Posts published in May 2019

花花酱 LeetCode 90. Subsets II

By zxi on May 15, 2019

Given a collection of integers that might contain duplicates, nums, return all possible subsets (the power set).

Note: The solution set must not contain duplicate subsets.

Example:

<strong>Input:</strong> [1,2,2]

<strong>Output:</strong>

[

[2],

[1],

[1,2,2],

[2,2],

[1,2],

[]

]

Solution: DFS

The key to this problem is how to remove/avoid duplicates efficiently.

For the same depth, among the same numbers, only the first number can be used.

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

C++

class Solution {

public:

vector<vector<int>> subsetsWithDup(vector<int>& nums) {

const int n = nums.size();

sort(begin(nums), end(nums));

vector<vector<int>> ans;

vector<int> cur;

function<void(int)> dfs = [&](int s) {

ans.push_back(cur);

if (cur.size() == n)

return;

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

if (i > s && nums[i] == nums[i - 1]) continue;

cur.push_back(nums[i]);

dfs(i + 1);

cur.pop_back();

}

};

dfs(0);

return ans;

}

};

花花酱 LeetCode 1043. Partition Array for Maximum Sum

By zxi on May 14, 2019

Given an integer array A, you partition the array into (contiguous) subarrays of length at most K. After partitioning, each subarray has their values changed to become the maximum value of that subarray.

Return the largest sum of the given array after partitioning.

Example 1:

Input: A = [1,15,7,9,2,5,10], K = 3
Output: 84
Explanation: A becomes [15,15,15,9,10,10,10]

Note:

1 <= K <= A.length <= 500
0 <= A[i] <= 10^6

Solution: DP

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

dp[i] := max sum of A[1] ~ A[i]
init: dp[0] = 0
transition: dp[i] = max{dp[i – k] + max(A[i-k:i]) * k}, 1 <= k <= min(i, K)
ans: dp[n]

A = | 2 | 1 | 4 | 3 |
K = 3
dp[0] = 0
dp[1] = max(dp[0] + 2 * 1) = 2
dp[2] = max(dp[0] + 2 * 2, dp[1] + 1 * 1) = max(4, 3) = 4
dp[3] = max(dp[0] + 4 * 3, dp[1] + 4 * 2, dp[2] + 4 * 1) = max(12, 10, 8) = 12
dp[4] = max(dp[1] + 4 * 3, dp[2] + 4 * 2, dp[3] + 3 * 1) = max(14, 12, 15) = 15
best = | 4 | 4 | 4 | 3 |

C++

class Solution {

public:

int maxSumAfterPartitioning(vector<int>& A, int K) {

int n = A.size();

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

for (int i = 1; i <= n; ++i) {

int m = INT_MIN;

for (int k = 1; k <= K && i - k >= 0; ++k) {

m = max(m, A[i - k]);

dp[i] = max(dp[i], dp[i - k] + m * k);

}

return dp[n];

}

};

花花酱 LeetCode 1041. Robot Bounded In Circle

By zxi on May 14, 2019

On an infinite plane, a robot initially stands at (0, 0) and faces north. The robot can receive one of three instructions:

"G": go straight 1 unit;
"L": turn 90 degrees to the left;
"R": turn 90 degress to the right.

The robot performs the instructions given in order, and repeats them forever.

Return true if and only if there exists a circle in the plane such that the robot never leaves the circle.

Example 1:

Input: "GGLLGG"
Output: true
Explanation: 
The robot moves from (0,0) to (0,2), turns 180 degrees, and then returns to (0,0).
When repeating these instructions, the robot remains in the circle of radius 2 centered at the origin.

Example 2:

Input: "GG"
Output: false
Explanation: 
The robot moves north indefinitely.

Example 3:

Input: "GL"
Output: true
Explanation: 
The robot moves from (0, 0) -> (0, 1) -> (-1, 1) -> (-1, 0) -> (0, 0) -> ...

