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package g1401_1500.s1488_avoid_flood_in_the_city;

// #Medium #Array #Hash_Table #Greedy #Binary_Search #Heap_Priority_Queue #Binary_Search_II_Day_18
// #2022_04_05_Time_82_ms_(75.08%)_Space_64.3_MB_(81.23%)

import java.util.HashMap;
import java.util.TreeSet;

/**
 * 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.
 *
 * **Constraints:**
 *
 * *   1 <= rains.length <= 10⁵
 * *   0 <= rains[i] <= 10⁹
**/
public class Solution {
    public int[] avoidFlood(int[] rains) {
        HashMap hm = new HashMap<>();
        TreeSet tree = new TreeSet<>();
        int[] ans = new int[rains.length];
        for (int i = 0; i < rains.length; i = i + 1) {
            int val = rains[i];
            if (val != 0) {
                if (hm.containsKey(val)) {
                    int mapVal = hm.get(val);
                    if (tree.ceiling(mapVal) != null) {
                        ans[tree.ceiling(mapVal)] = val;
                        hm.put(val, i);
                        tree.remove(tree.ceiling(mapVal));
                    } else {
                        return new int[0];
                    }
                } else {
                    hm.put(val, i);
                }
                ans[i] = -1;
            } else {
                tree.add(i);
            }
        }
        for (int tr : tree) {
            ans[tr] = 1;
        }
        return ans;
    }
}