Posts tagged as “DFS”

花花酱 LeetCode 488. Zuma Game

By zxi on October 9, 2017

https://leetcode.com/problems/zuma-game/description/

Problem:

Think about Zuma Game. You have a row of balls on the table, colored red(R), yellow(Y), blue(B), green(G), and white(W). You also have several balls in your hand.

Each time, you may choose a ball in your hand, and insert it into the row (including the leftmost place and rightmost place). Then, if there is a group of 3 or more balls in the same color touching, remove these balls. Keep doing this until no more balls can be removed.

Find the minimal balls you have to insert to remove all the balls on the table. If you cannot remove all the balls, output -1.

Examples:
Input: “WRRBBW”, “RB”
Output: -1
Explanation: WRRBBW -> WRR[R]BBW -> WBBW -> WBB[B]W -> WW

Input: “WWRRBBWW”, “WRBRW”
Output: 2
Explanation: WWRRBBWW -> WWRR[R]BBWW -> WWBBWW -> WWBB[B]WW -> WWWW -> empty

Input:“G”, “GGGGG”
Output: 2
Explanation: G -> G[G] -> GG[G] -> empty

Input: “RBYYBBRRB”, “YRBGB”
Output: 3
Explanation: RBYYBBRRB -> RBYY[Y]BBRRB -> RBBBRRB -> RRRB -> B -> B[B] -> BB[B] -> empty

Note:

You may assume that the initial row of balls on the table won’t have any 3 or more consecutive balls with the same color.
The number of balls on the table won’t exceed 20, and the string represents these balls is called “board” in the input.
The number of balls in your hand won’t exceed 5, and the string represents these balls is called “hand” in the input.
Both input strings will be non-empty and only contain characters ‘R’,’Y’,’B’,’G’,’W’.

Idea: Search

Solution1: C++ / Search

// Author: Huahua

// Runtime: 3 ms

class Solution {

public:

int findMinStep(string board, string hand) {

vector<int> h(128, 0);

for (char color : hand) ++h[color];

return dfs(board, h);

}

private:

// Return the min # of balls needed in hand to clear the board.

// Returns -1 if not possible.

int dfs(const string& board, vector<int>& hand) {

if (board.empty()) return 0;

int ans = INT_MAX;

int i = 0;

int j = 0;

while (i < board.size()) {

while (j < board.size() && board[i] == board[j]) ++j;

// board[i] ~ board[j - 1] have the same color

const char color = board[i];

// Number of balls needed to clear board[i] ~ board[j - 1]

const int b = 3 - (j - i);

// Have sufficient balls in hand

if (hand[color] >= b) {

// Remove board[i] ~ board[j - 1] and update the board

string nb = update(board.substr(0, i) + board.substr(j));

// Find the solution on new board with updated hand

hand[color] -= b;

int r = dfs(nb, hand);

if (r >= 0) ans = min(ans, r + b);

// Recover the balls in hand

hand[color] += b;

}

i = j;

}

return ans == INT_MAX ? -1 : ans;

}

// Update the board by removing all consecutive 3+ balls.

// "YWWRRRWWYY" -> "YWWWWYY" -> "YYY" -> ""

string update(string board) {

int i = 0;

while (i < board.size()) {

int j = i;

while (j < board.size() && board[i] == board[j]) ++j;

if (j - i >= 3) {

board = board.substr(0, i) + board.substr(j);

i = 0;

} else {

++i;

}

return board;

}

};

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 39. Combination Sum

By zxi on October 3, 2017

Problem:

Given a set of candidate numbers (C) (without duplicates) and a target number (T), find all unique combinations in C where the candidate numbers sums to T.

The same repeated number may be chosen from C unlimited number of times.

Note:

All numbers (including target) will be positive integers.
The solution set must not contain duplicate combinations.

