Huahua’s Tech Road

:= 海象运算符 Walrus Operator – Python Weekly EP2

By zxi on May 28, 2020

Python3

// Author: Huahua

import re

import io

# Example 1:

a = [0] * 100

if len(a) > 10:

print(f"List is too long ({len(a)} elements, expected <= 10)")

n = len(a)

if n > 10:

print(f"List is too long ({n} elements, expected <= 10)")

if (x := len(a)) > 10:

print(f"List is too long ({x} elements, expected <= 10)")

# Example 2, bad

ads = "Now 20% off till 6/18"

m1 = re.search(r'(\d+)% off', ads)

discount1 = float(m1.group(1)) / 100 if m1 else 0.0

discount2 = float(m2.group(1)) / 100 if (m2 := re.search(r'(\d+)% off', ads)) else 0.0

print(f'{discount1 = }, {discount2 = }')

# Example 3

f1 = io.StringIO("123456789")

x1 = f1.read(4)

while x1 != '':

print(x1)

x1 = f1.read(4)

f2 = io.StringIO("123456789")

while (x2 := f2.read(4)) != '':

print(x2)

# Example 4

prog_langs = {'c++', 'python', 'java'}

langs = ['C++', 'Java', 'PYthon', 'English', '中文']

l1 = [lang.lower() for lang in langs if lang.lower() in prog_langs]

print(l1)

l2 = [l for lang in langs if (l := lang.lower()) in prog_langs]

print(l2)

花花酱 LeetCode 1458. Max Dot Product of Two Subsequences

By zxi on May 24, 2020

Given two arrays nums1 and nums2.

Return the maximum dot product between non-empty subsequences of nums1 and nums2 with the same length.

A subsequence of a array is a new array which is formed from the original array by deleting some (can be none) of the characters without disturbing the relative positions of the remaining characters. (ie, [2,3,5] is a subsequence of [1,2,3,4,5] while [1,5,3] is not).

Example 1:

Input: nums1 = [2,1,-2,5], nums2 = [3,0,-6]
Output: 18
Explanation: Take subsequence [2,-2] from nums1 and subsequence [3,-6] from nums2.
Their dot product is (2*3 + (-2)*(-6)) = 18.

Example 2:

Input: nums1 = [3,-2], nums2 = [2,-6,7]
Output: 21
Explanation: Take subsequence [3] from nums1 and subsequence [7] from nums2.
Their dot product is (3*7) = 21.

Example 3:

Input: nums1 = [-1,-1], nums2 = [1,1]
Output: -1
Explanation: Take subsequence [-1] from nums1 and subsequence [1] from nums2.
Their dot product is -1.

Constraints:

1 <= nums1.length, nums2.length <= 500
-1000 <= nums1[i], nums2[i] <= 1000

Solution: DP

dp[i][j] := max product of nums1[0~i], nums2[0~j].

dp[i][j] = max(dp[i-1][j], dp[i][j -1], max(0, dp[i-1][j-1]) + nums1[i]*nums2[j])

Time complexity: O(n1*n2)
Space complexity: O(n1*n2)

C++

// Author: Huahua

class Solution {

public:

int maxDotProduct(vector<int>& nums1, vector<int>& nums2) {

const int n1 = nums1.size();

const int n2 = nums2.size();

// dp[i][j] = max dot product of two subsequences

// of nums1[0:i] and nums2[0:j]

vector<vector<int>> dp(n1 + 1, vector<int>(n2 + 1, INT_MIN / 2));

for (int i = 1; i <= n1; ++i)

for (int j = 1; j <= n2; ++j)

dp[i][j] = max({

dp[i - 1][j],

dp[i][j - 1],

max(0, dp[i - 1][j - 1]) + nums1[i - 1] * nums2[j - 1]

});

return dp[n1][n2];

}

};

C++

// Author: Huahua

class Solution {

public:

int maxDotProduct(vector<int>& nums1, vector<int>& nums2) {

const int n1 = nums1.size();

const int n2 = nums2.size();

// dp[i][j] = max dot product of two subsequences

// of nums1[0~i] and nums2[0~j]

vector<vector<int>> dp(n1, vector<int>(n2));

for (int i = 0; i < n1; ++i)

for (int j = 0; j < n2; ++j) {

dp[i][j] = nums1[i] * nums2[j];

if (i > 0 && j > 0) dp[i][j] += max(0, dp[i - 1][j - 1]);

if (i > 0) dp[i][j] = max(dp[i][j], dp[i - 1][j]);

if (j > 0) dp[i][j] = max(dp[i][j], dp[i][j - 1]);

}

return dp.back().back();

}

};

花花酱 LeetCode 1457. Pseudo-Palindromic Paths in a Binary Tree

By zxi on May 23, 2020

Given a binary tree where node values are digits from 1 to 9. A path in the binary tree is said to be pseudo-palindromic if at least one permutation of the node values in the path is a palindrome.

Return the number of pseudo-palindromic paths going from the root node to leaf nodes.

Example 1:

Input: root = [2,3,1,3,1,null,1]
Output: 2 
Explanation: The figure above represents the given binary tree. There are three paths going from the root node to leaf nodes: the red path [2,3,3], the green path [2,1,1], and the path [2,3,1]. Among these paths only red path and green path are pseudo-palindromic paths since the red path [2,3,3] can be rearranged in [3,2,3] (palindrome) and the green path [2,1,1] can be rearranged in [1,2,1] (palindrome).

Example 2:

Input: root = [2,1,1,1,3,null,null,null,null,null,1]
Output: 1 
Explanation: The figure above represents the given binary tree. There are three paths going from the root node to leaf nodes: the green path [2,1,1], the path [2,1,3,1], and the path [2,1]. Among these paths only the green path is pseudo-palindromic since [2,1,1] can be rearranged in [1,2,1] (palindrome).

