Posts published in “Graph”

花花酱 LeetCode 399. Evaluate Division

By zxi on December 1, 2017

题目大意：给你一些含有变量名的分式的值，让你计算另外一些分式的值，如果不能计算返回-1。

Problem:

Equations are given in the format A / B = k, where A and B are variables represented as strings, and k is a real number (floating point number). Given some queries, return the answers. If the answer does not exist, return -1.0.

Example:
Given a / b = 2.0, b / c = 3.0.
queries are: a / c = ?, b / a = ?, a / e = ?, a / a = ?, x / x = ? .
return [6.0, 0.5, -1.0, 1.0, -1.0 ].

The input is: vector<pair<string, string>> equations, vector<double>& values, vector<pair<string, string>> queries , where equations.size() == values.size(), and the values are positive. This represents the equations. Return vector<double>.

According to the example above:

equations = [ ["a", "b"], ["b", "c"] ],

values = [2.0, 3.0],

queries = [ ["a", "c"], ["b", "a"], ["a", "e"], ["a", "a"], ["x", "x"] ].

The input is always valid. You may assume that evaluating the queries will result in no division by zero and there is no contradiction.

Solution 1: DFS

C++

// Author: Huahua

// Runtime: 3 ms

class Solution {

public:

vector<double> calcEquation(vector<pair<string, string>> equations, vector<double>& values, vector<pair<string, string>> queries) {

// g[A][B] = k -> A / B = k

unordered_map<string, unordered_map<string, double>> g;

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

const string& A = equations[i].first;

const string& B = equations[i].second;

const double k = values[i];

g[A][B] = k;

g[B][A] = 1.0 / k;

}

vector<double> ans;

for (const auto& pair : queries) {

const string& X = pair.first;

const string& Y = pair.second;

if (!g.count(X) || !g.count(Y)) {

ans.push_back(-1.0);

continue;

}

unordered_set<string> visited;

ans.push_back(divide(X, Y, g, visited));

}

return ans;

}

private:

// get result of A / B

double divide(const string& A, const string& B,

unordered_map<string, unordered_map<string, double>>& g,

unordered_set<string>& visited) {

if (A == B) return 1.0;

visited.insert(A);

for (const auto& pair : g[A]) {

const string& C = pair.first;

if (visited.count(C)) continue;

double d = divide(C, B, g, visited); // d = C / B

// A / B = C / B * A / C

if (d > 0) return d * g[A][C];

}

return -1.0;

}

};

Java

// Updated: 2020/9/27

// Author: Huahua

// Running time: 1 ms

class Solution {

private Map<String, HashMap<String, Double>> g = new HashMap<>();

public double[] calcEquation(List<List<String>> equations, double[] values, List<List<String>> queries) {

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

String x = equations.get(i).get(0);

String y = equations.get(i).get(1);

double k = values[i];

g.computeIfAbsent(x, l -> new HashMap<String, Double>()).put(y, k);

g.computeIfAbsent(y, l -> new HashMap<String, Double>()).put(x, 1.0 / k);

}

double[] ans = new double[queries.size()];

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

String x = queries.get(i).get(0);

String y = queries.get(i).get(1);

if (!g.containsKey(x) || !g.containsKey(y)) {

ans[i] = -1.0;

} else {

ans[i] = divide(x, y, new HashSet<String>());

}

return ans;

}

private double divide(String x, String y, Set<String> visited) {

if (x.equals(y)) return 1.0;

visited.add(x);

if (!g.containsKey(x)) return -1.0;

for (String n : g.get(x).keySet()) {

if (visited.contains(n)) continue;

visited.add(n);

double d = divide(n, y, visited);

if (d > 0) return d * g.get(x).get(n);

}

return -1.0;

}

Python3

"""

Author: Huahua

Running time: 32 ms (beats 100%)

"""

class Solution:

def calcEquation(self, equations, values, queries):

def divide(x, y, visited):

if x == y: return 1.0

visited.add(x)

for n in g[x]:

if n in visited: continue

visited.add(n)

d = divide(n, y, visited)

if d > 0: return d * g[x][n]

return -1.0

g = collections.defaultdict(dict)

for (x, y), v in zip(equations, values):

g[x][y] = v

g[y][x] = 1.0 / v

ans = [divide(x, y, set()) if x in g and y in g else -1 for x, y in queries]

return ans

Solution 2: Union Find

C++

// Author: Huahua

// Runtime: 3 ms

class Solution {

public:

vector<double> calcEquation(const vector<pair<string, string>>& equations, vector<double>& values, const vector<pair<string, string>>& queries) {