For example, given candidate set [2, 3, 6, 7] and target 7,
A solution set is:

[

[7],

[2, 2, 3]

]

Idea:

DFS

Solution: C++

class Solution {

public:

vector<vector<int>> combinationSum(vector<int>& candidates, int target) {

vector<vector<int>> ans;

vector<int> cur;

std::sort(candidates.begin(), candidates.end());

dfs(candidates, target, 0, cur, ans);

return ans;

}

private:

void dfs(vector<int>& candidates, int target, int s, vector<int>& cur, vector<vector<int>>& ans) {

if (target == 0) {

ans.push_back(cur);

return;

}

for (int i = s; i < candidates.size(); ++i) {

if (candidates[i] > target) break;

cur.push_back(candidates[i]);

dfs(candidates, target - candidates[i], i, cur, ans);

cur.pop_back();

}

};

Python

"""

Author: Huahua

Runtime: 75 ms (batter than 97.04%)

"""

class Solution(object):

def combinationSum(self, candidates, target):

def dfs(candidates, target, s, curr, ans):

if target == 0:

ans.append(curr[:])

return

for i in xrange(s, len(candidates)):

if candidates[i] > target: return

curr.append(candidates[i])

dfs(candidates, target - candidates[i], i, curr, ans)

curr.pop()

ans = []

candidates.sort()

dfs(candidates, target, 0, [], ans);

return ans

花花酱 LeetCode 690. Employee Importance

By zxi on September 28, 2017

https://leetcode.com/problems/employee-importance/description/

Problem:

ou are given a data structure of employee information, which includes the employee’s unique id, his importance value and his direct subordinates’ id.

For example, employee 1 is the leader of employee 2, and employee 2 is the leader of employee 3. They have importance value 15, 10 and 5, respectively. Then employee 1 has a data structure like [1, 15, [2]], and employee 2 has [2, 10, [3]], and employee 3 has [3, 5, []]. Note that although employee 3 is also a subordinate of employee 1, the relationship is not direct.

Now given the employee information of a company, and an employee id, you need to return the total importance value of this employee and all his subordinates.

Example 1:

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

Output: 11

Explanation:

Employee 1 has importance value 5, and he has two direct subordinates: employee 2 and employee 3. They both have importance value 3.

So the total importance value of employee 1 is 5 + 3 + 3 = 11.

Note:

One employee has at most one direct leader and may have several subordinates.
The maximum number of employees won’t exceed 2000.

Idea:

BFS / DFS

Time complexity: O(n)

Space complexity: O(n)

Solution:

C++ / BFS

// Author: Huahua

// Running time: 13 ms

class Solution {

public:

int getImportance(vector<Employee*> employees, int id) {

unordered_map<int, Employee*> es;

for (auto e : employees)

es.emplace(e->id, e);

queue<int> q;

q.push(id);

int sum = 0;

while (!q.empty()) {

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

auto e = es[eid];

sum += e->importance;

for (auto rid : e->subordinates)

q.push(rid);

}

return sum;

}

};

C++ / DFS

class Solution {

public:

int getImportance(vector<Employee*> employees, int id) {

unordered_map<int, Employee*> es;

for (auto e : employees)

es.emplace(e->id, e);

return dfs(id, es);

}

private:

int dfs(int id, const unordered_map<int, Employee*>& es) {

const auto e = es.at(id);

int sum = e->importance;

for (int rid : e->subordinates)

sum += dfs(rid, es);

return sum;

}

};

花花酱 LeetCode 684. Redundant Connection

By zxi on September 28, 2017

https://leetcode.com/problems/redundant-connection/description/

Problem:

In this problem, a tree is an undirected graph that is connected and has no cycles.

The given input is a graph that started as a tree with N nodes (with distinct values 1, 2, …, N), with one additional 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] with u < v, that represents an undirected edge connecting nodes u and v.

Return an edge that can be removed so that the resulting graph is a tree of N nodes. If there are multiple answers, return the answer that occurs last in the given 2D-array. The answer edge [u, v] should be in the same format, with u < v.

Example 1:

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

Output: [2,3]

Explanation: The given undirected graph will be like this:

/ \

2 - 3

Example 2:

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

Output: [1,4]

Explanation: The given undirected graph will be like this:

5 - 1 - 2

| |

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:
DFS / Union-Find

Solutions:

C++ / DFS

// Author: Huahua

// Time Complexity: O(n^2)

// Running Time: 22 ms

class Solution {

public:

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

unordered_map<int, vector<int>> graph;

for (const auto& edge : edges) {

int u = edge[0];

int v = edge[1];

unordered_set<int> visited;

if (hasPath(u, v, graph, visited))

return edge;

graph[u].push_back(v);

graph[v].push_back(u);

