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- divcode1986 May 17, 2016/* Assumptions: Conference room data is provided as a list of sorted disjoint interval objects. Meeting time of interest (m) is an interval object. Brute-force: Iterate through each list of interval objects representing conference rooms. For each list, do a linear search for the presence of an interval that overalps with (m). Time complexity: O(nm) where n is the number of conference rooms and m is the size of the longest list of disjoint intervals. Better appraoach: Same as brute force but instead of linear search use binary search. Time complexity: O(nlogm)**/ public class Interval { int start; int end; } public int countOpenings(ArrayList<ArrayList<Interval>> ls, Interval m) { if(ls==null||ls.isEmpty()||m==null) { return 0; } int c=0; for(int i=0;i<ls.size();i++) { if(isAvailable(ls.get(i),m)) { c++; } } return c; } private boolean isAvailable(ArrayList<Interval> ls, Interval n) { int start=0; int end=ls.size()-1; while(start<=end) { int mid=(start+end)/2; if(isOverlap(ls.get(mid),n)) { return false; } if(ls.get(mid).end<=n.start) { start=mid+1; }else { end=mid-1; } } return true; } private boolean isOverlap(Interval m, Interval n) { return(m.start>=n.start && m.start<n.end)||(n.start>=m.start && n.start<m.end); }