README_EN.md

1353. Maximum Number of Events That Can Be Attended

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Description

You are given an array of events where events[i] = [startDayi, endDayi]. Every event i starts at startDayi and ends at endDayi.

You can attend an event i at any day d where startTimei <= d <= endTimei. You can only attend one event at any time d.

Return the maximum number of events you can attend.

 

Example 1:

Input: events = [[1,2],[2,3],[3,4]]
Output: 3
Explanation: You can attend all the three events.
One way to attend them all is as shown.
Attend the first event on day 1.
Attend the second event on day 2.
Attend the third event on day 3.

Example 2:

Input: events= [[1,2],[2,3],[3,4],[1,2]]
Output: 4

 

Constraints:

• 1 <= events.length <= 105
• events[i].length == 2
• 1 <= startDayi <= endDayi <= 105

Solutions

Python3

class Solution:
    def maxEvents(self, events: List[List[int]]) -> int:
        d = defaultdict(list)
        i, j = inf, 0
        for s, e in events:
            d[s].append(e)
            i = min(i, s)
            j = max(j, e)
        h = []
        ans = 0
        for s in range(i, j + 1):
            while h and h[0] < s:
                heappop(h)
            for e in d[s]:
                heappush(h, e)
            if h:
                ans += 1
                heappop(h)
        return ans

Java

class Solution {
    public int maxEvents(int[][] events) {
        Map<Integer, List<Integer>> d = new HashMap<>();
        int i = Integer.MAX_VALUE, j = 0;
        for (var v : events) {
            int s = v[0], e = v[1];
            d.computeIfAbsent(s, k -> new ArrayList<>()).add(e);
            i = Math.min(i, s);
            j = Math.max(j, e);
        }
        PriorityQueue<Integer> q = new PriorityQueue<>();
        int ans = 0;
        for (int s = i; s <= j; ++s) {
            while (!q.isEmpty() && q.peek() < s) {
                q.poll();
            }
            for (int e : d.getOrDefault(s, Collections.emptyList())) {
                q.offer(e);
            }
            if (!q.isEmpty()) {
                q.poll();
                ++ans;
            }
        }
        return ans;
    }
}

C++

class Solution {
public:
    int maxEvents(vector<vector<int>>& events) {
        unordered_map<int, vector<int>> d;
        int i = INT_MAX, j = 0;
        for (auto& v : events) {
            int s = v[0], e = v[1];
            d[s].push_back(e);
            i = min(i, s);
            j = max(j, e);
        }
        priority_queue<int, vector<int>, greater<int>> q;
        int ans = 0;
        for (int s = i; s <= j; ++s) {
            while (q.size() && q.top() < s) {
                q.pop();
            }
            for (int e : d[s]) {
                q.push(e);
            }
            if (q.size()) {
                ++ans;
                q.pop();
            }
        }
        return ans;
    }
};

Go

func maxEvents(events [][]int) int {
	d := map[int][]int{}
	i, j := math.MaxInt32, 0
	for _, v := range events {
		s, e := v[0], v[1]
		d[s] = append(d[s], e)
		i = min(i, s)
		j = max(j, e)
	}
	q := hp{}
	ans := 0
	for s := i; s <= j; s++ {
		for q.Len() > 0 && q.IntSlice[0] < s {
			heap.Pop(&q)
		}
		for _, e := range d[s] {
			heap.Push(&q, e)
		}
		if q.Len() > 0 {
			heap.Pop(&q)
			ans++
		}
	}
	return ans
}

func max(a, b int) int {
	if a > b {
		return a
	}
	return b
}

func min(a, b int) int {
	if a < b {
		return a
	}
	return b
}

type hp struct{ sort.IntSlice }

func (h *hp) Push(v interface{}) { h.IntSlice = append(h.IntSlice, v.(int)) }
func (h *hp) Pop() interface{} {
	a := h.IntSlice
	v := a[len(a)-1]
	h.IntSlice = a[:len(a)-1]
	return v
}
func (h *hp) Less(i, j int) bool { return h.IntSlice[i] < h.IntSlice[j] }

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