}

return {};

}

private:

bool hasPath(int curr,

int goal,

const unordered_map<int, vector<int>>& graph,

unordered_set<int>& visited) {

if (curr == goal) return true;

visited.insert(curr);

if (!graph.count(curr) || !graph.count(goal)) return false;

for (int next : graph.at(curr)) {

if (visited.count(next)) continue;

if (hasPath(next, goal, graph, visited)) return true;

}

return false;

}

};

C++ / Union Find

// Author: Huahua

// Running time: 6 ms

class UnionFindSet {

public:

UnionFindSet(int n) {

ranks_ = vector<int>(n + 1, 0);

parents_ = vector<int>(n + 1, 0);

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

parents_[i] = i;

}

// Merge sets that contains u and v.

// Return true if merged, false if u and v are already in one set.

bool Union(int u, int v) {

int pu = Find(u);

int pv = Find(v);

if (pu == pv) return false;

// Meger low rank tree into high rank tree

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

parents_[pu] = pv;

else if (ranks_[pu] > ranks_[pv])

parents_[pv] = pu;

else {

parents_[pv] = pu;

ranks_[pv] += 1;

}

return true;

}

// Get the root of u.

int Find(int u) {

// Compress the path during traversal

if (u != parents_[u])

parents_[u] = Find(parents_[u]);

return parents_[u];

}

private:

vector<int> parents_;

vector<int> ranks_;

};

class Solution {

public:

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

UnionFindSet s(edges.size());

for(const auto& edge: edges)

if (!s.Union(edge[0], edge[1]))

return edge;

return {};

}

};

Java / Union Find

// Author: Huahua

// Runtime: 6 ms

class Solution {

class UnionFindSet {

private int[] parents_;

private int[] ranks_;

public UnionFindSet(int n) {

parents_ = new int[n + 1];

ranks_ = new int[n + 1];

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

parents_[i] = i;

ranks_[i] = 1;

}

public boolean Union(int u, int v) {

int pu = Find(u);

int pv = Find(v);

if (pu == pv) return false;

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

parents_[pu] = pv;

else if (ranks_[pu] > ranks_[pv])

parents_[pv] = pu;

else {

parents_[pv] = pu;

ranks_[pu] += 1;

}

return true;

}

public int Find(int u) {

while (parents_[u] != u) {

parents_[u] = parents_[parents_[u]];

u = parents_[u];

}

return u;

}

public int[] findRedundantConnection(int[][] edges) {

UnionFindSet s = new UnionFindSet(edges.length);

for (int[] edge : edges)

if (!s.Union(edge[0], edge[1]))

return edge;

return null;

}

Python: Union Find

"""

Author: Huahua

Running Time: 66 ms

"""

class Solution:

def findRedundantConnection(self, edges):

p = [0]*(len(edges) + 1)

s = [1]*(len(edges) + 1)

def find(u):

while p[u] != u:

p[u] = p[p[u]]

u = p[u]

return u

for u, v in edges:

if p[u] == 0: p[u] = u

if p[v] == 0: p[v] = v

pu, pv = find(u), find(v)

if pu == pv: return [u, v]

if s[pv] > s[pu]: u, v = v, u

p[pv] = pu

s[pu] += s[pv]

return []

Python / Union Find V2

"""

Author: Huahua

Runtime: 59 ms (<92.42%)

"""

class UnionFindSet:

def __init__(self, n):

self._parents = [i for i in range(n + 1)]

self._ranks = [1 for i in range(n + 1)]

def find(self, u):

while u != self._parents[u]:

self._parents[u] = self._parents[self._parents[u]]

u = self._parents[u]

return u

def union(self, u, v):

pu, pv = self.find(u), self.find(v)

if pu == pv: return False

if self._ranks[pu] < self._ranks[pv]:

self._parents[pu] = pv

elif self._ranks[pu] > self._ranks[pv]:

self._parents[pv] = pu

else:

self._parents[pv] = pu

self._ranks[pu] += 1

return True

class Solution:

def findRedundantConnection(self, edges):

s = UnionFindSet(len(edges))

for edge in edges:

if not s.union(edge[0], edge[1]): return edge

return None

ndSet(len(edges))

for edge in edges:

if not s.union(edge[0], edge[1]): return edge

return None