// parents["A"] = {"B", 2.0} -> A = 2.0 * B

// parents["B"] = {"C", 3.0} -> B = 3.0 * C

unordered_map<string, pair<string, double>> parents;

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

const string& A = equations[i].first;

const string& B = equations[i].second;

const double k = values[i];

// Neighter is in the forrest

if (!parents.count(A) && !parents.count(B)) {

parents[A] = {B, k};

parents[B] = {B, 1.0};

} else if (!parents.count(A)) {

parents[A] = {B, k};

} else if (!parents.count(B)) {

parents[B] = {A, 1.0 / k};

} else {

auto& rA = find(A, parents);

auto& rB = find(B, parents);

parents[rA.first] = {rB.first, k / rA.second * rB.second};

}

vector<double> ans;

for (const auto& pair : queries) {

const string& X = pair.first;

const string& Y = pair.second;

if (!parents.count(X) || !parents.count(Y)) {

ans.push_back(-1.0);

continue;

}

auto& rX = find(X, parents); // {rX, X / rX}

auto& rY = find(Y, parents); // {rY, Y / rY}

if (rX.first != rY.first)

ans.push_back(-1.0);

else // X / Y = (X / rX / (Y / rY))

ans.push_back(rX.second / rY.second);

}

return ans;

}

private:

pair<string, double>& find(const string& C, unordered_map<string, pair<string, double>>& parents) {

if (C != parents[C].first) {

const auto& p = find(parents[C].first, parents);

parents[C].first = p.first;

parents[C].second *= p.second;

}

return parents[C];

}

};

Java

// Updated: 9/27/2020

// Author: Huahua

// Running time: 3 ms

class Solution {

class Node {

public String parent;

public double ratio;

public Node(String parent, double ratio) {

this.parent = parent;

this.ratio = ratio;

}

class UnionFindSet {

private Map<String, Node> parents = new HashMap<>();

public Node find(String s) {

if (!parents.containsKey(s)) return null;

Node n = parents.get(s);

if (!n.parent.equals(s)) {

Node p = find(n.parent);

n.parent = p.parent;

n.ratio *= p.ratio;

}

return n;

}

public void union(String A, String B, double ratio) {

boolean hasA = parents.containsKey(A);

boolean hasB = parents.containsKey(B);

if (!hasA && !hasB) {

parents.put(A, new Node(B, ratio));

parents.put(B, new Node(B, 1.0));

} else if (!hasA) {

parents.put(A, new Node(B, ratio));

} else if (!hasB) {

parents.put(B, new Node(A, 1.0 / ratio));

} else {

Node rA = find(A);

Node rB = find(B);

parents.put(rA.parent,

new Node(rB.parent, ratio / rA.ratio * rB.ratio));

}

public double[] calcEquation(List<List<String>> equations, double[] values, List<List<String>> queries) {

UnionFindSet u = new UnionFindSet();

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

u.union(equations.get(i).get(0), equations.get(i).get(1), values[i]);

double[] ans = new double[queries.size()];

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

Node rx = u.find(queries.get(i).get(0));

Node ry = u.find(queries.get(i).get(1));

if (rx == null || ry == null || !rx.parent.equals(ry.parent))

ans[i] = -1.0;

else

ans[i] = rx.ratio / ry.ratio;

}

return ans;

}

Python3

"""

Author: Huahua

Running time: 32 ms (beats 100%)

"""

class Solution:

def calcEquation(self, equations, values, queries):

def find(x):

if x != U[x][0]:

px, pv = find(U[x][0])

U[x] = (px, U[x][1] * pv)

return U[x]

def divide(x, y):

rx, vx = find(x)

ry, vy = find(y)

if rx != ry: return -1.0

return vx / vy

U = {}

for (x, y), v in zip(equations, values):

if x not in U and y not in U:

U[x] = (y, v)

U[y] = (y, 1.0)

elif x not in U:

U[x] = (y, v)

elif y not in U:

U[y] = (x, 1.0 / v)

else:

rx, vx = find(x)

ry, vy = find(y)

U[rx] = (ry, v / vx * vy)

ans = [divide(x, y) if x in U and y in U else -1 for x, y in queries]

return ans

Related Problems:

花花酱 LeetCode 733. Flood Fill

By zxi on November 28, 2017

Problem:

An image is represented by a 2-D array of integers, each integer representing the pixel value of the image (from 0 to 65535).

