Huahua’s Tech Road

花花酱 LeetCode 1912. Design Movie Rental System

By zxi on January 1, 2022

You have a movie renting company consisting of n shops. You want to implement a renting system that supports searching for, booking, and returning movies. The system should also support generating a report of the currently rented movies.

Each movie is given as a 2D integer array entries where entries[i] = [shop_i, movie_i, price_i] indicates that there is a copy of movie movie_i at shop shop_i with a rental price of price_i. Each shop carries at most one copy of a movie movie_i.

The system should support the following functions:

Search: Finds the cheapest 5 shops that have an unrented copy of a given movie. The shops should be sorted by price in ascending order, and in case of a tie, the one with the smaller shop_i should appear first. If there are less than 5 matching shops, then all of them should be returned. If no shop has an unrented copy, then an empty list should be returned.
Rent: Rents an unrented copy of a given movie from a given shop.
Drop: Drops off a previously rented copy of a given movie at a given shop.
Report: Returns the cheapest 5 rented movies (possibly of the same movie ID) as a 2D list res where res[j] = [shop_j, movie_j] describes that the j^th cheapest rented movie movie_j was rented from the shop shop_j. The movies in res should be sorted by price in ascending order, and in case of a tie, the one with the smaller shop_j should appear first, and if there is still tie, the one with the smaller movie_j should appear first. If there are fewer than 5 rented movies, then all of them should be returned. If no movies are currently being rented, then an empty list should be returned.

Implement the MovieRentingSystem class:

MovieRentingSystem(int n, int[][] entries) Initializes the MovieRentingSystem object with n shops and the movies in entries.
List<Integer> search(int movie) Returns a list of shops that have an unrented copy of the given movie as described above.
void rent(int shop, int movie) Rents the given movie from the given shop.
void drop(int shop, int movie) Drops off a previously rented movie at the given shop.
List<List<Integer>> report() Returns a list of cheapest rented movies as described above.

Note: The test cases will be generated such that rent will only be called if the shop has an unrented copy of the movie, and drop will only be called if the shop had previously rented out the movie.

Example 1:

Input
["MovieRentingSystem", "search", "rent", "rent", "report", "drop", "search"]
[[3, [[0, 1, 5], [0, 2, 6], [0, 3, 7], [1, 1, 4], [1, 2, 7], [2, 1, 5]]], [1], [0, 1], [1, 2], [], [1, 2], [2]]
Output
[null, [1, 0, 2], null, null, [[0, 1], [1, 2]], null, [0, 1]]

Explanation
MovieRentingSystem movieRentingSystem = new MovieRentingSystem(3, [[0, 1, 5], [0, 2, 6], [0, 3, 7], [1, 1, 4], [1, 2, 7], [2, 1, 5]]);
movieRentingSystem.search(1);  // return [1, 0, 2], Movies of ID 1 are unrented at shops 1, 0, and 2. Shop 1 is cheapest; shop 0 and 2 are the same price, so order by shop number.
movieRentingSystem.rent(0, 1); // Rent movie 1 from shop 0. Unrented movies at shop 0 are now [2,3].
movieRentingSystem.rent(1, 2); // Rent movie 2 from shop 1. Unrented movies at shop 1 are now [1].
movieRentingSystem.report();   // return [[0, 1], [1, 2]]. Movie 1 from shop 0 is cheapest, followed by movie 2 from shop 1.
movieRentingSystem.drop(1, 2); // Drop off movie 2 at shop 1. Unrented movies at shop 1 are now [1,2].
movieRentingSystem.search(2);  // return [0, 1]. Movies of ID 2 are unrented at shops 0 and 1. Shop 0 is cheapest, followed by shop 1.

Constraints:

1 <= n <= 3 * 10⁵
1 <= entries.length <= 10⁵
0 <= shop_i < n
1 <= movie_i, price_i <= 10⁴
Each shop carries at most one copy of a movie movie_i.
At most 10⁵ calls in total will be made to search, rent, drop and report.

