Huahua’s Tech Road

花花酱 LeetCode 795. Number of Subarrays with Bounded Maximum

By zxi on March 3, 2018

题目大意：问一个数组中有多少个子数组的最大元素值在[L, R]的范围里。

We are given an array A of positive integers, and two positive integers L and R (L <= R).

Return the number of (contiguous, non-empty) subarrays such that the value of the maximum array element in that subarray is at least L and at most R.

Example :

Input:

A = [2, 1, 4, 3]

L = 2

R = 3

Output: 3

Explanation: There are three subarrays that meet the requirements: [2], [2, 1], [3].

Solution 1:

C++

// Author: Huahua

// Running time: 42 ms

class Solution {

public:

int numSubarrayBoundedMax(vector<int>& A, int L, int R) {

return count(A, R) - count(A, L - 1);

}

private:

// Count # of sub arrays whose max element is <= N

int count(const vector<int>& A, int N) {

int ans = 0;

int cur = 0;

for (int a: A) {

if (a <= N)

ans += ++cur;

else

cur = 0;

}

return ans;

}

};

Solution 2: One pass

C++

// Author: Huahua

// Running time: 38 ms

class Solution {

public:

int numSubarrayBoundedMax(vector<int>& A, int L, int R) {

int ans = 0;

int left = -1;

int right = -1;

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

if (A[i] >= L) right = i;

if (A[i] > R) left = i;

ans += (right - left);

}

return ans;

}

};

花花酱 LeetCode 794. Valid Tic-Tac-Toe State

By zxi on March 3, 2018

题目大意：判断一个井字棋的棋盘是否有效。

A Tic-Tac-Toe board is given as a string array board. Return True if and only if it is possible to reach this board position during the course of a valid tic-tac-toe game.

The board is a 3 x 3 array, and consists of characters " ", "X", and "O". The ” ” character represents an empty square.

Here are the rules of Tic-Tac-Toe:

Players take turns placing characters into empty squares (” “).
The first player always places “X” characters, while the second player always places “O” characters.
“X” and “O” characters are always placed into empty squares, never filled ones.
The game ends when there are 3 of the same (non-empty) character filling any row, column, or diagonal.
The game also ends if all squares are non-empty.
No more moves can be played if the game is over.

Example 1:

Input: board = ["O ", " ", " "]

Output: false

Explanation: The first player always plays "X".

Example 2:

Input: board = ["XOX", " X ", " "]

Output: false

Explanation: Players take turns making moves.

Example 3:

Input: board = ["XXX", " ", "OOO"]

Output: false

Example 4:

Input: board = ["XOX", "O O", "XOX"]

Output: true

Note:

board is a length-3 array of strings, where each string board[i] has length 3.
Each board[i][j] is a character in the set {" ", "X", "O"}.

Idea: Verify all rules

C++

class Solution {

public:

bool validTicTacToe(vector<string>& board) {

int x = 0;

int o = 0;

int wins_x = getWins(board, 'X', x);

int wins_o = getWins(board, 'O', o);

return wins_x + wins_o <= 1 && x >= o + wins_x && x <= o + 1 - wins_o;

}

private:

int getWins(const vector<string>& board, char p, int& count) {

vector<string> rows(3);

string diag1, diag2;

for (int i = 0; i < 3; ++i) {

diag1 += board[i][i];

diag2 += board[i][2 - i];

for (int j = 0; j < 3; ++j) {

char c = board[i][j];

rows[j] += c;

count += c == p;

}

string win(3, p);

int wins = (diag1 == win) + (diag2 == win);

for (int i = 0; i < 3; ++i) {

wins += rows[i] == win;

wins += board[i] == win;

}

return wins;

}

};

花花酱 LeetCode 557 Reverse Words in a String III

By zxi on March 2, 2018

题目大意：独立反转字符串中的每个单词。

Given a string, you need to reverse the order of characters in each word within a sentence while still preserving whitespace and initial word order.

Example 1:

1 2	Input: "Let's take LeetCode contest" Output: "s'teL ekat edoCteeL tsetnoc"

Note: In the string, each word is separated by single space and there will not be any extra space in the string.

Idea: Brute Force

C++

// Author: Huahua

// Running time: 23 ms

class Solution {

public:

string reverseWords(string s) {

int index = 0;

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

if (i == s.length() || s[i] == ' ') {

std::reverse(s.begin() + index, s.begin() + i);

index = i + 1;

}

return s;

}

};

Python3

"""

Author: Huahua

Running time: 44 ms (beats 100%)

"""

class Solution:

def reverseWords(self, s):

return ' '.join(map(lambda x: x[::-1], s.split(' ')))

花花酱 LeetCode 508. Most Frequent Subtree Sum

By zxi on March 2, 2018

Given the root of a tree, you are asked to find the most frequent subtree sum. The subtree sum of a node is defined as the sum of all the node values formed by the subtree rooted at that node (including the node itself). So what is the most frequent subtree sum value? If there is a tie, return all the values with the highest frequency in any order.

