Posts tagged as “medium”

花花酱 LeetCode 672. Bulb Switcher II

By zxi on March 9, 2020

There is a room with n lights which are turned on initially and 4 buttons on the wall. After performing exactly m unknown operations towards buttons, you need to return how many different kinds of status of the n lights could be.

Suppose n lights are labeled as number [1, 2, 3 …, n], function of these 4 buttons are given below:

Flip all the lights.
Flip lights with even numbers.
Flip lights with odd numbers.
Flip lights with (3k + 1) numbers, k = 0, 1, 2, …

Example 1:

Input: n = 1, m = 1.
Output: 2
Explanation: Status can be: [on], [off]

Example 2:

Input: n = 2, m = 1.
Output: 3
Explanation: Status can be: [on, off], [off, on], [off, off]

Example 3:

Input: n = 3, m = 1.
Output: 4
Explanation: Status can be: [off, on, off], [on, off, on], [off, off, off], [off, on, on].

Note: n and m both fit in range [0, 1000].

Solution1: Bitmask + Simulation

The light pattern will be repeated if we have more than 6 lights, so n = n % 6, n = 6 if n == 0.

Time complexity: O(m*2^6)
Space complexity: O(2^6)

C++

// Author: Huahua

class Solution {

public:

int flipLights(int n, int m) {

if (n > 6) n = n % 6 + 6;

unordered_set<int> s1{(1 << n) - 1};

auto flip = [n](int s, int start, int step) {

for (int i = start; i < n; i += step)

s ^= (1 << i);

return s;

};

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

unordered_set<int> s2;

for (int s : s1) {

s2.insert(flip(s, 0, 1));

s2.insert(flip(s, 0, 2));

s2.insert(flip(s, 1, 2));

s2.insert(flip(s, 0, 3));

}

s1.swap(s2);

}

return s1.size();

}

};

花花酱 LeetCode 529. Minesweeper

By zxi on March 9, 2020

Let’s play the minesweeper game (Wikipedia, online game)!

You are given a 2D char matrix representing the game board. ‘M’ represents an unrevealed mine, ‘E’ represents an unrevealed empty square, ‘B’ represents a revealed blank square that has no adjacent (above, below, left, right, and all 4 diagonals) mines, digit (‘1’ to ‘8’) represents how many mines are adjacent to this revealed square, and finally ‘X’ represents a revealed mine.

Now given the next click position (row and column indices) among all the unrevealed squares (‘M’ or ‘E’), return the board after revealing this position according to the following rules:

If a mine (‘M’) is revealed, then the game is over – change it to ‘X’.
If an empty square (‘E’) with no adjacent mines is revealed, then change it to revealed blank (‘B’) and all of its adjacent unrevealed squares should be revealed recursively.
If an empty square (‘E’) with at least one adjacent mine is revealed, then change it to a digit (‘1’ to ‘8’) representing the number of adjacent mines.
Return the board when no more squares will be revealed.

Example 1:

Input: 

[['E', 'E', 'E', 'E', 'E'],
 ['E', 'E', 'M', 'E', 'E'],
 ['E', 'E', 'E', 'E', 'E'],
 ['E', 'E', 'E', 'E', 'E']]

Click : [3,0]

Output: 

[['B', '1', 'E', '1', 'B'],
 ['B', '1', 'M', '1', 'B'],
 ['B', '1', '1', '1', 'B'],
 ['B', 'B', 'B', 'B', 'B']]

Explanation:

Example 2:

Input: 

[['B', '1', 'E', '1', 'B'],
 ['B', '1', 'M', '1', 'B'],
 ['B', '1', '1', '1', 'B'],
 ['B', 'B', 'B', 'B', 'B']]

Click : [1,2]

Output: 

[['B', '1', 'E', '1', 'B'],
 ['B', '1', 'X', '1', 'B'],
 ['B', '1', '1', '1', 'B'],
 ['B', 'B', 'B', 'B', 'B']]

Explanation:

Note:

The range of the input matrix’s height and width is [1,50].
The click position will only be an unrevealed square (‘M’ or ‘E’), which also means the input board contains at least one clickable square.
The input board won’t be a stage when game is over (some mines have been revealed).
For simplicity, not mentioned rules should be ignored in this problem. For example, you don’t need to reveal all the unrevealed mines when the game is over, consider any cases that you will win the game or flag any squares.