Solution: Hashtable + TreeSet

We need three containers:
1. movies tracks the {movie -> price} of each shop. This is readonly to get the price of a movie for generating keys for treesets.
2. unrented tracks unrented movies keyed by movie id, value is a treeset ordered by {price, shop}.
3. rented tracks rented movies, a treeset ordered by {price, shop, movie}

Note: By using array<int, 3> we can unpack values like below:
array<int, 3> entries; // {price, shop, movie}
for (const auto [price, shop, moive] : entries)
…

Time complexity:
Init: O(nlogn)
rent / drop: O(logn)
search / report: O(1)

Space complexity: O(n)

C++

// Author: Huahua

class MovieRentingSystem {

public:

MovieRentingSystem(int n, vector<vector<int>>& entries): movies(n) {

for (const auto& e : entries) {

const int shop = e[0];

const int movie = e[1];

const int price = e[2];

movies[shop].emplace(movie, price);

unrented[movie].emplace(price, shop);

}

vector<int> search(int movie) {

vector<int> shops;

for (const auto [price, shop] : unrented[movie]) {

shops.push_back(shop);

if (shops.size() == 5) break;

}

return shops;

}

void rent(int shop, int movie) {

const int price = movies[shop][movie];

unrented[movie].erase({price, shop});

rented.insert({price, shop, movie});

}

void drop(int shop, int movie) {

const int price = movies[shop][movie];

rented.erase({price, shop, movie});

unrented[movie].emplace(price, shop);

}

vector<vector<int>> report() {

vector<vector<int>> ans;

for (const auto& [price, shop, movie] : rented) {

ans.push_back({shop, movie});

if (ans.size() == 5) break;

}

return ans;

}

private:

vector<unordered_map<int, int>> movies; // shop -> {movie -> price}

unordered_map<int, set<pair<int, int>>> unrented; // movie -> {price, shop}

set<array<int, 3>> rented; // {price, shop, movie}

};

花花酱 LeetCode 1998. GCD Sort of an Array

By zxi on December 31, 2021

You are given an integer array nums, and you can perform the following operation any number of times on nums:

Swap the positions of two elements nums[i] and nums[j] if gcd(nums[i], nums[j]) > 1 where gcd(nums[i], nums[j]) is the greatest common divisor of nums[i] and nums[j].

Return true if it is possible to sort nums in non-decreasing order using the above swap method, or false otherwise.

Example 1:

Input: nums = [7,21,3]
Output: true
Explanation: We can sort [7,21,3] by performing the following operations:
- Swap 7 and 21 because gcd(7,21) = 7. nums = [21,7,3]
- Swap 21 and 3 because gcd(21,3) = 3. nums = [3,7,21]

Example 2:

Input: nums = [5,2,6,2]
Output: false
Explanation: It is impossible to sort the array because 5 cannot be swapped with any other element.

Example 3:

Input: nums = [10,5,9,3,15]
Output: true
We can sort [10,5,9,3,15] by performing the following operations:
- Swap 10 and 15 because gcd(10,15) = 5. nums = [15,5,9,3,10]
- Swap 15 and 3 because gcd(15,3) = 3. nums = [3,5,9,15,10]
- Swap 10 and 15 because gcd(10,15) = 5. nums = [3,5,9,10,15]

Constraints:

1 <= nums.length <= 3 * 10⁴
2 <= nums[i] <= 10⁵

Solution: Union-Find

Let nums[j]’s target position be i. In order to put nums[j] to pos i by swapping. nums[i] and nums[j] must be in the same connected component. There is an edge between two numbers if they have gcd > 1.

We union two numbers if their have gcd > 1. However, it will be TLE if we do all pairs . Thus, for each number, we union it with its divisors instead.

