Posts tagged as “priority queue”

花花酱 LeetCode 871. Minimum Number of Refueling Stops

By zxi on July 14, 2018

Problem

A car travels from a starting position to a destination which is target miles east of the starting position.

Along the way, there are gas stations. Each station[i] represents a gas station that is station[i][0] miles east of the starting position, and has station[i][1] liters of gas.

The car starts with an infinite tank of gas, which initially has startFuel liters of fuel in it. It uses 1 liter of gas per 1 mile that it drives.

When the car reaches a gas station, it may stop and refuel, transferring all the gas from the station into the car.

What is the least number of refueling stops the car must make in order to reach its destination? If it cannot reach the destination, return -1.

Note that if the car reaches a gas station with 0 fuel left, the car can still refuel there. If the car reaches the destination with 0 fuel left, it is still considered to have arrived.

Example 1:

Input: target = 1, startFuel = 1, stations = []
Output: 0
Explanation: We can reach the target without refueling.

Example 2:

Input: target = 100, startFuel = 1, stations = [[10,100]]
Output: -1
Explanation: We can't reach the target (or even the first gas station).

Example 3:

Input: target = 100, startFuel = 10, stations = [[10,60],[20,30],[30,30],[60,40]]
Output: 2
Explanation: 
We start with 10 liters of fuel.
We drive to position 10, expending 10 liters of fuel.  We refuel from 0 liters to 60 liters of gas.
Then, we drive from position 10 to position 60 (expending 50 liters of fuel),
and refuel from 10 liters to 50 liters of gas.  We then drive to and reach the target.
We made 2 refueling stops along the way, so we return 2.

Note:

1 <= target, startFuel, stations[i][1] <= 10^9
0 <= stations.length <= 500
0 < stations[0][0] < stations[1][0] < ... < stations[stations.length-1][0] < target

Solution1: DP

Time complexity: O(n^2)

Space complexity: O(n)

C++

// Author: Huahua

// Running time: 24 ms

class Solution {

public:

int minRefuelStops(int target, int startFuel, vector<vector<int>>& stations) {

const int n = stations.size();

// dp[i]: max distance to go with i stops.

vector<long> dp(n + 1, startFuel);

// For each station

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

// for j stops, start from high to low to avoid reusing station i

for (int j = i + 1; j >= 1; --j)

if (dp[j - 1] >= stations[i][0])

// if we can reach station i with j - 1 stops,

// we can reach dp[j - 1] + station[i][1] with j stops.

dp[j] = max(dp[j], dp[j - 1] + stations[i][1]);

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

if (dp[i] >= target) return i;

return -1;

}

};

Solution2: Priority Queue

Time complexity: O(nlogn)

Space complexity: O(n)

// Author: Huahua

// Running time: 16 ms

class Solution {

public:

int minRefuelStops(int target, int startFuel, vector<vector<int>>& stations) {

int cur = startFuel;

int stops = 0;

int i = 0;

priority_queue<int> q; // gas (high to low) of reachable stations.

while (true) {

if (cur >= target) return stops;

while (i < stations.size() && stations[i][0] <= cur)

q.push(stations[i++][1]);

if (q.empty()) break;

cur += q.top(); q.pop();

++stops;

}

return -1;

}

};

V2: Iterator

// Author: Huahua

// Running time: 12 ms

class Solution {

public:

int minRefuelStops(int target, int startFuel, vector<vector<int>>& stations) {

int cur = startFuel;

int stops = 0;

priority_queue<int> q; // gas (high to low) of reachable stations.

auto it = begin(stations);

while (true) {

if (cur >= target) return stops;

while (it != end(stations) && it->at(0) <= cur)

q.push(it++->at(1));

if (q.empty()) break;

cur += q.top(); q.pop();

++stops;

}

return -1;

}

};

Solution 2: Priority queue / max heap

Time complexity: O(n) + O(nlogk)

Space complexity: O(n)

C++

// Author: Huahuas

// Running time: 16 ms

class Solution {

public:

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

unordered_map<int, int> count;

for (const int num : nums)

