Contained Ranges - Coding in a Smart Way


Contained Ranges - Coding in a Smart Way
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.

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).
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].
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;
}
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