Posts tagged as “hard”

花花酱 LeetCode 1697. Checking Existence of Edge Length Limited Paths

By zxi on December 20, 2020

An undirected graph of n nodes is defined by edgeList, where edgeList[i] = [u_i, v_i, dis_i] denotes an edge between nodes u_i and v_i with distance dis_i. Note that there may be multiple edges between two nodes.

Given an array queries, where queries[j] = [p_j, q_j, limit_j], your task is to determine for each queries[j] whether there is a path between p_j and q_jsuch that each edge on the path has a distance strictly less than limit_j .

Return a boolean array answer, where answer.length == queries.length and the j^th value of answer is true if there is a path for queries[j] is true, and false otherwise.

Example 1:

Input: n = 3, edgeList = [[0,1,2],[1,2,4],[2,0,8],[1,0,16]], queries = [[0,1,2],[0,2,5]]
Output: [false,true]
Explanation: The above figure shows the given graph. Note that there are two overlapping edges between 0 and 1 with distances 2 and 16.
For the first query, between 0 and 1 there is no path where each distance is less than 2, thus we return false for this query.
For the second query, there is a path (0 -> 1 -> 2) of two edges with distances less than 5, thus we return true for this query.

Example 2:

Input: n = 5, edgeList = [[0,1,10],[1,2,5],[2,3,9],[3,4,13]], queries = [[0,4,14],[1,4,13]]
Output: [true,false]
Exaplanation: The above figure shows the given graph.

Constraints:

2 <= n <= 10⁵
1 <= edgeList.length, queries.length <= 10⁵
edgeList[i].length == 3
queries[j].length == 3
0 <= u_i, v_i, p_j, q_j <= n - 1
u_i != v_i
p_j != q_j
1 <= dis_i, limit_j <= 10⁹
There may be multiple edges between two nodes.

Solution: Union Find

Since queries are offline, we can reorder them to optimize time complexity. Answer queries by their limits in ascending order while union edges by weights up to the limit. In this case, we just need to go through the entire edge list at most once.

Time complexity: O(QlogQ + ElogE)
Space complexity: O(Q + E)

C++

// Author: Huahua

class Solution {

public:

vector<bool> distanceLimitedPathsExist(int n, vector<vector<int>>& E, vector<vector<int>>& Q) {

vector<int> parents(n);

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

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

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

};

const int m = Q.size();

for (int i = 0; i < m; ++i) Q[i].push_back(i);

// Sort edges by weight in ascending order.

sort(begin(E), end(E), [](const auto& a, const auto& b) { return a[2] < b[2]; });

// Sort queries by limit in ascending order

sort(begin(Q), end(Q), [](const auto& a, const auto& b) { return a[2] < b[2]; });

vector<bool> ans(m);

int i = 0;

for (const auto& q : Q) {

while (i < E.size() && E[i][2] < q[2])

parents[find(E[i++][0])] = find(E[i][1]);

ans[q[3]] = find(q[0]) == find(q[1]);

}

return ans;

}

};

花花酱 LeetCode 1691. Maximum Height by Stacking Cuboids

By zxi on December 13, 2020

Given n cuboids where the dimensions of the i^th cuboid is cuboids[i] = [width_i, length_i, height_i] (0-indexed). Choose a subset of cuboids and place them on each other.

You can place cuboid i on cuboid j if width_i <= width_j and length_i <= length_j and height_i <= height_j. You can rearrange any cuboid’s dimensions by rotating it to put it on another cuboid.

Return the maximum height of the stacked cuboids.

Example 1:

Input: cuboids = [[50,45,20],[95,37,53],[45,23,12]]
Output: 190
Explanation:
Cuboid 1 is placed on the bottom with the 53x37 side facing down with height 95.
Cuboid 0 is placed next with the 45x20 side facing down with height 50.
Cuboid 2 is placed next with the 23x12 side facing down with height 45.
The total height is 95 + 50 + 45 = 190.

Example 2:

Input: cuboids = [[38,25,45],[76,35,3]]
Output: 76
Explanation:
You can't place any of the cuboids on the other.
We choose cuboid 1 and rotate it so that the 35x3 side is facing down and its height is 76.

Example 3:

Input: cuboids = [[7,11,17],[7,17,11],[11,7,17],[11,17,7],[17,7,11],[17,11,7]]
Output: 102
Explanation:
After rearranging the cuboids, you can see that all cuboids have the same dimension.
You can place the 11x7 side down on all cuboids so their heights are 17.
The maximum height of stacked cuboids is 6 * 17 = 102.

