Expedia Interview Question for Software Developers


Country: United States
Interview Type: In-Person




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/**
 * [1,3],[2,6],[8,12],[10,18] => [1,6],[8,18]
 * Definition for an interval.
**/
 struct Interval {
     int start;
     int end;
     Interval() : start(0), end(0) {}
     Interval(int s, int e) : start(s), end(e) {}
 };
 
bool compare(Interval a, Interval b) {
    return a.start < b.start;
}

vector<Interval> merge(vector<Interval> &A) {
    vector<Interval> result;
    vector<Interval>::iterator it;
    sort(A.begin(), A.end(), compare);
    if (A.empty()) {
        return result;
    }
    else {  
        result.push_back(*A.begin());
        for (it = (A.begin() + 1); it != A.end(); ++it) {
            if ((*it).start > result.back().end)
                result.push_back(*it);
            else
                result.back().end = max(result.back().end, (*it).end);
        }
    }    
}

- PraTrick July 24, 2017 | Flag Reply
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// Time Complexity O(n)
// Space Complexity O(n)

public class OverlappingProblem {

        static class Interval {

                int start;
                int end;

                public Interval() {
                        this.start = 0;
                        this.end = 0;
                }

                public Interval(int start, int end) {
                        this.start = start;
                        this.end = end;
                }

        }

        static ArrayList<Interval> getOverlapingInterval(List<Interval> intervals) {

                ArrayList<Interval> results = new ArrayList<Interval>();

                Collections.sort(intervals, new MyComparator());

                int start = intervals.get(0).start;
                int end = intervals.get(0).end;

                for (int index = 0; index <= intervals.size(); index++) {
                        if (index == 0) {
                                start = intervals.get(index).start;
                                end = intervals.get(index).end;
                        } else if (index != intervals.size()) {
                                if (intervals.get(index).start <= end) {
                                        end = (intervals.get(index).end > end) ? intervals.get(index).end : end;
                                } else {
                                        results.add(new Interval(start, end));
                                        start = intervals.get(index).start;
                                        end = intervals.get(index).end;
                                }
                        } else {
                                results.add(new Interval(start, end));
                        }

                }

                return results;
        }

        public static void main(String args[]) {
                
                List<Interval> listOfIntervals = new ArrayList<Interval>();
                listOfIntervals.add(new Interval(1,3));
                listOfIntervals.add(new Interval(2,6));
                listOfIntervals.add(new Interval(8, 12));
                listOfIntervals.add(new Interval(10, 18));
                
                List<Interval> results = getOverlapingInterval(listOfIntervals);
                for(Interval result : results) {
                        System.out.println(result.start+"   "+result.end);
                }

        }

        static class MyComparator implements Comparator<Interval> {

                @Override
                public int compare(Interval a, Interval b) {

                        int first = a.start - b.start;
                        return (first == 0) ? (a.end - b.end) : first;
                }

        }

}

- Kapil July 24, 2017 | Flag Reply
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/*
Merge Intervals in-place without using the 'new' operator.
Time: O(nlogn)
Space: O(1)
*/

private static void mergeIntervals(List<Interval> interval) {
		if (interval == null)	throw new IllegalArgumentException();
		if (interval.size() == 0)	return;
 
		Collections.sort(interval, new Comparator<Interval>(){
			@Override
			public int compare(Interval a, Interval b) {
				if (a == b)	return b.end - a.end;
				return a.start - b.start;
			}
		});
 
		int i = 0;
		while (i < interval.size() - 1) {
			if (interval.get(i).end >= interval.get(i + 1).start) {
				interval.get(i).end = Math.max(interval.get(i).end, interval.get(i + 1).end);
				interval.remove(i + 1);
			}
			else	i++;
		}
	}

- raitGroup1007 July 25, 2017 | Flag Reply
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0
of 0 vote

