C++ solution with comments. Time complexity: O(nlog(n)) and space: O(n)
\nclass Solution {\npublic:\n    vector<int> avoidFlood(vector<int>& rains) {\n\t\n        vector<int> result(rains.size(), 1);\n        \n\t\t// Create a hash_
```\nclass Solution {\npublic:\n    vector<int> avoidFlood(vector<int>& rains) {\n\t\n        vector<int> result(rains.size(), 1);\n        \n\t\t// Create a hash_map with lake and corresponding position.\n\t\t// When we see the same lake again, we need to \n\t\t// find a dry day (0) between current appearance and previous appearance (map will have index of previous appearance).\n        unordered_map<int, int> lakes;\n\t\t\n\t\t// Keep track of dry days position in a sorted order, so that we can binary-search between\n\t\t// lakes appearing twice (between two lakes postion).\n        std::set<int> dry_days_pos;\n        \n        // time: O(nlog(n)), space: O(n)\n        for (int i = 0; i < rains.size(); i++)\n        {\n            if (rains[i] == 0)\n            {\n                // insert takes log(n)\n                dry_days_pos.insert(i); // Save position of dry days.\n                // Will update the result array with appropriate lake to be dried out as we move forward.\n                continue;\n            }\n            \n            result[i] = -1; // fill -1 to result array when there is a rainy day;\n            \n\t\t\t// Check if this lake was already filled before.\n            auto it = lakes.find(rains[i]);\n            if (it == lakes.end()) // If lake was not filled before, no fludding \n            {\n                lakes[rains[i]] = i; // => save the lake and the day it got filled (day refers to position)\n            }\n            else\n            {\n                // upper_bound() takes log(n) as it uses binary search to find an element.\n                auto dry_day_it = dry_days_pos.upper_bound(it->second); // remember, we saved the lake\'s day when it was filled before.\n                if (dry_day_it == dry_days_pos.end())\n                    return {}; // Could not find any dry day between the days, the lake was previously filled and currently being filled\n                \n\t\t\t\t// When we could find a dry day, fill the result array at dry day position with current lake being dried out.\n                result[*dry_day_it] = rains[i]; // Update the result array with appropriate lake to dry it out\n                dry_days_pos.erase(dry_day_it); // Remove dry day from set as we already used it.\n                \n                // Update the dictionary to the current lake\'s index, \n                // as the previously filled lake is already dried out.\n                lakes[rains[i]] = i;\n            }\n        }\n        return result;\n    }\n};\n\n```
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