Constraints:

n == cuboids.length
1 <= n <= 100
1 <= width_i, length_i, height_i <= 100

Solution: Math/Greedy + DP

Direct DP is very hard, since there is no orders.

We have to find some way to sort the cuboids such that cuboid i can NOT stack on cuboid j if i > j. Then dp[i] = max(dp[j]) + height[i], 0 <= j < i, for each i, find the best base j and stack on top of it.
ans = max(dp)

We can sort the cuboids by their sorted dimensions, and cuboid i can stack stack onto cuboid j if and only if w[i] <= w[j] and l[i] <= l[j] and h[i] <= h[j].

First of all, we need to prove that all heights must come from the largest dimension of each cuboid.

1. If the top of the stack is A1*A2*A3, A3 < max(A1, A2), we can easily swap A3 with max(A1, A2), it’s still stackable but we get larger heights.
e.g. 3x5x4, base is 3×5, height is 4, we can rotate to get base of 3×4 with height of 5.

2. If a middle cuboid A of size A1*A2*A3, assuming A1 >= A2, A1 > A3, on top of A we have another cuboid B of size B1*B2*B3, B1 <= B2 <= B3.
We have A1 >= B1, A2 >= B2, A3 >= B3, by rotating A we have A3*A2*A1
A3 >= B3 >= B1, A2 >= B2, A1 > A3 >= B3, so B can be still on top of A, and we get larger height.

e.g. A: 3x5x4, B: 2x3x4
A -> 3x4x5, B is still stackable.

…

Time complexity: O(n^2)
Space complexity: O(n^2)

C++

// Author: Huahua

class Solution {

public:

int maxHeight(vector<vector<int>>& A) {

A.push_back({0, 0, 0});

const int n = A.size();

for (auto& box : A) sort(begin(box), end(box));

sort(begin(A), end(A));

vector<int> dp(n);

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

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

if (A[i][0] >= A[j][0] && A[i][1] >= A[j][1] && A[i][2] >= A[j][2])

dp[i] = max(dp[i], dp[j] + A[i][2]);

return *max_element(begin(dp), end(dp));

}

};

花花酱 LeetCode 1687. Delivering Boxes from Storage to Ports

By zxi on December 13, 2020

You have the task of delivering some boxes from storage to their ports using only one ship. However, this ship has a limit on the number of boxes and the total weight that it can carry.

You are given an array boxes, where boxes[i] = [ports_i, weight_i], and three integers portsCount, maxBoxes, and maxWeight.

ports_i is the port where you need to deliver the i^th box and weights_i is the weight of the i^th box.
portsCount is the number of ports.
maxBoxes and maxWeight are the respective box and weight limits of the ship.

The boxes need to be delivered in the order they are given. The ship will follow these steps:

The ship will take some number of boxes from the boxes queue, not violating the maxBoxes and maxWeight constraints.
For each loaded box in order, the ship will make a trip to the port the box needs to be delivered to and deliver it. If the ship is already at the correct port, no trip is needed, and the box can immediately be delivered.
The ship then makes a return trip to storage to take more boxes from the queue.

The ship must end at storage after all the boxes have been delivered.

Return the minimum number of trips the ship needs to make to deliver all boxes to their respective ports.

Example 1:

Input: boxes = [[1,1],[2,1],[1,1]], portsCount = 2, maxBoxes = 3, maxWeight = 3
Output: 4
Explanation: The optimal strategy is as follows: 
- The ship takes all the boxes in the queue, goes to port 1, then port 2, then port 1 again, then returns to storage. 4 trips.
So the total number of trips is 4.
Note that the first and third boxes cannot be delivered together because the boxes need to be delivered in order (i.e. the second box needs to be delivered at port 2 before the third box).

Example 2:

Input: boxes = [[1,2],[3,3],[3,1],[3,1],[2,4]], portsCount = 3, maxBoxes = 3, maxWeight = 6
Output: 6
Explanation: The optimal strategy is as follows: 
- The ship takes the first box, goes to port 1, then returns to storage. 2 trips.
- The ship takes the second, third and fourth boxes, goes to port 3, then returns to storage. 2 trips.
- The ship takes the fifth box, goes to port 3, then returns to storage. 2 trips.
So the total number of trips is 2 + 2 + 2 = 6.

