花花酱 LeetCode 47. Permutations II

By zxi on July 26, 2018

Problem

Given a collection of numbers that might contain duplicates, return all possible unique permutations.

Example:

Input: [1,1,2]
Output:
[
  [1,1,2],
  [1,2,1],
  [2,1,1]
]

Solution

Time complexity: O(n!)

Space complexity: O(n + k)

// Author: Huahua

// Running time: 20 ms

class Solution {

public:

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

sort(begin(nums), end(nums));

vector<vector<int>> ans;

vector<int> used(nums.size());

vector<int> cur;

dfs(nums, cur, used, ans);

return ans;

}

private:

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

if (cur.size() == nums.size()) {

ans.push_back(cur);

return;

}

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

if (used[i]) continue;

// Same number can be only used once at each depth.

if (i > 0 && nums[i] == nums[i - 1] && !used[i - 1]) continue;

used[i] = 1;

cur.push_back(nums[i]);

dfs(nums, cur, used, ans);

cur.pop_back();

used[i] = 0;

}

};

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();

}

};

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 864. Shortest Path to Get All Keys

By zxi on July 8, 2018

Problem

We are given a 2-dimensional grid. "." is an empty cell, "#" is a wall, "@" is the starting point, ("a", "b", …) are keys, and ("A", "B", …) are locks.

We start at the starting point, and one move consists of walking one space in one of the 4 cardinal directions. We cannot walk outside the grid, or walk into a wall. If we walk over a key, we pick it up. We can’t walk over a lock unless we have the corresponding key.

For some 1 <= K <= 6, there is exactly one lowercase and one uppercase letter of the first K letters of the English alphabet in the grid. This means that there is exactly one key for each lock, and one lock for each key; and also that the letters used to represent the keys and locks were chosen in the same order as the English alphabet.

Return the lowest number of moves to acquire all keys. If it’s impossible, return -1.

Example 1:

Input: ["@.a.#","###.#","b.A.B"]
Output: 8

Example 2:

Input: ["@..aA","..B#.","....b"]
Output: 6

Note:

1 <= grid.length <= 30
1 <= grid[0].length <= 30
grid[i][j] contains only '.', '#', '@', 'a'-'f' and 'A'-'F'
The number of keys is in [1, 6]. Each key has a different letter and opens exactly one lock.

Solution: BFS

Time complexity: O(m*n*64)

Space complexity: O(m*n*64)

C++

// Author: Huahua

// Running time: 8 ms

class Solution {

public:

int shortestPathAllKeys(vector<string>& grid) {

int m = grid.size();

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

int all_keys = 0;

queue<int> q;

vector<vector<vector<int>>> seen(m, vector<vector<int>>(n, vector<int>(64, 0)));

// Init

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

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

const char c = grid[i][j];

if (c == '@') {

q.push((j << 16) | (i << 8));

seen[i][j][0] = 1;

} else if (c >= 'a' && c <= 'f') {

all_keys |= (1 << (c - 'a'));

}

const vector<int> dirs{-1, 0, 1, 0, -1};

int steps = 0;

while (!q.empty()) {

int size = q.size();

while (size--) {

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

int x = s >> 16;

int y = (s >> 8) & 0xFF;

int keys = s & 0xFF;

if (keys == all_keys) return steps;

for (int i = 0; i < 4; ++i) {

int nkeys = keys;

int nx = x + dirs[i];

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

if (nx < 0 || nx >= n || ny < 0 || ny >= m) continue;

const char c = grid[ny][nx];

if (c == '#') continue; // Wall

// Do not have the key.

if (c >= 'A' && c <= 'F' && !(keys & (1 << (c - 'A')))) continue;

// Update the keys we have.

if (c >= 'a' && c <= 'f') nkeys |= (1 << (c - 'a'));

if (seen[ny][nx][nkeys]) continue;

q.push((nx << 16) | (ny << 8) | nkeys);

seen[ny][nx][nkeys] = 1;

}

++steps;

}

return -1;

}

};

Python3

# Author: Huahua, 390 ms

class Solution:

def shortestPathAllKeys(self, grid):

m, n = len(grid), len(grid[0])

all_keys = 0

seen = [[[None]* 64 for _ in range(n)] for _ in range(m)]

q = collections.deque()

for i in range(m):

for j in range(n):

c = grid[i][j]

if c == '@':

q.append((j << 16) | (i << 8))

seen[i][j][0] = 1

elif c >= 'a' and c <= 'f':

all_keys |= (1 << (ord(c) - ord('a')))

dirs = [-1, 0, 1, 0, -1]

steps = 0

while q:

size = len(q)

while size > 0:

size -= 1

s = q.popleft()

x = s >> 16

y = (s >> 8) & 0xff

keys = s & 0xff

if keys == all_keys: return steps

for i in range(4):

nx, ny, nkeys = x + dirs[i], y + dirs[i + 1], keys

if nx < 0 or nx >= n or ny < 0 or ny >= m: continue

c = grid[ny][nx]

if c == '#': continue

if c in string.ascii_uppercase and keys & (1 << (ord(c) - ord('A'))) == 0: continue

if c in string.ascii_lowercase: nkeys |= (1 << (ord(c) - ord('a')))

if seen[ny][nx][nkeys]: continue

q.append((nx << 16) | (ny << 8) | nkeys)

seen[ny][nx][nkeys] = 1

steps += 1

return -1

Posts published in “Search”

花花酱 LeetCode 47. Permutations II

Problem

Solution

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花花酱 LeetCode 22. Generate Parentheses

Problem

Solution: DFS

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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 864. Shortest Path to Get All Keys

Problem

Solution: BFS

C++

Python3

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