++count[num];

priority_queue<pair<int, int>> q;

for (const auto& pair : count) {

q.emplace(-pair.second, pair.first);

if (q.size() > k) q.pop();

}

vector<int> ans;

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

ans.push_back(q.top().second);

q.pop();

}

return ans;

}

};

Java

// Author: Huahua

// Running time: 56 ms

class Solution {

public List<Integer> topKFrequent(int[] nums, int k) {

Map<Integer, Integer> counts = new HashMap<>();

for (int num : nums)

counts.put(num, counts.getOrDefault(num, 0) + 1);

PriorityQueue<int[]> queue = new PriorityQueue<>((x, y) -> x[0] - y[0]);

for (Integer num : counts.keySet()) {

queue.offer(new int[]{counts.get(num), num});

if (queue.size() > k) queue.poll();

}

List<Integer> ans = new ArrayList<>();

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

ans.add(queue.poll()[1]);

return ans;

}

Python

"""

Author: Huahua

Running time: 56 ms

"""

class Solution:

def topKFrequent(self, nums, k):

counts = collections.Counter(nums)

h = []

for num in counts.keys():

heapq.heappush(h, (counts[num], num))

if len(h) > k: heapq.heappop(h)

return [heapq.heappop(h)[1] for i in range(k)]

Solution 3: Bucket Sort

Time complexity: O(n)

Space complexity: O(n)

C++/hashmap

class Solution {

public:

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

unordered_map<int, int> count;

int max_freq = 1;

for (const int num : nums)

max_freq = max(max_freq, ++count[num]);

map<int, vector<int>> buckets;

for (const auto& kv : count)

buckets[kv.second].push_back(kv.first);

vector<int> ans;

for (int i = max_freq; i >= 1; --i) {

auto it = buckets.find(i);

if (it == buckets.end()) continue;

ans.insert(ans.end(), it->second.begin(), it->second.end());

if (ans.size() == k) return ans;

}

return ans;

}

};

C++/array

class Solution {

public:

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

unordered_map<int, int> count;

for (const int num : nums)

++count[num];

vector<vector<int>> buckets(nums.size() + 1);

for (const auto& kv : count)

buckets[kv.second].push_back(kv.first);

vector<int> ans;

for (auto it = buckets.rbegin(); it != buckets.rend(); ++it) {

const vector<int>& keys = *it;

if (keys.empty()) continue;

ans.insert(ans.begin(), keys.begin(), keys.end());

if (ans.size() == k) return ans;

}

return ans;

}

};

Java

// Author: Huahua

// Running time: 23 ms

class Solution {

public List<Integer> topKFrequent(int[] nums, int k) {

List[] buckets = new List[nums.length + 1];

Map<Integer, Integer> counts = new HashMap<>();

for (int num : nums)

counts.put(num, counts.getOrDefault(num, 0) + 1);

for (int num : counts.keySet()) {

int count = counts.get(num);

if (buckets[count] == null)

buckets[count] = new ArrayList<Integer>();

buckets[count].add(num);

}

List<Integer> ans = new ArrayList<>();

for (int i = buckets.length - 1; i > 0 && ans.size() < k; --i) {

if (buckets[i] != null) ans.addAll(buckets[i]);

}

return ans;

}

Python

"""

Author: Huahua

Running time: 64 ms

"""

class Solution:

def topKFrequent(self, nums, k):

counts = collections.Counter(nums)

buckets = [[] for _ in range(len(nums) + 1)]

for num in counts.keys():

buckets[counts[num]].append(num)

ans = []

for i in range(len(nums), 0, -1):

ans += buckets[i]

if len(ans) == k: return ans

return ans

Posts tagged as “priority queue”

花花酱 LeetCode 871. Minimum Number of Refueling Stops

Problem

Solution1: DP

Solution2: Priority Queue

Related Problems

花花酱 LeetCode 347. Top K Frequent Elements

Solution 2: Priority queue / max heap

C++

Java

Python

Solution 3: Bucket Sort

C++/hashmap

C++/array

Java

Python

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