Given a coordinate (sr, sc) representing the starting pixel (row and column) of the flood fill, and a pixel value newColor, “flood fill” the image.

To perform a “flood fill”, consider the starting pixel, plus any pixels connected 4-directionally to the starting pixel of the same color as the starting pixel, plus any pixels connected 4-directionally to those pixels (also with the same color as the starting pixel), and so on. Replace the color of all of the aforementioned pixels with the newColor.

At the end, return the modified image.

Example 1:

Input:

image = [[1,1,1],[1,1,0],[1,0,1]]

sr = 1, sc = 1, newColor = 2

Output: [[2,2,2],[2,2,0],[2,0,1]]

Explanation:

From the center of the image (with position (sr, sc) = (1, 1)), all pixels connected

by a path of the same color as the starting pixel are colored with the new color.

Note the bottom corner is not colored 2, because it is not 4-directionally connected

to the starting pixel.

Note:

The length of image and image[0] will be in the range [1, 50].
The given starting pixel will satisfy 0 <= sr < image.length and 0 <= sc < image[0].length.
The value of each color in image[i][j] and newColor will be an integer in [0, 65535].

Idea

DFS

Solution:

C++ / DFS

Time complexity: O(mn)

Space complexity: O(1)

// Author: Huahua

// Runtime: 33 ms

class Solution {

public:

vector<vector<int>> floodFill(vector<vector<int>>& image,

int sr, int sc, int newColor) {

if (image[sr][sc] == newColor) return image;

int m = image.size();

int n = image[0].size();

floodFill(image, sc, sr, n, m, image[sr][sc], newColor);

return image;

}

private:

void floodFill(vector<vector<int>>& image,

int x, int y, int n, int m, int orgColor, int newColor) {

if (x < 0 || x >= n || y < 0 || y >= m) return;

if (image[y][x] != orgColor) return;

image[y][x] = newColor;

floodFill(image, x + 1, y, n, m, orgColor, newColor);

floodFill(image, x - 1, y, n, m, orgColor, newColor);

floodFill(image, x, y + 1, n, m, orgColor, newColor);

floodFill(image, x, y - 1, n, m, orgColor, newColor);

}

};

Related Problems:

花花酱 LeetCode 207. Course Schedule

By zxi on October 20, 2017

Problem:

There are a total of n courses you have to take, labeled from 0 to n - 1.

Some courses may have prerequisites, for example to take course 0 you have to first take course 1, which is expressed as a pair: [0,1]

Given the total number of courses and a list of prerequisite pairs, is it possible for you to finish all courses?

For example:

1	2, [[1,0]]

There are a total of 2 courses to take. To take course 1 you should have finished course 0. So it is possible.

1	2, [[1,0],[0,1]]

There are a total of 2 courses to take. To take course 1 you should have finished course 0, and to take course 0 you should also have finished course 1. So it is impossible.

Idea:

Finding cycles O(n^2) -> Topological sort O(n)

Solution 1: Topological Sort

C++

// Author: Huahua

// Time complexity: O(n)

// Runtime: 12 ms

class Solution {

public:

bool canFinish(int numCourses, vector<pair<int, int>>& prerequisites) {

graph_ = vector<vector<int>>(numCourses);

for(const auto& p : prerequisites)

graph_[p.first].push_back(p.second);

// states: 0 = unkonwn, 1 == visiting, 2 = visited

vector<int> v(numCourses, 0);

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

if(dfs(i, v)) return false;

return true;

}

private:

vector<vector<int>> graph_;

bool dfs(int cur, vector<int>& v) {

if(v[cur] == 1) return true;

if(v[cur] == 2) return false;

v[cur] = 1;

for(const int t : graph_[cur])

if(dfs(t, v)) return true;

v[cur] = 2;

return false;

}

};

Java

// Author: Huahua

// Runtime: 6 ms

// HashMap is slower than ArrayList in this problem.