Note:

1 <= instructions.length <= 100
instructions[i] is in {'G', 'L', 'R'}

Solution: Simulation

When instructions end, if the robot is back to (0,0) or is not facing north (which guarantees it will come back to 0, 0 for another 1 or 3 rounds)

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

C++

// Author: Huahua

class Solution {

public:

bool isRobotBounded(string instructions) {

int x = 0;

int y = 0;

int d = 0;

vector<int> dx{0, 1, 0, -1};

vector<int> dy{-1, 0, 1, 0};

for (char c : instructions) {

if (c == 'G') {

x += dx[d];

y += dy[d];

} else {

d = (d + (c == 'L' ? 3 : 1)) % 4;

}

return x == 0 && y == 0 || d;

}

};

花花酱 LeetCode 1042. Flower Planting With No Adjacent

By zxi on May 14, 2019

You have N gardens, labelled 1 to N. In each garden, you want to plant one of 4 types of flowers.

paths[i] = [x, y] describes the existence of a bidirectional path from garden x to garden y.

Also, there is no garden that has more than 3 paths coming into or leaving it.

Your task is to choose a flower type for each garden such that, for any two gardens connected by a path, they have different types of flowers.

Return any such a choice as an array answer, where answer[i] is the type of flower planted in the (i+1)-th garden. The flower types are denoted 1, 2, 3, or 4. It is guaranteed an answer exists.

Example 1:

Input: N = 3, paths = [[1,2],[2,3],[3,1]]
Output: [1,2,3]

Example 2:

Input: N = 4, paths = [[1,2],[3,4]]
Output: [1,2,1,2]

Example 3:

Input: N = 4, paths = [[1,2],[2,3],[3,4],[4,1],[1,3],[2,4]]
Output: [1,2,3,4]

Note:

1 <= N <= 10000
0 <= paths.size <= 20000
No garden has 4 or more paths coming into or leaving it.
It is guaranteed an answer exists.

Solution: Graph coloring, choose any available color

Time complexity: O(|V|+|E|)
Space complexity: O(|V|+|E|)

C++

class Solution {

public:

vector<int> gardenNoAdj(int N, vector<vector<int>>& paths) {

vector<int> ans(N, 0);

vector<vector<int>> G(N);

for (const auto& p : paths) {

G[p[0] - 1].push_back(p[1] - 1);

G[p[1] - 1].push_back(p[0] - 1);

}

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

int mask = 0;

for (const auto& j : G[i])

mask |= (1 << ans[j]);

for (int c = 1; c <= 4 && ans[i] == 0; ++c)

if (!(mask & (1 << c))) ans[i] = c;

}

return ans;

}

};

花花酱 LeetCode 630. Course Schedule III

By zxi on May 10, 2019

There are n different online courses numbered from 1 to n. Each course has some duration(course length) t and closed on d_th day. A course should be taken continuouslyfor t days and must be finished before or on the d_th day. You will start at the 1_st day.

Given n online courses represented by pairs (t,d), your task is to find the maximal number of courses that can be taken.

Example:

Input: [[100, 200], [200, 1300], [1000, 1250], [2000, 3200]]
Output: 3
Explanation: 
There're totally 4 courses, but you can take 3 courses at most:
First, take the 1st course, it costs 100 days so you will finish it on the 100th day, and ready to take the next course on the 101st day.
Second, take the 3rd course, it costs 1000 days so you will finish it on the 1100th day, and ready to take the next course on the 1101st day. 
Third, take the 2nd course, it costs 200 days so you will finish it on the 1300th day. 
The 4th course cannot be taken now, since you will finish it on the 3300th day, which exceeds the closed date.

Note:

The integer 1 <= d, t, n <= 10,000.
You can’t take two courses simultaneously.

Solution: Priority queue

Sort courses by end date
Use a priority queue (Max-Heap) to store the course lengths or far
Swap with a longer course if we could not take the current one

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

C++

// Author: Huahua, running time: 196 ms, 40.5 MB

class Solution {

public:

int scheduleCourse(vector<vector<int>>& courses) {

sort(begin(courses), end(courses), [](const auto& a, const auto& b) {

return a[1] < b[1];

});

priority_queue<int> q;

int d = 0;

for (const auto& c: courses) {

if (d + c[0] <= c[1]) {

d += c[0];

q.push(c[0]);

} else if (q.size() && q.top() > c[0]) {

d += c[0] - q.top(); q.pop();

q.push(c[0]);

}

return q.size();

}

};