Examples 1
Input:

/ \

2 -3

return [2, -3, 4], since all the values happen only once, return all of them in any order.

Examples 2
Input:

/ \

2 -5

return [2], since 2 happens twice, however -5 only occur once.

Note: You may assume the sum of values in any subtree is in the range of 32-bit signed integer.

// Author: Huahua

// Running time: 18 ms

class Solution {

public:

vector<int> findFrequentTreeSum(TreeNode* root) {

unordered_map<int, int> freqs;

int max_freq = -1;

vector<int> ans;

(void)treeSum(root, freqs, max_freq, ans);

return ans;

}

private:

int treeSum(TreeNode* root, unordered_map<int, int>& freqs, int& max_freq, vector<int>& ans) {

if (!root) return 0;

int sum = root->val +

treeSum(root->left, freqs, max_freq, ans) +

treeSum(root->right, freqs, max_freq, ans);

int freq = ++freqs[sum];

if (freq > max_freq) {

max_freq = freq;

ans.clear();

}

if (freq == max_freq)

ans.push_back(sum);

return sum;

}

};

Python

"""

Author: Huahua

Running time: 72 ms (beats 100%)

"""

class Solution:

def findFrequentTreeSum(self, root):

if not root: return []

def treeSum(root):

if not root: return 0

s = root.val + treeSum(root.left) + treeSum(root.right)

f[s] += 1

return s

f = collections.Counter()

treeSum(root)

max_freq = max(f.values())

return [s for s in f.keys() if f[s] == max_freq]

花花酱 LeetCode 518. Coin Change 2

By zxi on March 2, 2018

题目大意：给你一些硬币的面值，问使用这些硬币（无限多块）能够组成amount的方法有多少种。

You are given coins of different denominations and a total amount of money. Write a function to compute the number of combinations that make up that amount. You may assume that you have infinite number of each kind of coin.

Note: You can assume that

0 <= amount <= 5000
1 <= coin <= 5000
the number of coins is less than 500
the answer is guaranteed to fit into signed 32-bit integer

Example 1:

Input: amount = 5, coins = [1, 2, 5]

Output: 4

Explanation: there are four ways to make up the amount:

5=5

5=2+2+1

5=2+1+1+1

5=1+1+1+1+1

Example 2:

Input: amount = 3, coins = [2]

Output: 0

Explanation: the amount of 3 cannot be made up just with coins of 2.

Example 3:

1 2	Input: amount = 10, coins = [10] Output: 1

Idea: DP

Transition 1:

Let us use dp[i][j] to denote the number of ways to sum up to amount j using first i kind of coins.

dp[i][j] = dp[i – 1][j – coin] + dp[i – 1][j – 2* coin] + …

Time complexity: O(n*amount^2) TLE

Space complexity: O(n*amount) -> O(amount)

Transition 2:

Let us use dp[i] to denote the number of ways to sum up to amount i.

dp[i + coin] += dp[i]

Time complexity: O(n*amount)

Space complexity: O(amount)

C++

// Author: Huahua

// Running time: 4 ms

class Solution {

public:

int change(int amount, vector<int>& coins) {

vector<int> dp(amount + 1, 0);

dp[0] = 1;

for (const int coin : coins)

for (int i = 0; i <= amount - coin; ++i)

dp[i + coin] += dp[i];

return dp[amount];

}

};

Java

// Author: Huahua

// Running time: 7 ms

class Solution {

public int change(int amount, int[] coins) {

int[] dp = new int[amount + 1];

dp[0] = 1;

for (int coin : coins)

for (int i = 0; i <= amount - coin; ++i)

dp[i + coin] += dp[i];

return dp[amount];

}

Python

"""

Author: Huahua

Running time: 107 ms (beats 100%)

"""

class Solution(object):

def change(self, amount, coins):

dp = [1] + [0] * (amount);

for coin in coins:

for i in xrange(amount - coin + 1):

if dp[i]:

dp[i + coin] += dp[i]

return dp[amount]

Related Problems:

花花酱 LeetCode 322. Coin Change

Huahua's Tech Road

花花酱 LeetCode 795. Number of Subarrays with Bounded Maximum

花花酱 LeetCode 794. Valid Tic-Tac-Toe State

花花酱 LeetCode 557 Reverse Words in a String III

花花酱 LeetCode 508. Most Frequent Subtree Sum

花花酱 LeetCode 518. Coin Change 2