# Accepted and easy to understand Java solution with HashMaps

• This solution passed all the test cases on LeetCode. This may not be the best solution out there but definitely a good one to start with and then iterate on it. Does not need any sorting. Uses HashMaps. Please follow comments.

``````public int minMeetingRooms(Interval[] intervals) {

if (intervals == null || intervals.length == 0) {
return 0;
}

if (intervals.length == 1) {
return 1;
}

// how many meetings start at a given time
Map<Integer, Integer> startToCount = new HashMap<>();
// how many meetings end at a given time
Map<Integer, Integer> endToCount = new HashMap<>();

// min of all the meetings start times
int minStart = intervals[0].start;
// max of all the meetings end times
int maxEnd = intervals[0].end;

for (Interval i : intervals) {

if (!startToCount.containsKey(i.start)) {
startToCount.put(i.start, 1);
} else {
startToCount.put(i.start, startToCount.get(i.start) + 1);
}

if (!endToCount.containsKey(i.end)) {
endToCount.put(i.end, 1);
} else {
endToCount.put(i.end, endToCount.get(i.end) + 1);
}

if (i.start < minStart) {
minStart = i.start;
}

if (i.end > maxEnd) {
maxEnd = i.end;
}
}

int minRooms = 0;
int rooms = 0;

for (int i = minStart; i <= maxEnd; i++) {
if (endToCount.containsKey(i)) {
// endToCount.get(i) number of meeting rooms were vacated at this time
rooms -= endToCount.get(i);
}
if (startToCount.containsKey(i)) {
// startToCount.get(i) number of meeting rooms occupied at this time
rooms += startToCount.get(i);
// record the max of the occupied rooms at any given time.
minRooms = Math.max(rooms, minRooms);
}
}

return minRooms;
}
``````

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