class Solution {

public boolean canFinish(int numCourses, int[][] prerequisites) {

ArrayList<ArrayList<Integer>> graph = new ArrayList<>();

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

graph.add(new ArrayList<Integer>());

for (int i = 0; i < prerequisites.length; ++i) {

int course = prerequisites[i][0];

int prerequisite = prerequisites[i][1];

graph.get(course).add(prerequisite);

}

int[] visited = new int[numCourses];

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

if (dfs(i, graph, visited)) return false;

return true;

}

private boolean dfs(int curr, ArrayList<ArrayList<Integer>> graph, int[] visited) {

if (visited[curr] == 1) return true;

if (visited[curr] == 2) return false;

visited[curr] = 1;

for (int next : graph.get(curr))

if (dfs(next, graph, visited)) return true;

visited[curr] = 2;

return false;

}

Solution 2: DFS Finding cycles

Time complexity: O(n^2)

Space complexity: O(n)

C++

// Author: Huahua

// Time complexity: O(n^2)

// Runtime: 179 ms

class Solution {

public:

bool canFinish(int numCourses, vector<pair<int, int>>& prerequisites) {

graph_.clear();

for(const auto& p : prerequisites)

graph_[p.first].insert(p.second);

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

vector<int> visited(numCourses, 0);

if (cycle(i, i, visited)) return false;

}

return true;

}

private:

unordered_map<int, unordered_set<int>> graph_;

bool cycle(int start, int curr, vector<int>& visited) {

if (curr == start && visited[start]) return true;

if (!graph_.count(curr)) return false;

for (const int next : graph_.at(curr)) {

if (visited[next]) continue;

visited[next] = true;

if (cycle(start, next, visited)) return true;

}

return false;

}

};

Related Pr0blems:

花花酱 LeetCode 695. Max Area of Island

By zxi on October 7, 2017

Problem:

Given a non-empty 2D array grid of 0’s and 1’s, an island is a group of 1‘s (representing land) connected 4-directionally (horizontal or vertical.) You may assume all four edges of the grid are surrounded by water.

Find the maximum area of an island in the given 2D array. (If there is no island, the maximum area is 0.)

Example 1:

[[0,0,1,0,0,0,0,1,0,0,0,0,0],

[0,0,0,0,0,0,0,1,1,1,0,0,0],

[0,1,1,0,1,0,0,0,0,0,0,0,0],

[0,1,0,0,1,1,0,0,1,0,1,0,0],

[0,1,0,0,1,1,0,0,1,1,1,0,0],

[0,0,0,0,0,0,0,0,0,0,1,0,0],

[0,0,0,0,0,0,0,1,1,1,0,0,0],

[0,0,0,0,0,0,0,1,1,0,0,0,0]]

Given the above grid, return 6. Note the answer is not 11, because the island must be connected 4-directionally.

Example 2:

1	[[0,0,0,0,0,0,0,0]]

Given the above grid, return 0.

Note: The length of each dimension in the given grid does not exceed 50.

Idea:

Use DFS to find the connected components

Solution:

C++

// Author: Huahua

// Runtime: 32 ms

class Solution {

public:

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

int h = grid.size();

if (h == 0) return 0;

int w = grid[0].size();

int ans = 0;

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

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

ans = max(ans, area(grid, j, i, w, h));

return ans;

}

private:

int area(vector<vector<int>>& grid, int x, int y, int w, int h) {

if (x < 0 || y < 0 || x >= w || y >= h || grid[y][x] == 0) return 0;

grid[y][x] = 0;

return area(grid, x - 1, y, w, h)

+ area(grid, x + 1, y, w, h)

+ area(grid, x, y - 1, w, h)

+ area(grid, x, y + 1, w, h)

+ 1;

}

};

Related problems:

花花酱 LeetCode 685. Redundant Connection II

By zxi on October 7, 2017

Problem:

In this problem, a rooted tree is a directed graph such that, there is exactly one node (the root) for which all other nodes are descendants of this node, plus every node has exactly one parent, except for the root node which has no parents.

The given input is a directed graph that started as a rooted tree with N nodes (with distinct values 1, 2, …, N), with one additional directed edge added. The added edge has two different vertices chosen from 1 to N, and was not an edge that already existed.