Time complexity: O(n²) TLE -> O(sum(sqrt(nums[i]))) <= O(n*sqrt(m))
Space complexity: O(n)

C++

// Author: Huahua

class Solution {

public:

bool gcdSort(vector<int>& nums) {

const int m = *max_element(begin(nums), end(nums));

const int n = nums.size();

vector<int> p(m + 1);

iota(begin(p), end(p), 0);

function<int(int)> find = [&](int x) {

return p[x] == x ? x : (p[x] = find(p[x]));

};

for (int x : nums)

for (int d = 2; d <= sqrt(x); ++d)

if (x % d == 0)

p[find(x)] = p[find(x / d)] = find(d);

vector<int> sorted(nums);

sort(begin(sorted), end(sorted));

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

if (find(sorted[i]) != find(nums[i]))

return false;

return true;

}

};

花花酱 LeetCode 1995. Count Special Quadruplets

By zxi on December 31, 2021

Given a 0-indexed integer array nums, return the number of distinct quadruplets (a, b, c, d) such that:

nums[a] + nums[b] + nums[c] == nums[d], and
a < b < c < d

Example 1:

Input: nums = [1,2,3,6]
Output: 1
Explanation: The only quadruplet that satisfies the requirement is (0, 1, 2, 3) because 1 + 2 + 3 == 6.

Example 2:

Input: nums = [3,3,6,4,5]
Output: 0
Explanation: There are no such quadruplets in [3,3,6,4,5].

Example 3:

Input: nums = [1,1,1,3,5]
Output: 4
Explanation: The 4 quadruplets that satisfy the requirement are:
- (0, 1, 2, 3): 1 + 1 + 1 == 3
- (0, 1, 3, 4): 1 + 1 + 3 == 5
- (0, 2, 3, 4): 1 + 1 + 3 == 5
- (1, 2, 3, 4): 1 + 1 + 3 == 5

Constraints:

4 <= nums.length <= 50
1 <= nums[i] <= 100

Solution 1: Brute force (224ms)

Enumerate a, b, c, d.

Time complexity: O(C(n, 4)) = O(n⁴/24)
Space complexity: O(1)

C++

// Author: Huahua

class Solution {

public:

int countQuadruplets(vector<int>& nums) {

const int n = nums.size();

int ans = 0;

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

for (int b = a + 1; b < n; ++b)

for (int c = b + 1; c < n; ++c)

for (int d = c + 1; d < n; ++d)

ans += nums[a] + nums[b] + nums[c] == nums[d];

return ans;

}

};

Solution 2: Static frequency table + binary search (39ms)

For each element, we store its indices (sorted).

Given a, b, c, target t = nums[a] + nums[b] + nums[c], we check the hashtable and use binary search to find how many times it occurred after index c.

Time complexity: O(n³/6*logn)
Space complexity: O(n)

C++

// Author: Huahua

class Solution {

public:

int countQuadruplets(vector<int>& nums) {

constexpr int kMax = 100;

const int n = nums.size();

int ans = 0;

vector<vector<int>> m(kMax + 1);

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

m[nums[i]].push_back(i);

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

for (int b = a + 1; b < n; ++b)

for (int c = b + 1; c < n; ++c) {

const int t = nums[a] + nums[b] + nums[c];

if (t > kMax) continue;

ans += end(m[t]) - lower_bound(begin(m[t]), end(m[t]), c + 1);

}

return ans;

}

};

Solution 3: Dynamic frequency table (29ms)

Similar to 花花酱 LeetCode 1. Two Sum, we dynamically add elements (from right to left) into the hashtable.

Time complexity: O(n³/6)
Space complexity: O(n)

C++

// Author: Huahua

class Solution {

public:

int countQuadruplets(vector<int>& nums) {

constexpr int kMax = 100;

const int n = nums.size();

int ans = 0;

vector<int> m(kMax + 1);

// for every c we had seen, we can use it as target (nums[d]).

for (int c = n - 1; c >= 0; ++m[nums[c--]])

for (int a = 0; a < c; ++a)

for (int b = a + 1; b < c; ++b) {

const int t = nums[a] + nums[b] + nums[c];

if (t > kMax) continue;

ans += m[t];

}

return ans;

}

};

花花酱 LeetCode 1991. Find the Middle Index in Array

By zxi on December 31, 2021

Given a 0-indexed integer array nums, find the leftmost middleIndex (i.e., the smallest amongst all the possible ones).

