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MergeOverlappingIntervals.java
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MergeOverlappingIntervals.java
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package arrays;
import java.util.ArrayList;
import java.util.List;
public class MergeOverlappingIntervals {
/*
* Given an array of intervals as input where each interval has a start and end timestamps.
* Input array is sorted by starting timestamps.
* Merge overlapping intervals and return output array (list).
* Eg:
* Input: Intervals [(1, 5), (3, 7), (4, 6), (6, 8)]
* Output: [(1, 8)]
*
* */
/*
* Solution: Linear Scan
* Runtime Complexity - Linear, O(n)
* Memory Complexity - Constant O(1)
* */
/*
* For each interval in the input list,
* Step 1: check if the current element value < the output interval y value. If it is, then it is overlapping
* Step 2: Update last interval of output list with merged interval i.e. the max value between y1 & y2
* Step 3: If it is not overlapping, add the current input interval element to the output list
* */
public static List<Tuple<Integer, Integer>> mergeIntervals(List<Tuple<Integer, Integer>> v1) {
List<Tuple<Integer, Integer>> v2 = new ArrayList<>();
v2.add(new Tuple<>(v1.get(0).x, v1.get(0).y));
for(int i=1; i<v1.size(); i++) {
int x1 = v1.get(i).x;
int y1 = v1.get(i).y;
int x2 = v2.get(v2.size()-1).getX();
int y2 = v2.get(v2.size()-1).getY();
if(y2 >= x1) {
v2.get(v2.size()-1).setY(Math.max(y1,y2));
} else {
v2.add(new Tuple<>(x1,y1));
}
}
return v2;
}
public static void main(String[] args) {
List<Tuple<Integer, Integer>> v1 = new ArrayList<>();
v1.add(new Tuple(10, 12));
v1.add(new Tuple(11, 15));
// v1.add(new Tuple(4, 6));
// v1.add(new Tuple(6, 8));
List<Tuple<Integer, Integer>> v2 = mergeIntervals(v1);
for(Tuple tuple: v2) {
System.out.println(tuple.x + " " + tuple.y);
}
}
public static class Tuple<X, Y> {
private X x;
private Y y;
public Tuple(X x, Y y) {
this.x = x;
this.y = y;
}
public X getX() {
return x;
}
public void setX(X x) {
this.x = x;
}
public Y getY() {
return y;
}
public void setY(Y y) {
this.y = y;
}
}
}