Solution: DFS

Time complexity: O(m*n)
Space complexity: O(m* n)

C++

// Author: Huahua

class Solution {

public:

vector<vector<char>> updateBoard(vector<vector<char>>& board, vector<int>& click) {

const int m = board.size();

const int n = board[0].size();

function<void(int, int)> dfs = [&](int x, int y) {

if (board[y][x] != 'E') return;

int c = 0;

for (int tx = x - 1; tx <= x + 1; ++tx)

for (int ty = y - 1; ty <= y + 1; ++ty)

if (tx >= 0 && tx < n && ty >= 0 && ty < m)

c += board[ty][tx] == 'M';

if (c > 0) {

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

return;

}

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

for (int tx = x - 1; tx <= x + 1; ++tx)

for (int ty = y - 1; ty <= y + 1; ++ty)

if (tx >= 0 && tx < n && ty >= 0 && ty < m)

dfs(tx, ty);

};

int x = click[1];

int y = click[0];

dfs(x, y);

if (board[y][x] == 'M') board[y][x] = 'X';

return board;

}

};

Solution 2: BFS

Python3

# Author: Huahua

class Solution:

def updateBoard(self, board: List[List[str]], click: List[int]) -> List[List[str]]:

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

cx, cy = click[1], click[0]

q = [(cx, cy)]

while q:

q2 = []

for x, y in q:

if board[y][x] != 'E': continue

c = 0

for tx in range(x - 1, x + 2):

for ty in range(y - 1, y + 2):

if 0 <= tx < n and 0 <= ty < m and board[ty][tx] == 'M':

c += 1

if c > 0:

board[y][x] = chr(ord('0') + c)

continue

board[y][x] = 'B'

for tx in range(x - 1, x + 2):

for ty in range(y - 1, y + 2):

if 0 <= tx < n and 0 <= ty < m and board[ty][tx] == 'E':

q2.append((tx, ty))

q = q2

if board[cy][cx] == 'M': board[y][x] = 'X'

return board

花花酱 LeetCode 227. Basic Calculator II

By zxi on March 9, 2020

Implement a basic calculator to evaluate a simple expression string.

The expression string contains only non-negative integers, +, -, *, / operators and empty spaces . The integer division should truncate toward zero.

Example 1:

Input: "3+2*2"
Output: 7

Example 2:

Input: " 3/2 "
Output: 1

Example 3:

Input: " 3+5 / 2 "
Output: 5

Note:

You may assume that the given expression is always valid.
Do not use the eval built-in library function.

Solution: Stack

if operator is ‘+’ or ‘-’, push the current num * sign onto stack.
if operator ‘*’ or ‘/’, pop the last num from stack and * or / by the current num and push it back to stack.

The answer is the sum of numbers on stack.

3+2*2 => {3}, {3,2}, {3, 2*2} = {3, 4} => ans = 7
3 +5/2 => {3}, {3,5}, {3, 5/2} = {3, 2} => ans = 5
1 + 2*3 – 5 => {1}, {1,2}, {1,2*3} = {1,6}, {1, 6, -5} => ans = 2

Time complexity: O(n)
Space complexity: O(n)

C++

// Author: Huahua

class Solution {

public:

int calculate(string s) {

vector<int> nums;

char op = '+';

int cur = 0;

int pos = 0;

while (pos < s.size()) {

if (s[pos] == ' ') {

++pos;

continue;

}

while (isdigit(s[pos]) && pos < s.size())

cur = cur * 10 + (s[pos++] - '0');

if (op == '+' || op == '-') {

nums.push_back(cur * (op == '+' ? 1 : -1));

} else if (op == '*') {

nums.back() *= cur;

} else if (op == '/') {

nums.back() /= cur;

}

cur = 0;

op = s[pos++];

}

return accumulate(begin(nums), end(nums), 0);

}

};

python3

# Author: Huahua

class Solution:

def calculate(self, s: str) -> int:

nums = []

op = '+'

cur = 0

i = 0

while i < len(s):

if s[i] == ' ':

i += 1

continue

while i < len(s) and s[i].isdigit():

cur = cur * 10 + ord(s[i]) - ord('0')

i += 1

if op in '+-':

nums.append(cur * (1 if op == '+' else -1))

elif op == '*':

nums[-1] *= cur

elif op == '/':

sign = -1 if nums[-1] < 0 or cur < 0 else 1

nums[-1] = abs(nums[-1]) // abs(cur) * sign

cur = 0

if (i < len(s)): op = s[i]

i += 1

return sum(nums)

花花酱 LeetCode 165. Compare Version Numbers

By zxi on March 9, 2020

Compare two version numbers version1 and version2.
If version1 > version2 return 1; if version1 < version2 return -1;otherwise return 0.

