Create solution.java

The Number of the Smallest Unoccupied Chair/solution.java

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+class Item {
+    int friend;
+    int chair = -1;
+    int startTime;
+    int endTime;
+
+    public Item(int friend, int startTime, int endTime) {
+        this.friend = friend;
+        this.startTime = startTime;
+        this.endTime = endTime;
+    }
+
+    public void setChair(int chair) {
+        this.chair = chair;
+    }
+}
+
+class Solution {
+    public int smallestChair(int[][] times, int targetFriend) {
+        Item[] arrival = new Item[times.length];
+
+        PriorityQueue<Integer> available = new PriorityQueue<>();
+        PriorityQueue<Item> assigned = new PriorityQueue<>((a, b) -> a.endTime - b.endTime);
+
+        for (int i = 0; i < times.length; i++) {
+            arrival[i] = new Item(i, times[i][0], times[i][1]);
+            available.add(i);
+        }
+
+        // here i am Sorting friends by their arrival times
+        Arrays.sort(arrival, (a, b) -> a.startTime - b.startTime);
+
+        for (int i = 0; i < arrival.length; i++) {
+            Item item = arrival[i];
+
+            //here i am Freeing up chairs for people who have left by this time
+            while (!assigned.isEmpty() && assigned.peek().endTime <= item.startTime) {
+                available.add(assigned.poll().chair);
+            }
+
+            // now Assign the smallest available chair to the current friend
+            item.setChair(available.poll());
+            assigned.add(item);
+
+            // If this is the target friend, return their chair number
+            if (item.friend == targetFriend) {
+                return item.chair;
+            }
+        }
+
+        return -1;
+    }
+}