Formatted question description: https://leetcode.ca/all/1488.html

1488. Avoid Flood in The City (Medium)

Your country has an infinite number of lakes. Initially, all the lakes are empty, but when it rains over the nth lake, the nth lake becomes full of water. If it rains over a lake which is full of water, there will be a flood. Your goal is to avoid the flood in any lake.

Given an integer array rains where:

  • rains[i] > 0 means there will be rains over the rains[i] lake.
  • rains[i] == 0 means there are no rains this day and you can choose one lake this day and dry it.

Return an array ans where:

  • ans.length == rains.length
  • ans[i] == -1 if rains[i] > 0.
  • ans[i] is the lake you choose to dry in the ith day if rains[i] == 0.

If there are multiple valid answers return any of them. If it is impossible to avoid flood return an empty array.

Notice that if you chose to dry a full lake, it becomes empty, but if you chose to dry an empty lake, nothing changes. (see example 4)

 

Example 1:

Input: rains = [1,2,3,4]
Output: [-1,-1,-1,-1]
Explanation: After the first day full lakes are [1]
After the second day full lakes are [1,2]
After the third day full lakes are [1,2,3]
After the fourth day full lakes are [1,2,3,4]
There's no day to dry any lake and there is no flood in any lake.

Example 2:

Input: rains = [1,2,0,0,2,1]
Output: [-1,-1,2,1,-1,-1]
Explanation: After the first day full lakes are [1]
After the second day full lakes are [1,2]
After the third day, we dry lake 2. Full lakes are [1]
After the fourth day, we dry lake 1. There is no full lakes.
After the fifth day, full lakes are [2].
After the sixth day, full lakes are [1,2].
It is easy that this scenario is flood-free. [-1,-1,1,2,-1,-1] is another acceptable scenario.

Example 3:

Input: rains = [1,2,0,1,2]
Output: []
Explanation: After the second day, full lakes are  [1,2]. We have to dry one lake in the third day.
After that, it will rain over lakes [1,2]. It's easy to prove that no matter which lake you choose to dry in the 3rd day, the other one will flood.

Example 4:

Input: rains = [69,0,0,0,69]
Output: [-1,69,1,1,-1]
Explanation: Any solution on one of the forms [-1,69,x,y,-1], [-1,x,69,y,-1] or [-1,x,y,69,-1] is acceptable where 1 <= x,y <= 10^9

Example 5:

Input: rains = [10,20,20]
Output: []
Explanation: It will rain over lake 20 two consecutive days. There is no chance to dry any lake.

 

Constraints:

  • 1 <= rains.length <= 10^5
  • 0 <= rains[i] <= 10^9

Related Topics:
Array, Hash Table

Solution 1. Greedy + Heap

Consider input [1, 2, 0, 2, 1, ...]. At the first 0, I’d greedily dry 2 instead of 1 because that’s the first positive number following this 0.

So my idea is that:

  1. For each A[i], compute next[i] which is the index of the next element equals A[i]. If there is no such element, next[i] = -1.
  2. Maintain a set s storing the active lakes (i.e. the lakes waiting to be dried) and a min-root heap pq storing the next indexes of active lakes.
    1. If A[i] > 0:
      1. If A[i] is in s, it means that A[i] is already an active lake. We should return {}.
      2. If next[i] == -1, it means that it won’t be filled again later, we can safely ignore it when drying, so just skip it.
      3. Otherwise, we need to add A[i] to s and add next[i] to pq.
    2. If A[i] == 0:
      1. If pq.empty(), it means that there is no active lakes, we can just arbitrarily dry some lake, say 1.
      2. Otherwise, we should dry the active lake whose next index is smallest, i.e. A[pq.top()]. So ans[i] = A[pq.top()] and we can remove A[pq.top()] from s and pop the top from pq.
// OJ: https://leetcode.com/problems/avoid-flood-in-the-city/