Elegant and simple - Time => O(2N)

var intervals = function (array) {
    var result = [];
    var minIndex = Number.MAX_VALUE;
    var maxIndex = Number.MIN_VALUE;
    var totalResult = [];
    for (var i = 0; i < array.length; i++) {
        for (var start = array[i][0]; start <= array[i][1]; start++) {
            result[start] = 1;
        }
    }
    for (var i = 0; i < result.length; i++) {
        if (result[i] === 1) {
            minIndex = Math.min(minIndex, i);
            maxIndex = Math.max(maxIndex, i);
        }
        else {
            if (minIndex !== Number.MAX_VALUE && maxIndex !== Number.MIN_VALUE) {
                totalResult.push([minIndex, maxIndex]);
                minIndex = Number.MAX_VALUE;
                maxIndex = Number.MIN_VALUE;
            }
        }
    }
    if (minIndex !== Number.MAX_VALUE && maxIndex !== Number.MIN_VALUE) {
        totalResult.push([minIndex, maxIndex]);
    }
    return totalResult;
}

$(intervals([[1, 3], [2, 6], [8, 12], [10, 18]]));

- Anonymous July 26, 2017 | Flag Reply
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0
of 0 vote

Time O(2N)

/**
 * [1,3],[2,6],[8,12],[10,18] => [1,6],[8,18]
 * Definition for an interval.
**/


var intervals = function (array) {
    var result = [];
    var minIndex = Number.MAX_VALUE;
    var maxIndex = Number.MIN_VALUE;
    var totalResult = [];
    for (var i = 0; i < array.length; i++) {
        for (var start = array[i][0]; start <= array[i][1]; start++) {
            result[start] = 1;
        }
    }
    for (var i = 0; i < result.length; i++) {
        if (result[i] === 1) {
            minIndex = Math.min(minIndex, i);
            maxIndex = Math.max(maxIndex, i);
        }
        else {
            if (minIndex !== Number.MAX_VALUE && maxIndex !== Number.MIN_VALUE) {
                totalResult.push([minIndex, maxIndex]);
                minIndex = Number.MAX_VALUE;
                maxIndex = Number.MIN_VALUE;
            }
        }
    }
    if (minIndex !== Number.MAX_VALUE && maxIndex !== Number.MIN_VALUE) {
        totalResult.push([minIndex, maxIndex]);
    }
    return totalResult;
}

$(intervals([[1, 3], [2, 6], [8, 12], [10, 18]]));

- maksymas July 26, 2017 | Flag Reply
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of 0 vote

public class Interval
        {
            public int start;
            public int end;
            public Interval()
            {
                start =0;
                end = 0;
            }

            public Interval(int s, int e)
            {
                start = s;
                end = e;
            }
        }

        public IList<Interval> Merge(IList<Interval> intervals)
        {
            IList<Interval> res = new List<Interval>();
            intervals = intervals.OrderBy(x => x.start).ToList();

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

            for (int i = 0; i < intervals.Count; i++)
            {
                var newInterval = new Interval(intervals[i].start, intervals[i].end);
                while(i< intervals.Count -1 && newInterval.end >= intervals[i+1].start)
                {
                    if(intervals[i+1].end > newInterval.end)
                    {
                        newInterval.end = intervals[i+1].end;
                    }
                }
                res.Add(newInterval);
            }
            return res;
        }

- asraf November 12, 2017 | Flag Reply
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0
of 0 vote

We can solve this in C# with O(n) complexity

public class Interval
        {
            public int start;
            public int end;
            public Interval()
            {
                start =0;
                end = 0;
            }

            public Interval(int s, int e)
            {
                start = s;
                end = e;
            }
        }

        public IList<Interval> Merge(IList<Interval> intervals)
        {
            IList<Interval> res = new List<Interval>();
            intervals = intervals.OrderBy(x => x.start).ToList();

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

            for (int i = 0; i < intervals.Count; i++)
            {
                var newInterval = new Interval(intervals[i].start, intervals[i].end);
                while(i< intervals.Count -1 && newInterval.end >= intervals[i+1].start)
                {
                    if(intervals[i+1].end > newInterval.end)
                    {
                        newInterval.end = intervals[i+1].end;
                    }
                }
                res.Add(newInterval);
            }
            return res;

}

- asraf November 12, 2017 | Flag Reply


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