Example 3:

Input: boxes = [[1,4],[1,2],[2,1],[2,1],[3,2],[3,4]], portsCount = 3, maxBoxes = 6, maxWeight = 7
Output: 6
Explanation: The optimal strategy is as follows:
- The ship takes the first and second boxes, goes to port 1, then returns to storage. 2 trips.
- The ship takes the third and fourth boxes, goes to port 2, then returns to storage. 2 trips.
- The ship takes the fifth and sixth boxes, goes to port 3, then returns to storage. 2 trips.
So the total number of trips is 2 + 2 + 2 = 6.

Example 4:

Input: boxes = [[2,4],[2,5],[3,1],[3,2],[3,7],[3,1],[4,4],[1,3],[5,2]], portsCount = 5, maxBoxes = 5, maxWeight = 7
Output: 14
Explanation: The optimal strategy is as follows:
- The ship takes the first box, goes to port 2, then storage. 2 trips.
- The ship takes the second box, goes to port 2, then storage. 2 trips.
- The ship takes the third and fourth boxes, goes to port 3, then storage. 2 trips.
- The ship takes the fifth box, goes to port 3, then storage. 2 trips.
- The ship takes the sixth and seventh boxes, goes to port 3, then port 4, then storage. 3 trips. 
- The ship takes the eighth and ninth boxes, goes to port 1, then port 5, then storage. 3 trips.
So the total number of trips is 2 + 2 + 2 + 2 + 3 + 3 = 14.

Constraints:

1 <= boxes.length <= 10⁵
1 <= portsCount, maxBoxes, maxWeight <= 10⁵
1 <= ports_i <= portsCount
1 <= weights_i <= maxWeight

Solution: Sliding Window

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

C++

// Author: Huahua

class Solution {

public:

int boxDelivering(vector<vector<int>>& boxes, int portsCount, int maxBoxes, int maxWeight) {

const int n = boxes.size();

vector<int> dp(n + 1); // dp[i] := min trips to deliver boxes[0:i]

for (int i = 0, j = 0, b = 0, w = 0, t = 1; j < n; ++j) {

// Different ports.

if (j == 0 || boxes[j][0] != boxes[j - 1][0]) ++t;

// Load the box.

w += boxes[j][1];

while (w > maxWeight // Too heavy.

|| (j - i + 1) > maxBoxes // Too many boxes.

|| (i < j && dp[i + 1] == dp[i])) { // Same cost more boxes.

w -= boxes[i++][1]; // 'Unload' the box.

if (boxes[i][0] != boxes[i - 1][0]) --t; // Different ports.

}

// Takes t trips to deliver boxes[i~j].

dp[j + 1] = dp[i] + t;

}

return dp[n];

}

};

花花酱 LeetCode 1681. Minimum Incompatibility

By zxi on December 6, 2020

You are given an integer array nums and an integer k. You are asked to distribute this array into k subsets of equal size such that there are no two equal elements in the same subset.

A subset’s incompatibility is the difference between the maximum and minimum elements in that array.

Return the minimum possible sum of incompatibilities of the k subsets after distributing the array optimally, or return -1 if it is not possible.

A subset is a group integers that appear in the array with no particular order.

Example 1:

Input: nums = [1,2,1,4], k = 2
Output: 4
Explanation: The optimal distribution of subsets is [1,2] and [1,4].
The incompatibility is (2-1) + (4-1) = 4.
Note that [1,1] and [2,4] would result in a smaller sum, but the first subset contains 2 equal elements.

Example 2:

Input: nums = [6,3,8,1,3,1,2,2], k = 4
Output: 6
Explanation: The optimal distribution of subsets is [1,2], [2,3], [6,8], and [1,3].
The incompatibility is (2-1) + (3-2) + (8-6) + (3-1) = 6.

Example 3:

Input: nums = [5,3,3,6,3,3], k = 3
Output: -1
Explanation: It is impossible to distribute nums into 3 subsets where no two elements are equal in the same subset.

Constraints:

1 <= k <= nums.length <= 16
nums.length is divisible by k
1 <= nums[i] <= nums.length

Solution: TSP

dp[s][i] := min cost to distribute a set of numbers represented as a binary mask s, the last number in the set is the i-th number.

Init:
1.dp[*][*] = inf
2. dp[1 <<i][i] = 0, cost of selecting a single number is zero.

