@@ -40,5 +40,32 @@ public int minMeetingRooms(int[][] intervals) {
4040return heap .size ();
4141 }
4242 }
43+
44+ public static class Solution2 {
45+ /**
46+ * I'm so glad to have come up with this solution completely on my own on 10/13/2021.
47+ * Drawing on a piece of paper helps A LOT! It helps visualize your thoughts and clear the ambiguity up!
48+ */
49+ public int minMeetingRooms (int [][]intervals ) {
50+ //I use the meeting's end time as the room indicate and put them into a heap
51+ PriorityQueue <Integer >rooms =new PriorityQueue <>();
52+ Arrays .sort (intervals , (a ,b ) ->a [0 ] -b [0 ]);
53+ for (int i =0 ;i <intervals .length ;i ++) {
54+ if (rooms .isEmpty ()) {
55+ rooms .add (intervals [i ][1 ]);
56+ }else {
57+ if (rooms .peek () >intervals [i ][0 ]) {
58+ //if the room that becomes available the earliest still cannot accommodate this new meeting, then we'll have to add a new room
59+ rooms .add (intervals [i ][1 ]);
60+ }else {
61+ //otherwise, we'll just update the room that finishes the earliest with the new finish time.
62+ rooms .poll ();
63+ rooms .add (intervals [i ][1 ]);
64+ }
65+ }
66+ }
67+ return rooms .size ();
68+ }
69+ }
4370}
4471