Huahua’s Tech Road

花花酱 LeetCode 1462. Course Schedule IV

By zxi on May 30, 2020

There are a total of n courses you have to take, labeled from 0 to n-1.

Some courses may have direct prerequisites, for example, to take course 0 you have first to take course 1, which is expressed as a pair: [1,0]

Given the total number of courses n, a list of direct prerequisite pairs and a list of queries pairs.

You should answer for each queries[i] whether the course queries[i][0] is a prerequisite of the course queries[i][1] or not.

Return a list of boolean, the answers to the given queries.

Please note that if course a is a prerequisite of course b and course b is a prerequisite of course c, then, course a is a prerequisite of course c.

Example 1:

Input: n = 2, prerequisites = [[1,0]], queries = [[0,1],[1,0]]
Output: [false,true]
Explanation: course 0 is not a prerequisite of course 1 but the opposite is true.

Example 2:

Input: n = 2, prerequisites = [], queries = [[1,0],[0,1]]
Output: [false,false]
Explanation: There are no prerequisites and each course is independent.

Example 3:

Input: n = 3, prerequisites = [[1,2],[1,0],[2,0]], queries = [[1,0],[1,2]]
Output: [true,true]

Example 4:

Input: n = 3, prerequisites = [[1,0],[2,0]], queries = [[0,1],[2,0]]
Output: [false,true]

Example 5:

Input: n = 5, prerequisites = [[0,1],[1,2],[2,3],[3,4]], queries = [[0,4],[4,0],[1,3],[3,0]]
Output: [true,false,true,false]

Constraints:

2 <= n <= 100
0 <= prerequisite.length <= (n * (n - 1) / 2)
0 <= prerequisite[i][0], prerequisite[i][1] < n
prerequisite[i][0] != prerequisite[i][1]
The prerequisites graph has no cycles.
The prerequisites graph has no repeated edges.
1 <= queries.length <= 10^4
queries[i][0] != queries[i][1]

Solution: Floyd-Warshall Algorithm (All pairs shortest paths)

Time complexity: O(n^3 + q)
Space complexity: O(n^2)

C++

// Author: Huahua

class Solution {

public:

vector<bool> checkIfPrerequisite(int n,

vector<vector<int>>& prerequisites, vector<vector<int>>& queries) {

vector<vector<int>> g(n, vector<int>(n));

for (const auto& p : prerequisites)

g[p[0]][p[1]] = 1;

for (int k = 0; k < n; ++k)

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

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

g[i][j] |= g[i][k] & g[k][j];

vector<bool> ans;

for (const auto& q : queries)

ans.push_back(g[q[0]][q[1]]);

return ans;

}

};

花花酱 LeetCode 1461. Check If a String Contains All Binary Codes of Size K

By zxi on May 30, 2020

Given a binary string s and an integer k.

Return True if any binary code of length k is a substring of s. Otherwise, return False.

Example 1:

Input: s = "00110110", k = 2
Output: true
Explanation: The binary codes of length 2 are "00", "01", "10" and "11". They can be all found as substrings at indicies 0, 1, 3 and 2 respectively.

Example 2:

Input: s = "00110", k = 2
Output: true

Example 3:

Input: s = "0110", k = 1
Output: true
Explanation: The binary codes of length 1 are "0" and "1", it is clear that both exist as a substring.

Example 4:

Input: s = "0110", k = 2
Output: false
Explanation: The binary code "00" is of length 2 and doesn't exist in the array.

Example 5:

Input: s = "0000000001011100", k = 4
Output: false

Constraints:

1 <= s.length <= 5 * 10^5
s consists of 0’s and 1’s only.
1 <= k <= 20

Solution: Hashtable

Insert all possible substrings into a hashtable, the size of the hashtable should be 2^k.

Time complexity: O(n*k)
Space complexity: O(2^k*k) -> O(2^k)

std::string_view: 484 ms, 40.1MB
std::string 644 ms, 58.6MB

C++

// Author: Huahua

// std::string_view: 468 ms, 37.3MB

// std::string: 636 ms, 58.6MB

class Solution {

public:

bool hasAllCodes(string_view s, int k) {

const int n = s.length();

if ((n - k + 1) * k < (1 << k)) return false;

unordered_set<string_view> c;

for (int i = 0; i + k <= n; ++i)

c.insert(s.substr(i, k));

return c.size() == (1 << k);

}

};

花花酱 LeetCode 1460. Make Two Arrays Equal by Reversing Sub-arrays

By zxi on May 30, 2020

Given two integer arrays of equal length target and arr.

In one step, you can select any non-empty sub-array of arr and reverse it. You are allowed to make any number of steps.

Return True if you can make arr equal to target, or False otherwise.

Example 1:

Input: target = [1,2,3,4], arr = [2,4,1,3]
Output: true
Explanation: You can follow the next steps to convert arr to target:
1- Reverse sub-array [2,4,1], arr becomes [1,4,2,3]
2- Reverse sub-array [4,2], arr becomes [1,2,4,3]
3- Reverse sub-array [4,3], arr becomes [1,2,3,4]
There are multiple ways to convert arr to target, this is not the only way to do so.

Example 2:

Input: target = [7], arr = [7]
Output: true
Explanation: arr is equal to target without any reverses.

Example 3:

Input: target = [1,12], arr = [12,1]
Output: true

Example 4:

Input: target = [3,7,9], arr = [3,7,11]
Output: false
Explanation: arr doesn't have value 9 and it can never be converted to target.

Example 5:

Input: target = [1,1,1,1,1], arr = [1,1,1,1,1]
Output: true

Constraints:

target.length == arr.length
1 <= target.length <= 1000
1 <= target[i] <= 1000
1 <= arr[i] <= 1000

Solution: Counting

target and arr must have same elements.

Time complexity: O(n)
Space complexity: O(1001)

C++

// Author: Huahua

class Solution {

public:

bool canBeEqual(vector<int>& target, vector<int>& arr) {

vector<int> t(1001);

vector<int> a(1001);

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

++t[target[i]];

++a[arr[i]];

}

return t == a;

}

};

Python3

# Author: Huahua

class Solution:

def canBeEqual(self, target: List[int], arr: List[int]) -> bool:

return Counter(target) == Counter(arr)

:= 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();

}

};

Huahua's Tech Road

花花酱 LeetCode 1462. Course Schedule IV

Solution: Floyd-Warshall Algorithm (All pairs shortest paths)

C++

花花酱 LeetCode 1461. Check If a String Contains All Binary Codes of Size K

Solution: Hashtable

C++

花花酱 LeetCode 1460. Make Two Arrays Equal by Reversing Sub-arrays

Solution: Counting

C++

Python3

:= 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++