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Leetcode #1488: Avoid Flood in The City (
class Solution {
vector<int> avoidFlood(vector<int>& rains) {
vector<int> ans; // Store the final answer here.
int n = rains.size();
unordered_map<int, int> fulllakes; // Lake number -> day on which it became full.
set<int> drydays; // Set of available days that can be used for drying a full lake.
for (int i=0; i<n; i++) { // For each day -
if (rains[i] == 0) { // No rain on this day.
drydays.insert(i); // This day can be used as a day to dry some lake.
// We don't know which lake to prioritize for drying yet.
ans.push_back(1); // Any number would be ok. This will get overwritten
// eventually. If it doesn't get overwritten, its totally
// ok to dry a lake irrespective of whether it is full or
// empty.
} else { // Rained in rains[i]-th lake.
int lake = rains[i];
if (fulllakes.find(lake) != fulllakes.end()) { // If it is already full -
// We must dry this lake before it rains in this lake.
// So find a day in "drydays" to dry this lake. Obviously, that day must
// be a day that is after the day on which the lake was full.
// i.e. if the lake got full on 7th day, we must find a dry day that is
// greater than 7.
auto it = drydays.lower_bound(fulllakes[lake]);
if (it == drydays.end()) { // If there is no available dry day to dry
// the lake, flooding is inevitable.
return {}; // Sorry, couldn't stop flooding.
int dryday = *it; // Great, found a day which we can use to dry the lake.
ans[dryday] = lake; // Overwrite the "1" and dry "lake"-th lake instead.
drydays.erase(dryday); // We dried "lake"-th lake on "dryday", and we
// can't use the same day to dry any other lake,
// so remove the day from the set of available
// drydays.
fulllakes[lake] = i; // Update that the "lake" became full on "i"-th day.
ans.push_back(-1); // As the problem statement expects.
return ans; // Congratualtions, you avoided flooding.
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