[l][1], (uint8)(m[l][0] + 1));

m[t][1] = n + 1 + MIN(m[l][0], (uint8)(m[l][1] + 1));

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

const int mi = MIN(m[i][0] + 2, m[i][1] + 1);

m[t][0] = MIN(m[t][0], (uint8)(m[t - i][0] + mi));

m[t][1] = MIN(m[t][1], (uint8)(m[t - i][1] + mi));

}

return min(m[target][0], m[target][1]);

}

};

Java

// Author: Huahua

// Running time: 562 ms

class Solution {

private static int[][] m;

public int racecar(int target) {

if (m == null) {

final int kMaxT = 10000;

m = new int[kMaxT + 1][2];

for (int t = 1; t <= kMaxT; ++t) {

int n = (int)Math.ceil(Math.log(t + 1) / Math.log(2));

if (1 << n == t + 1) {

m[t][0] = n;

m[t][1] = n + 1;

continue;

}

int l = (1 << n) - 1 - t;

m[t][0] = n + 1 + Math.min(m[l][1], m[l][0] + 1);

m[t][1] = n + 1 + Math.min(m[l][0], m[l][1] + 1);

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

for (int d = 0; d <= 1; d++)

m[t][d] = Math.min(m[t][d], Math.min(

m[i][0] + 2 + m[t - i][d],

m[i][1] + 1 + m[t - i][d]));

}

return Math.min(m[target][0], m[target][1]);

}

花花酱 LeetCode 817. Linked List Components

By zxi on April 14, 2018

Problem

题目大意：给你一个链表，再给你一些合法的节点，问你链表中有多少个连通分量（所有节点必须合法）。

https://leetcode.com/problems/linked-list-components/description/

We are given head, the head node of a linked list containing unique integer values.

We are also given the list G, a subset of the values in the linked list.

Return the number of connected components in G, where two values are connected if they appear consecutively in the linked list.

Example 1:

Input: 
head: 0->1->2->3
G = [0, 1, 3]
Output: 2
Explanation: 
0 and 1 are connected, so [0, 1] and [3] are the two connected components.

Example 2:

Input: 
head: 0->1->2->3->4
G = [0, 3, 1, 4]
Output: 2
Explanation: 
0 and 1 are connected, 3 and 4 are connected, so [0, 1] and [3, 4] are the two connected components.

Note:

If N is the length of the linked list given by head, 1 <= N <= 10000.
The value of each node in the linked list will be in the range [0, N - 1].
1 <= G.length <= 10000.
G is a subset of all values in the linked list.

Solution1: Graph Traversal using DFS

// Author: Huahua

// Running time: 76 ms

class Solution {

public:

int numComponents(ListNode* head, vector<int>& G) {

unordered_set<int> f(G.begin(), G.end());

unordered_map<int, vector<int>> g;

int u = head->val;

while (head->next) {

head = head->next;

int v = head->val;

if (f.count(v) && f.count(u)) {

g[u].push_back(v);

g[v].push_back(u);

}

u = v;

}

int ans = 0;

unordered_set<int> visited;

for (int u : G) {

if (visited.count(u)) continue;

++ans;

dfs(u, g, visited);

}

return ans;

}

private:

void dfs(int cur, unordered_map<int, vector<int>>& g, unordered_set<int>& visited) {

if (visited.count(cur)) return;

visited.insert(cur);

for (const int next : g[cur])

dfs(next, g, visited);

}

};

Solution 2: Count tail node in sub graph

// Author: Huahua

// Running time: 35 ms

class Solution {

public:

int numComponents(ListNode* head, vector<int>& G) {

unordered_set<int> g(G.begin(), G.end());

int ans = 0;

while (head) {

if (g.count(head->val) && (!head->next || !g.count(head->next->val)))

++ans;

head = head->next;

}

return ans;

}

};

Posts tagged as “graph”

花花酱 LeetCode 863. All Nodes Distance K in Binary Tree

Problem

Solution1: DFS + BFS

Solution 2: Recursion

花花酱 LeetCode 847. Shortest Path Visiting All Nodes

Problem

Solution: BFS

Related Problems

花花酱 LeetCode 827. Making A Large Island

Problem

Solution

花花酱 LeetCode 818. Race Car

Problem

Visualization of the Solution

Solution 1: BFS

C++/Str

C++/Int

Solution 2: DP O(TlogT)

C++

Solution 3: DP O(T^2)

C++

C++/opt

Java