**Problem**: Contained Ranges

A range [x1, x2] where x1 (start of range) and x2 (end of range) are integer and x1 ≤ x2 includes all numbers x such that x1 ≤ x ≤ x2. A range [x1, x2] is contained in a range [y1, y2] if y1 ≤ x1 ≤ x2 ≤ y2. Given a list of ranges A, check whether there exists a pair of ranges such that one range is contained in the other range.

**Discussion**

A naive solution which runs in O(N^{2}) where N is the number of ranges is that we verify all pairs of ranges and check whether one range is contained in the other range.

Since the condition of range [x1, x2] is contained in range [y1, y2] is that y1 ≤ x1 ≤ x2 ≤ y2, we want to sort ranges by start of range in increasing order. Sorting only takes O(NlogN).

For example, assume after sorting, we have list of ranges A = [[1, 3], [2, 4], [5, 9], [6, 10], [7, 8]]. Since the ranges are sorted, we just check whether A[i] is contained in A[j] where i > j. However, this still requires O(N^{2}).

Intuitively, we will iterate on ranges and given a new range, we check whether it is contained in any previous range. For example, we check [5, 9] is contained by [1, 3] or [2, 4]. Thus, verifying range A[i] requires checking on i previous ranges. Can we do it better?

Yes we can. Given A[i], do we need to check all A[j] where j < i whether A[i] is contained in A[j]? Actually, it is sufficient to check whether A[i] is contained in A[i-1]. Given any range A[j] before A[i-1], since A[i-1] is not contained by A[j], the end of range A[i-1] must greater than the end of range A[j]. Therefore, if A[i] is contained in A[j], it is also contained in A[i-1]. This is similar to the range [7, 8] is contained in [5, 9] and also [6, 10].

Below is the implementation which runs in O(NlogN).

public class Range{ public int start; public int end; public Range(int s, int e){ start = s; end = e; } @Override public String toString() { return "[" + start + "," + end + "]"; } @Override public boolean equals(Object o) { if (o == this) { return true; } if (!(o instanceof Range)) { return false; } Range c = (Range) o; return start == c.start && end == c.end; } } //compare two range by start, used to sort list of ranges by their starts public class RangeComparator implements Comparator<Range> { public int compare(Range range1, Range range2) { return Integer.compare(range1.start, range2.start); } } public boolean hasContainedRange(List<Range> ranges){ if (ranges.size() < 2){ return false; } ranges.sort(new RangeComparator()); for(int i = 1; i < ranges.size(); i++) { if (ranges.get(i).end <= ranges.get(i - 1).end){ return true; } } return false; }