You may assume that the version strings are non-empty and contain only digits and the . character.

The . character does not represent a decimal point and is used to separate number sequences.

For instance, 2.5 is not “two and a half” or “half way to version three”, it is the fifth second-level revision of the second first-level revision.

You may assume the default revision number for each level of a version number to be 0. For example, version number 3.4 has a revision number of 3 and 4 for its first and second level revision number. Its third and fourth level revision number are both 0.

Example 1:

Input: version1 = "0.1", version2 = "1.1"
Output: -1

Example 2:

Input: version1 = "1.0.1", version2 = "1"
Output: 1

Example 3:

Input: version1 = "7.5.2.4", version2 = "7.5.3"
Output: -1

Example 4:

Input: version1 = "1.01", version2 = "1.001"
Output: 0
Explanation: Ignoring leading zeroes, both “01” and “001" represent the same number “1”

Example 5:

Input: version1 = "1.0", version2 = "1.0.0"
Output: 0
Explanation: The first version number does not have a third level revision number, which means its third level revision number is default to "0"

Note:

Version strings are composed of numeric strings separated by dots . and this numeric strings may have leading zeroes.
Version strings do not start or end with dots, and they will not be two consecutive dots.

Solution: String

Split the version string to a list of numbers, and compare two lists.

Time complexity: O(l1 + l2)
Space complexity: O(l1 + l2)

C++

// Author: Huahua

class Solution {

public:

int compareVersion(string version1, string version2) {

const auto& v1 = parseVersion(version1);

const auto& v2 = parseVersion(version2);

int i1 = 0, i2 = 0;

while (i1 < v1.size() || i2 < v2.size()) {

int n1 = i1 < v1.size() ? v1[i1++] : 0;

int n2 = i2 < v2.size() ? v2[i2++] : 0;

if (n1 < n2) return -1;

else if (n1 > n2) return 1;

}

return 0;

}

private:

vector<int> parseVersion(const string& version) {

vector<int> v;

int s = 0;

for (char c : version) {

if (c == '.') {

v.push_back(s);

s = 0;

} else {

s = s * 10 + (c - '0');

}

v.push_back(s);

return v;

}

};

花花酱 LeetCode 365. Water and Jug Problem

By zxi on March 9, 2020

You are given two jugs with capacities x and y litres. There is an infinite amount of water supply available. You need to determine whether it is possible to measure exactly z litres using these two jugs.

If z liters of water is measurable, you must have z liters of water contained within one or both buckets by the end.

Operations allowed:

Fill any of the jugs completely with water.
Empty any of the jugs.
Pour water from one jug into another till the other jug is completely full or the first jug itself is empty.

Example 1: (From the famous “Die Hard” example)

Input: x = 3, y = 5, z = 4
Output: True

Example 2:

Input: x = 2, y = 6, z = 5
Output: False

Solution: Math

special case 1: x == z or y == z or x + y == z: return True
special case 2: x + y < z: return False
normal case: z must be a factor of gcd(x, y)

Time complexity: O(log(min(x, y))
Space complexity: O(1)

C++

// Author: Huahua

class Solution {

public:

bool canMeasureWater(int x, int y, int z) {

if (x == z || y == z || x + y == z) return true;

if (x + y < z) return false;

return z % gcd(x, y) == 0;

}

};

Posts tagged as “medium”

花花酱 LeetCode 672. Bulb Switcher II

Solution1: Bitmask + Simulation

C++

花花酱 LeetCode 529. Minesweeper

Solution: DFS

C++

Solution 2: BFS

Python3

花花酱 LeetCode 227. Basic Calculator II

Solution: Stack

C++

python3

Related Problems

花花酱 LeetCode 165. Compare Version Numbers

C++

花花酱 LeetCode 365. Water and Jug Problem

Solution: Math

C++