Transition:
1. dp[s | (1 << j)][j] = dp[s][i] if s % (n / k) == 0, we start a new group, no extra cost.
2. dp[s | (1 << j)][j] = dp[s][i] + nums[j] – nums[i] if nums[j] > nums[i]. In the same group, we require the selected numbers are monotonically increasing. Each cost is nums[j] – nums[i].
e.g. 1, 3, 7, 20, cost is (3 – 1) + (7 – 3) + (20 – 7) = 20 – 1 = 19.

Ans: min(dp[(1 << n) – 1])

Time complexity: O(2^n * n^2)
Space complexity: O(2^n * n)

C++

// Author: Huahua

class Solution {

public:

int minimumIncompatibility(vector<int>& nums, int k) {

const int n = nums.size();

const int c = n / k;

int dp[1 << 16][16];

memset(dp, 0x7f, sizeof(dp));

for (int i = 0; i < n; ++i) dp[1 << i][i] = 0;

for (int s = 0; s < 1 << n; ++s)

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

if ((s & (1 << i)) == 0) continue;

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

if ((s & (1 << j))) continue;

const int t = s | (1 << j);

if (__builtin_popcount(s) % c == 0) {

dp[t][j] = min(dp[t][j], dp[s][i]);

} else if (nums[j] > nums[i]) {

dp[t][j] = min(dp[t][j],

dp[s][i] + nums[j] - nums[i]);

}

int ans = *min_element(begin(dp[(1 << n) - 1]),

end(dp[(1 << n) - 1]));

return ans > 1e9 ? - 1 : ans;

}

};

花花酱 LeetCode 1515. Best Position for a Service Centre

By zxi on November 30, 2020

A delivery company wants to build a new service centre in a new city. The company knows the positions of all the customers in this city on a 2D-Map and wants to build the new centre in a position such that the sum of the euclidean distances to all customers is minimum.

Given an array positions where positions[i] = [x_i, y_i] is the position of the ith customer on the map, return the minimum sum of the euclidean distances to all customers.

In other words, you need to choose the position of the service centre [x_centre, y_centre] such that the following formula is minimized:

Answers within 10^-5 of the actual value will be accepted.

Example 1:

Input: positions = [[0,1],[1,0],[1,2],[2,1]]
Output: 4.00000
Explanation: As shown, you can see that choosing [x_centre, y_centre] = [1, 1] will make the distance to each customer = 1, the sum of all distances is 4 which is the minimum possible we can achieve.

Example 2:

Input: positions = [[1,1],[3,3]]
Output: 2.82843
Explanation: The minimum possible sum of distances = sqrt(2) + sqrt(2) = 2.82843

Example 3:

Input: positions = [[1,1]]
Output: 0.00000

Example 4:

Input: positions = [[1,1],[0,0],[2,0]]
Output: 2.73205
Explanation: At the first glance, you may think that locating the centre at [1, 0] will achieve the minimum sum, but locating it at [1, 0] will make the sum of distances = 3.
Try to locate the centre at [1.0, 0.5773502711] you will see that the sum of distances is 2.73205.
Be careful with the precision!

Example 5:

Input: positions = [[0,1],[3,2],[4,5],[7,6],[8,9],[11,1],[2,12]]
Output: 32.94036
Explanation: You can use [4.3460852395, 4.9813795505] as the position of the centre.

Constraints:

1 <= positions.length <= 50
positions[i].length == 2
0 <= positions[i][0], positions[i][1] <= 100

Solution: Weiszfeld’s algorithm

Use Weiszfeld’s algorithm to compute geometric median of the samples.

Time complexity: O(f(epsilon) * O)
Space complexity: O(1)

C++

template<typename T1, typename T2>

double dist(const vector<T1>& a, const vector<T2>& b) {

return sqrt((a[0] - b[0]) * (a[0] - b[0]) + (a[1] - b[1]) * (a[1] - b[1]));

};

class Solution {

public:

double getMinDistSum(vector<vector<int>>& positions) {

constexpr double kDelta = 1e-7;

constexpr double kEpsilon = 1e-15;

vector<double> c{-1, -1};

double ans = 0;

while (true) {

ans = 0;

double nx = 0, ny = 0;

double denominator = 0;

for (const auto& p : positions) {

double d = dist(c, p) + kEpsilon;

ans += d;

nx += p[0] / d;

ny += p[1] / d;

denominator += 1.0 / d;

}

vector<double> t{nx / denominator, ny / denominator};

if (dist(c, t) < kDelta) return ans;

c = t;

}

return -1;

}

};