The resulting graph is given as a 2D-array of edges. Each element of edges is a pair [u, v] that represents a directed edge connecting nodes u and v, where u is a parent of child v.

Return an edge that can be removed so that the resulting graph is a rooted tree of N nodes. If there are multiple answers, return the answer that occurs last in the given 2D-array.

Example 1:

Input: [[1,2], [1,3], [2,3]]

Output: [2,3]

Explanation: The given directed graph will be like this:

/ \

v v

2-->3

Example 2:

Input: [[1,2], [2,3], [3,4], [4,1], [1,5]]

Output: [4,1]

Explanation: The given directed graph will be like this:

5 <- 1 -> 2

^ |

| v

4 <- 3

Note:

The size of the input 2D-array will be between 3 and 1000.
Every integer represented in the 2D-array will be between 1 and N, where N is the size of the input array.

Idea:

Union Find

Time complexity: O(nlog*n) ~ O(n)

Space complexity: O(n)

Solution:

C++

// Author: Huahua

// Runtime: 6 ms

class Solution {

public:

vector<int> findRedundantDirectedConnection(vector<vector<int>>& edges) {

vector<int> parents(edges.size() + 1, 0);

vector<int> roots(edges.size() + 1, 0);

vector<int> sizes(edges.size() + 1, 1);

vector<int> ans1;

vector<int> ans2;

for(auto& edge: edges) {

int u = edge[0];

int v = edge[1];

// A node has two parents

if (parents[v] > 0) {

ans1 = {parents[v], v};

ans2 = edge;

// Delete the later edge

edge[0] = edge[1] = -1;

}

parents[v] = u;

}

for(const auto& edge: edges) {

int u = edge[0];

int v = edge[1];

// Invalid edge (we deleted in step 1)

if (u < 0 || v < 0) continue;

if (!roots[u]) roots[u] = u;

if (!roots[v]) roots[v] = v;

int pu = find(u, roots);

int pv = find(v, roots);

// Both u and v are already in the tree

if (pu == pv)

return ans1.empty() ? edge : ans1;

// Unoin, always merge smaller set (pv) to larger set (pu)

if (sizes[pv] > sizes[pu])

swap(pu, pv);

roots[pv] = pu;

sizes[pu] += sizes[pv];

}

return ans2;

}

private:

int find(int node, vector<int>& roots) {

while (roots[node] != node) {

roots[node] = roots[roots[node]];

node = roots[node];

}

return node;

}

};

C++ / without using Union find

// Author: Huahua

// Runtime: 6 ms

class Solution {

public:

vector<int> findRedundantDirectedConnection(vector<vector<int>>& edges) {

vector<int> parents(edges.size() + 1, 0);

vector<int> ans1;

vector<int> ans2;

bool dup_parents = false;

for(auto& edge: edges) {

int u = edge[0];

int v = edge[1];

// A node has two parents

if (parents[v] > 0) {

ans1 = {parents[v], v};

ans2 = edge;

dup_parents = true;

// Delete the later edge

edge[0] = edge[1] = -1;

} else {

parents[v] = u;

}

// Reset parents

parents = vector<int>(edges.size() + 1, 0);

for(const auto& edge: edges) {

int u = edge[0];

int v = edge[1];

// Invalid edge (we deleted in step 1)

if (u < 0 || v < 0) continue;

parents[v] = u;

if (cycle(v, parents))

return dup_parents ? ans1 : edge;

}

return ans2;

}

private:

bool cycle(int v, const vector<int>& parents) {

int u = parents[v];

while (u) {

if (u == v) return true;

u = parents[u];

}

return false;

}

};

Python

"""

Author: Huahua

Runtime: 62 ms

"""

class Solution:

def findRedundantDirectedConnection(self, edges):

def cycle(v, p):

u = p[v]

while u != 0:

if u == v: return True

u = p[u]

return False

n = len(edges)

p = [0] * (n + 1)

ans1 = []

ans2 = []

dup_p = False

for e in edges:

u, v = e

if p[v] > 0:

ans1 = [p[v], v]

ans2 = [u, v]

dup_p = True

e[0] = e[1] = -1

else:

p[v] = u

p = [0] * (n + 1)

for u, v in edges:

if u < 0: continue

p[v] = u

if cycle(v, p):

return ans1 if dup_p else [u, v]

return ans2