Java Exercises: Merge all overlapping Intervals from a given a collection of intervals
Java Basic: Exercise-140 with Solution
Write a Java program to merge all overlapping Intervals from a given a collection of intervals.
Sample Solution:
Java Code:
#// Source: https://bit.ly/2PjMfds
import java.util.*;
public class Solution
{
public static void main (String[] args) throws java.lang.Exception
{
ArrayList<Interval> x = new ArrayList<>();
x.add(new Interval(1, 3));
x.add(new Interval(2, 6));
x.add(new Interval(8, 10));
x.add(new Interval(15, 18));
x.add(new Interval(17, 20));
x = merge(x);
for(Interval i : x)
{
System.out.println(i.getStart() + " " + i.getEnd());
}
}
public static ArrayList<Interval> merge(ArrayList<Interval> intervals) {
if(intervals.size() == 0 || intervals.size() == 1)
return intervals;
Collections.sort(intervals, new IntervalComparator());
Interval first = intervals.get(0);
int start = first.getStart();
int end = first.getEnd();
ArrayList<Interval> result = new ArrayList<Interval>();
for (int i = 1; i < intervals.size(); i++) {
Interval current = intervals.get(i);
if (current.getStart() <= end) {
end = Math.max(current.getEnd(), end);
} else {
result.add(new Interval(start, end));
start = current.getStart();
end = current.getEnd();
}
}
result.add(new Interval(start, end));
return result;
}
}
class Interval
{
private int start;
private int end;
Interval() {
start = 0;
end = 0;
}
Interval(int s, int e)
{
start = s;
end = e;
}
public int getStart() {
return start;
}
public int getEnd() {
return end;
}
}
class IntervalComparator implements Comparator<Interval>
{
public int compare(Interval i1, Interval i2)
{
return i1.getStart() - i2.getStart();
}
}
Sample Output:
1 6 8 10 15 20
Flowchart:

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Java: Tips of the Day
countOccurrences
Counts the occurrences of a value in an array.
Use Arrays.stream().filter().count() to count total number of values that equals the specified value.
public static long countOccurrences(int[] numbers, int value) { return Arrays.stream(numbers) .filter(number -> number == value) .count(); }
Ref: https://bit.ly/3kCAgLb
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