// Time: O(NlogN)
// Space: O(N)
class Solution {
public:
    vector<int> avoidFlood(vector<int>& A) {
        int N = A.size();
        vector<int> ans(N, -1), next(N, -1);
        unordered_map<int, int> m;
        for (int i = 0; i < N; ++i) {
            if (A[i] == 0) continue;
            if (m.count(A[i])) next[m[A[i]]] = i;
            m[A[i]] = i;
        }
        unordered_set<int> s;
        priority_queue<int, vector<int>, greater<int>> pq;
        for (int i = 0; i < N; ++i) {
            if (A[i] > 0) {
                if (s.count(A[i])) return {};
                if (next[i] == -1) continue;
                pq.push(next[i]);
                s.insert(A[i]);
            } else {
                if (pq.empty()) ans[i] = 1;
                else {
                    int j = pq.top();
                    pq.pop();
                    ans[i] = A[j];
                    s.erase(A[j]);
                }
            }
        }
        return ans;
    }
};

Solution 2. Greedy

I came with this solution first in the context but instead of using set::upper_bound I used std::upper_bound which degrades to O(N) complexity on set :(

So remember! When doing upper_bound on set, must use set::upper_bound!

Use a set s to store the index of available 0s. Use map to store the mapping from a active lake to its index.

If A[i] == 0, we just add it to s and set a dummy value ans[i] = 1. If might get overwritten later. If it’s not overwritten, leaving this dummy value here is also correct.

If A[i] > 0, we first check if it exists in m:

  • Only when the answer is yes, we need to find the smallest index in s that is greater than m[A[i]]. If there is no such index, we don’t have available day to dry this lake, we should return {}. Otherwise, we dry A[i] using this index and erase this used index from s.
  • No matter if it’s yes or no, we need to set m[A[i]] = i to add this active lake.
// OJ: https://leetcode.com/problems/avoid-flood-in-the-city/

// Time: O(NlogN)
// Space: O(N)
class Solution {
public:
    vector<int> avoidFlood(vector<int>& A) {
        int N = A.size();
        vector<int> ans(N, -1);
        unordered_map<int, int> m;
        set<int> s;
        for (int i = 0; i < N; ++i) {
            if (A[i] > 0) {
                if (m.count(A[i])) {
                    auto it = s.upper_bound(m[A[i]]);
                    if (it == s.end()) return {};
                    ans[*it] = A[i];
                    s.erase(it);
                }
                m[A[i]] = i;
            } else {
                s.insert(i);
                ans[i] = 1;
            }
        }
        return ans;
    }
};

Java

class Solution {
    public int[] avoidFlood(int[] rains) {
        if (rains == null || rains.length == 0)
            return rains;
        int length = rains.length;
        int[] ans = new int[length];
        Queue<Integer> lakeQueue = new LinkedList<Integer>();
        TreeSet<Integer> drySet = new TreeSet<Integer>();
        for (int i = 0; i < length; i++) {
            if (rains[i] > 0) {
                ans[i] = -1;
                lakeQueue.offer(i);
            } else 
                drySet.add(i);
        }
        while (!drySet.isEmpty() && !lakeQueue.isEmpty() && drySet.first() < lakeQueue.peek()) {
            int dryIndex = drySet.first();
            ans[dryIndex] = 1;
            drySet.remove(dryIndex);
        }
        Map<Integer, Integer> map = new HashMap<Integer, Integer>();
        while (!lakeQueue.isEmpty()) {
            int index = lakeQueue.poll();
            int lake = rains[index];
            if (!map.containsKey(lake))
                map.put(lake, index);
            else {
                int prevIndex = map.get(lake);
                Integer dryIndex = drySet.ceiling(prevIndex);
                if (dryIndex == null || dryIndex > index)
                    return new int[0];
                else {
                    ans[dryIndex] = lake;
                    drySet.remove(dryIndex);
                    map.put(lake, index);
                }
            }
        }
        for (int dryIndex : drySet)
            ans[dryIndex] = 1;
        return ans;
    }
}

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