A middleIndex is an index where nums[0] + nums[1] + ... + nums[middleIndex-1] == nums[middleIndex+1] + nums[middleIndex+2] + ... + nums[nums.length-1].

If middleIndex == 0, the left side sum is considered to be 0. Similarly, if middleIndex == nums.length - 1, the right side sum is considered to be 0.

Return the leftmost middleIndex that satisfies the condition, or -1 if there is no such index.

Example 1:

Input: nums = [2,3,-1,8,4]
Output: 3
Explanation: The sum of the numbers before index 3 is: 2 + 3 + -1 = 4
The sum of the numbers after index 3 is: 4 = 4

Example 2:

Input: nums = [1,-1,4]
Output: 2
Explanation: The sum of the numbers before index 2 is: 1 + -1 = 0
The sum of the numbers after index 2 is: 0

Example 3:

Input: nums = [2,5]
Output: -1
Explanation: There is no valid middleIndex.

Constraints:

1 <= nums.length <= 100
-1000 <= nums[i] <= 1000

Solution: Pre-compute + prefix sum

Pre-compute the sum of entire array. We scan the array and accumulate prefix sum and we can compute the sum of the rest of array.

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

C++

// Author: Huahua

class Solution {

public:

int findMiddleIndex(vector<int>& nums) {

int sum = accumulate(begin(nums), end(nums), 0);

for (int i = 0, cur = 0; i < nums.size(); ++i) {

if (cur == sum - cur - nums[i]) return i;

cur += nums[i];

}

return -1;

}

};

花花酱 LeetCode 1985. Find the Kth Largest Integer in the Array

By zxi on December 31, 2021

You are given an array of strings nums and an integer k. Each string in nums represents an integer without leading zeros.

Return the string that represents the k^th largest integer in nums.

Note: Duplicate numbers should be counted distinctly. For example, if nums is ["1","2","2"], "2" is the first largest integer, "2" is the second-largest integer, and "1" is the third-largest integer.

Example 1:

Input: nums = ["3","6","7","10"], k = 4
Output: "3"
Explanation:
The numbers in nums sorted in non-decreasing order are ["3","6","7","10"].
The 4^th largest integer in nums is "3".

Example 2:

Input: nums = ["2","21","12","1"], k = 3
Output: "2"
Explanation:
The numbers in nums sorted in non-decreasing order are ["1","2","12","21"].
The 3^rd largest integer in nums is "2".

Example 3:

Input: nums = ["0","0"], k = 2
Output: "0"
Explanation:
The numbers in nums sorted in non-decreasing order are ["0","0"].
The 2^nd largest integer in nums is "0".

Constraints:

1 <= k <= nums.length <= 10⁴
1 <= nums[i].length <= 100
nums[i] consists of only digits.
nums[i] will not have any leading zeros.

Solution: nth_element / quick selection

Use std::nth_element to find the k-th largest element. When comparing two strings, compare their lengths first and compare their content if they have the same length.

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

C++

// Author: Huahua

class Solution {

public:

string kthLargestNumber(vector<string>& nums, int k) {

nth_element(begin(nums), begin(nums) + k - 1, end(nums), [](const auto& a, const auto& b) {

return a.length() == b.length() ? a > b : a.length() > b.length();

});

return nums[k - 1];

}

};

Huahua's Tech Road

花花酱 LeetCode 1912. Design Movie Rental System

Solution: Hashtable + TreeSet

C++

花花酱 LeetCode 1998. GCD Sort of an Array

Solution: Union-Find

C++

花花酱 LeetCode 1995. Count Special Quadruplets

Solution 1: Brute force (224ms)

C++

Solution 2: Static frequency table + binary search (39ms)

C++

Solution 3: Dynamic frequency table (29ms)

C++

花花酱 LeetCode 1991. Find the Middle Index in Array

Solution: Pre-compute + prefix sum

C++

花花酱 LeetCode 1985. Find the Kth Largest Integer in the Array

Solution: nth_element / quick selection

C++