Posts tagged as “DFS”

花花酱 LeetCode 22. Generate Parentheses

By zxi on July 26, 2018

Problem

Given n pairs of parentheses, write a function to generate all combinations of well-formed parentheses.

For example, given n = 3, a solution set is:

[
  "((()))",
  "(()())",
  "(())()",
  "()(())",
  "()()()"
]

Solution: DFS

Time complexity: O(2^n)

Space complexity: O(k + n)

C++

// Author: Huahua

// Running time: 0 ms

class Solution {

public:

vector<string> generateParenthesis(int n) {

vector<string> ans;

string cur;

if (n > 0) dfs(n, n, cur, ans);

return ans;

}

private:

void dfs(int l, int r, string& s, vector<string>& ans) {

if (l + r == 0) {

ans.push_back(s);

return;

}

if (r < l) return;

if (l > 0) {

dfs(l - 1, r, s += "(", ans);

s.pop_back();

}

if (r > 0) {

dfs(l, r - 1, s += ")", ans);

s.pop_back();

}

};

Problem

https://leetcode.com/problems/is-graph-bipartite/

Given an undirected graph, return true if and only if it is bipartite.

Recall that a graph is bipartite if we can split it’s set of nodes into two independent subsets A and B such that every edge in the graph has one node in A and another node in B.

The graph is given in the following form: graph[i] is a list of indexes j for which the edge between nodes i and j exists. Each node is an integer between 0 and graph.length - 1. There are no self edges or parallel edges: graph[i] does not contain i, and it doesn’t contain any element twice.

Example 1:
Input: [[1,3], [0,2], [1,3], [0,2]]
Output: true
Explanation: 
The graph looks like this:
0----1
|    |
|    |
3----2
We can divide the vertices into two groups: {0, 2} and {1, 3}.

Example 2:
Input: [[1,2,3], [0,2], [0,1,3], [0,2]]
Output: false
Explanation: 
The graph looks like this:
0----1
| \  |
|  \ |
3----2
We cannot find a way to divide the set of nodes into two independent subsets.

Note:

graph will have length in range [1, 100].
graph[i] will contain integers in range [0, graph.length - 1].
graph[i] will not contain i or duplicate values.
The graph is undirected: if any element j is in graph[i], then i will be in graph[j].

Solution: Graph Coloring

For each node

If has not been colored, color it to RED(1).
Color its neighbors with a different color RED(1) to BLUE(-1) or BLUE(-1) to RED(-1).

If we can finish the coloring then the graph is bipartite. All red nodes on the left no connections between them and all blues nodes on the right, again no connections between them. red and blue nodes are neighbors.

Time complexity: O(V+E)

Space complexity: O(V)

C++ / DFS

// Author: Huahua

// Running time: 12 ms

class Solution {

public:

bool isBipartite(vector<vector<int>>& graph) {

const int n = graph.size();

vector<int> colors(n);

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

if (!colors[i] && !coloring(graph, colors, 1, i))

return false;

return true;

}

private:

bool coloring(const vector<vector<int>>& graph, vector<int>& colors, int color, int node) {

if (colors[node]) return colors[node] == color;

colors[node] = color;

for (int nxt : graph[node])

if (!coloring(graph, colors, -color, nxt)) return false;

return true;

}

};

Solution: DFS

Time complexity: O(2^n)

Space complexity: O(n)

C++

// Author: Huahua

// Running time: 140 ms (<99.88%)

class Solution {

public:

vector<vector<int>> findSubsequences(vector<int>& nums) {

vector<vector<int>> ans;

vector<int> cur;

dfs(nums, 0, cur, ans);

return ans;

}

private:

void dfs(const vector<int>& nums, int s, vector<int>& cur, vector<vector<int>>& ans) {

unordered_set<int> seen;

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

if (!cur.empty() && nums[i] < cur.back()) continue;

// each number can be used only once at the same depth.

if (seen.count(nums[i])) continue;

seen.insert(nums[i]);

cur.push_back(nums[i]);

if (cur.size() > 1)

ans.push_back(cur);

dfs(nums, i + 1, cur, ans);

cur.pop_back();

}

};

花花酱 LeetCode 698. Partition to K Equal Sum Subsets

By zxi on July 13, 2018

Problem

Given an array of integers nums and a positive integer k, find whether it’s possible to divide this array into knon-empty subsets whose sums are all equal.

Example 1:

Input: nums = [4, 3, 2, 3, 5, 2, 1], k = 4
Output: True
Explanation: It's possible to divide it into 4 subsets (5), (1, 4), (2,3), (2,3) with equal sums.

Note:

1 <= k <= len(nums) <= 16.
0 < nums[i] < 10000.

Solution: Search

Time complexity: O(n!)

Space complexity: O(n)

// Author: Huahua

// Running time: 4 ms (<99.21%)

class Solution {

public:

bool canPartitionKSubsets(vector<int>& nums, int k) {

const int sum = accumulate(nums.begin(), nums.end(), 0);

if (sum % k != 0) return false;

sort(nums.rbegin(), nums.rend());

return dfs(nums, sum / k, 0, k, 0);

}

private:

bool dfs(const vector<int>& nums, int target, int cur, int k, int used) {

if (k == 0) return used == (1 << nums.size()) - 1;

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

if (used & (1 << i)) continue;

int t = cur + nums[i];

if (t > target) break; // Important

int new_used = used | (1 << i);

if (t == target && dfs(nums, target, 0, k - 1, new_used)) return true;

else if (dfs(nums, target, t, k, new_used)) return true;

}

return false;

}

};

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);

}

};

Posts tagged as “DFS”

花花酱 LeetCode 22. Generate Parentheses

Problem

Solution: DFS

Related Problems

花花酱 LeetCode 785. Is Graph Bipartite?

Problem

Solution: Graph Coloring

Related Problem

花花酱 LeetCode 491. Increasing Subsequences

Problem

Solution: DFS

花花酱 LeetCode 698. Partition to K Equal Sum Subsets

Problem

Solution: Search

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

Problem

Solution1: DFS + BFS

Solution 2: Recursion