Example 3:

Input: root = [9]
Output: 1

Constraints:

The given binary tree will have between 1 and 10^5 nodes.
Node values are digits from 1 to 9.

Solution: Counting

At most one number can occur odd times.

Time complexity: O(n)
Space complexity: O(n) / stack size

C++

// Author: Huahua

class Solution {

public:

int pseudoPalindromicPaths (TreeNode* root) {

vector<int> counts(10);

function<int(TreeNode*)> dfs = [&](TreeNode* node) {

if (!node) return 0;

++counts[node->val];

int c = 0;

if (!node->left && !node->right) {

int odds = 0;

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

if (counts[i] & 1) ++odds;

if (odds <= 1) c = 1;

}

int l = dfs(node->left);

int r = dfs(node->right);

--counts[node->val];

return c + l + r;

};

return dfs(root);

}

};

Use a binary string to represent occurrences of each number (even: 0 / odd: 1), we can use xor to flip the bit.

C++

// Author: Huahua

class Solution {

public:

int pseudoPalindromicPaths(TreeNode* root, int s = 0) {

if (!root) return 0;

s ^= (1 << root->val);

if (!root->left && !root->right)

return __builtin_popcount(s) <= 1;

return pseudoPalindromicPaths(root->left, s)

+ pseudoPalindromicPaths(root->right, s);

}

};

花花酱 LeetCode 1456. Maximum Number of Vowels in a Substring of Given Length

By zxi on May 23, 2020

Given a string s and an integer k.

Return the maximum number of vowel letters in any substring of s with length k.

Vowel letters in English are (a, e, i, o, u).

Example 1:

Input: s = "abciiidef", k = 3
Output: 3
Explanation: The substring "iii" contains 3 vowel letters.

Example 2:

Input: s = "aeiou", k = 2
Output: 2
Explanation: Any substring of length 2 contains 2 vowels.

Example 3:

Input: s = "leetcode", k = 3
Output: 2
Explanation: "lee", "eet" and "ode" contain 2 vowels.

Example 4:

Input: s = "rhythms", k = 4
Output: 0
Explanation: We can see that s doesn't have any vowel letters.

Example 5:

Input: s = "tryhard", k = 4
Output: 1

Constraints:

1 <= s.length <= 10^5
s consists of lowercase English letters.
1 <= k <= s.length

Solution: Sliding Window

Keep tracking the number of vows in a window of size k.

Time complexity: O(n)
Space complexity: O(1)

C++

// Author: Huahua

class Solution {

public:

int maxVowels(string s, int k) {

vector<int> v(128);

v['a'] = v['e'] = v['i'] = v['o'] = v['u'] = 1;

int total = 0;

int ans = 0;

for (int i = 1; i <= s.length(); ++i) {

total += v[s[i - 1]];

if (i >= k) {

ans = max(ans, total);

total -= v[s[i - k]];

}

return ans;

}

};

花花酱 LeetCode 1455. Check If a Word Occurs As a Prefix of Any Word in a Sentence

By zxi on May 23, 2020

Given a sentence that consists of some words separated by a single space, and a searchWord.

You have to check if searchWord is a prefix of any word in sentence.

Return the index of the word in sentence where searchWord is a prefix of this word (1-indexed).

If searchWord is a prefix of more than one word, return the index of the first word (minimum index). If there is no such word return -1.

A prefix of a string S is any leading contiguous substring of S.

Example 1:

Input: sentence = "i love eating burger", searchWord = "burg"
Output: 4
Explanation: "burg" is prefix of "burger" which is the 4th word in the sentence.

Example 2:

Input: sentence = "this problem is an easy problem", searchWord = "pro"
Output: 2
Explanation: "pro" is prefix of "problem" which is the 2nd and the 6th word in the sentence, but we return 2 as it's the minimal index.

Example 3:

Input: sentence = "i am tired", searchWord = "you"
Output: -1
Explanation: "you" is not a prefix of any word in the sentence.

Example 4:

Input: sentence = "i use triple pillow", searchWord = "pill"
Output: 4

Example 5:

Input: sentence = "hello from the other side", searchWord = "they"
Output: -1

Constraints:

1 <= sentence.length <= 100
1 <= searchWord.length <= 10
sentence consists of lowercase English letters and spaces.
searchWord consists of lowercase English letters.

Solution 1: Brute Force

Time complexity: O(n)
Space complexity: O(1)

C++

class Solution {

public:

int isPrefixOfWord(string sentence, string searchWord) {

const int n = sentence.size();

const int l = searchWord.size();

int word = 0;

for (int i = 0; i <= n - l; ++i) {

if (i == 0 || sentence[i - 1] == ' ') {

++word;

bool valid = true;

for (int j = 0; j < l && valid; ++j)

valid = valid && (sentence[i + j] == searchWord[j]);

if (valid) return word;

}

return -1;

}

};

Huahua's Tech Road

:= 海象运算符 Walrus Operator – Python Weekly EP2

Python3

花花酱 LeetCode 1458. Max Dot Product of Two Subsequences

Solution: DP

C++

C++

花花酱 LeetCode 1457. Pseudo-Palindromic Paths in a Binary Tree

Solution: Counting

C++

C++

花花酱 LeetCode 1456. Maximum Number of Vowels in a Substring of Given Length

Solution: Sliding Window

C++

花花酱 LeetCode 1455. Check If a Word Occurs As a Prefix of Any Word in a Sentence

Solution 1: Brute Force

C++