Merge pull request #433 from Krishna200608/feature/issue-379-room-allocation

Solved Issue #379: Room Allocation (CSES 1164)

Author: Dheeraj-06

Problems/Data-Structures/Day-05/sol/Krishna200608/Solution2.cpp

@@ -0,0 +1,113 @@
+/*
+Problem: Room Allocation
+Link: https://cses.fi/problemset/task/1164
+Author: Krishna Sikheriya (Krishna200608)
+
+Short Problem Statement:
+There is a hotel and n customers with specific arrival and departure times. 
+We need to assign rooms to customers such that the total number of rooms used is minimized.
+Two customers can stay in the same room if the first customer leaves before the second arrives.
+Print the minimum number of rooms needed and the room assignment for each customer.
+
+Approach:
+1. Store customer details (arrival, departure, original_index).
+2. Sort customers based on arrival times.
+3. Use a Min-Priority Queue (Min-Heap) to store pairs of {departure_time, room_number} for currently occupied rooms.
+   The heap allows us to efficiently find the room that becomes free earliest.
+4. Iterate through the sorted customers:
+   - If the earliest departure time in the heap is less than the current customer's arrival time, 
+     it means that room is free. We reuse that room (pop from heap).
+   - If no room is free, we allocate a new room number.
+   - Push the current customer's departure time and assigned room number into the heap.
+   - Store the assigned room number in an answer array using the original index.
+5. The maximum size of the heap reached (or the max room ID generated) gives the minimum rooms required.
+
+Time Complexity: O(N log N) - dominated by sorting the customers. Heap operations take O(log N).
+Space Complexity: O(N) - to store customer details, the priority queue, and the answer array.
+
+Example I/O:
+Input:
+3
+1 2
+2 4
+4 4
+
+Output:
+2
+1 2 1
+*/
+
+#include <iostream>
+#include <vector>
+#include <algorithm>
+#include <queue>
+
+using namespace std;
+
+// Struct to keep track of customer details and their original order
+struct Customer {
+    int id;
+    int arrival;
+    int departure;
+};
+
+// Comparator to sort customers by arrival time
+bool compareCustomers(const Customer& a, const Customer& b) {
+    return a.arrival < b.arrival;
+}
+
+int main() {
+    // Fast I/O
+    ios_base::sync_with_stdio(false);
+    cin.tie(NULL);
+
+    int n;
+    if (!(cin >> n)) return 0;
+
+    vector<Customer> customers(n);
+    for (int i = 0; i < n; ++i) {
+        customers[i].id = i;
+        cin >> customers[i].arrival >> customers[i].departure;
+    }
+
+    sort(customers.begin(), customers.end(), compareCustomers);
+
+    // Min-priority queue to store {departure_time, room_number}
+    // We want the room that frees up earliest to be at the top
+    priority_queue<pair<int, int>, vector<pair<int, int>>, greater<pair<int, int>>> pq;
+
+    vector<int> ans(n);
+    int last_room_id = 0;
+
+    for (int i = 0; i < n; ++i) {
+        int assigned_room;
+
+        // Check if the earliest departing room is free before current arrival
+        if (!pq.empty() && pq.top().first < customers[i].arrival) {
+            assigned_room = pq.top().second;
+            pq.pop();
+        } else {
+            // Allocate a new room
+            last_room_id++;
+            assigned_room = last_room_id;
+        }
+
+        // Push current customer's departure and room info
+        pq.push({customers[i].departure, assigned_room});
+        ans[customers[i].id] = assigned_room;
+    }
+
+    // Output results
+    cout << last_room_id << endl;
+    for (int i = 0; i < n; ++i) {
+        cout << ans[i] << " ";
+    }
+    cout << endl;
+
+    return 0;
+}
+
+/*
+SUBMISSION LINK:
+https://cses.fi/paste/d01d5f0ab0c6ae25f11225/
+*/