Java solution, beats 92%

• walk thru the intervals, find the biggest amount of meetings going on at the same time.

``````/**
* Definition for an interval.
* public class Interval {
*     int start;
*     int end;
*     Interval() { start = 0; end = 0; }
*     Interval(int s, int e) { start = s; end = e; }
* }
*/
public class Solution {
public int minMeetingRooms(Interval[] intervals) {
int[] starts = new int[intervals.length];
int[] ends = new int[intervals.length];
for(int i = 0; i < intervals.length; i++) {
starts[i] = intervals[i].start;
ends[i] = intervals[i].end;
}

Arrays.sort(starts);
Arrays.sort(ends);

int min = 0;
int idxStarts = 0;
int idxEnds = 0;
int count = 0;
while(idxStarts < intervals.length && idxEnds < intervals.length) {
if(starts[idxStarts] < ends[idxEnds]) {
count++;
idxStarts++;
} else {
count--;
idxEnds++;
}

if(count > min) {
min = count;
}
}

return min;
}
}
``````

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