Posts published in “Search”

花花酱 LeetCode 37. Sudoku Solver

By zxi on October 16, 2017

Problem:

Write a program to solve a Sudoku puzzle by filling the empty cells.

Empty cells are indicated by the character '.'.

You may assume that there will be only one unique solution.

A sudoku puzzle…

…and its solution numbers marked in red.

Idea:

DFS + backtracking

Solution:

C++

// Author: Huahua

// Runtime: 3 ms

class Solution {

public:

void solveSudoku(vector<vector<char>>& board) {

rows_ = vector<vector<int>>(9, vector<int>(10));

cols_ = vector<vector<int>>(9, vector<int>(10));

boxes_ = vector<vector<int>>(9, vector<int>(10));

for(int i = 0; i < 9; i++)

for(int j = 0; j < 9; j++) {

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

if ( c != '.') {

int n = c - '0';

int bx = j / 3;

int by = i / 3;

rows_[i][n] = 1;

cols_[j][n] = 1;

boxes_[by * 3 + bx][n] = 1;

}

fill(board, 0, 0);

}

private:

vector<vector<int>> rows_, cols_, boxes_;

bool fill(vector<vector<char>>& board, int x, int y) {

if (y == 9)

return true;

int nx = (x + 1) % 9;

int ny = (nx == 0) ? y + 1 : y;

if (board[y][x] != '.') return fill(board, nx, ny);

for (int i = 1; i <= 9; i++) {

int bx = x / 3;

int by = y / 3;

int box_key = by * 3 + bx;

if (!rows_[y][i] && !cols_[x][i] && !boxes_[box_key][i]) {

rows_[y][i] = 1;

cols_[x][i] = 1;

boxes_[box_key][i] = 1;

board[y][x] = i + '0';

if (fill(board, nx, ny)) return true;

board[y][x] = '.';

boxes_[box_key][i] = 0;

cols_[x][i] = 0;

rows_[y][i] = 0;

}

return false;

}

};

Related Problems:

[解题报告] LeetCode 51. N-Queens

花花酱 LeetCode 40. Combination Sum II

By zxi on October 15, 2017

Problem:

Given a collection of candidate numbers (C) and a target number (T), find all unique combinations in C where the candidate numbers sums to T.

Each number in C may only be used once in the combination.

Note:

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

For example, given candidate set [10, 1, 2, 7, 6, 1, 5] and target 8,
A solution set is:

[

[1, 7],

[1, 2, 5],

[2, 6],

[1, 1, 6]

]

Idea:

DFS

Solution:

C++ / set

// Author: Huahua

// Runtime: 9 ms

class Solution {

public:

vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {

set<vector<int>> ans;

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

vector<int> curr;

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

return vector<vector<int>>(ans.begin(), ans.end());

}

private:

void dfs(const vector<int>& candidates,

int target, int s,

set<vector<int>>& ans,

vector<int>& curr) {

if (target == 0) {

ans.insert(curr);

return;

}

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

int num = candidates[i];

if (num > target) return;

curr.push_back(num);

dfs(candidates, target - num, i + 1, ans, curr);

curr.pop_back();

}

};

C++ / vector

// Author: Huahua

// Runtime: 9ms

class Solution {

public:

vector<vector<int>> combinationSum2(vector<int>& candidates, int target) {

vector<vector<int>> ans;

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

vector<int> curr;

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

return ans;

}

private:

void dfs(const vector<int>& candidates,

int target, int s,

vector<vector<int>>& ans,

vector<int>& curr) {

if (target == 0) {

ans.push_back(curr);

return;

}

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

int num = candidates[i];

if (num > target) return;

if (i > s && candidates[i] == candidates[i - 1]) continue;

curr.push_back(num);

dfs(candidates, target - num, i + 1, ans, curr);

curr.pop_back();

}

};

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 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;

}

};