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MergeIntervals.java
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package com.leetcode;
import java.util.ArrayList;
import java.util.List;
import java.util.PriorityQueue;
public class MergeIntervals {
//https://leetcode.com/problems/merge-intervals/
/*
Instead of keeping in PQ sort the array in place.
*/
class SolutionMemOptimized {
/*
case 1 :
|----------|
|-----------|
case 2 :
|-----------|
|----|
case 3 :
|-----------|
|-------------|
case 4 :
|-----------|
|----|
case 5 :
|-----------|
|----|
*/
public int[][] merge(int[][] intervals) {
int n = intervals.length;
if(n <= 1) return intervals;
Arrays.sort(intervals , (a,b) -> (a[0]-b[0]));//ascend by start
//PriorityQueue<int[]> minHeap = new PriorityQueue<>((a,b) -> (a[0]-b[0]));
int[] prev = intervals[0];
List<int[]> resultList = new ArrayList<>();
for(int i = 1 ; i < n ; i++){
int[] next = intervals[i];
//merge
if(next[0] <= prev[1]){
prev[0] = Math.min(prev[0],next[0]);
prev[1] = Math.max(prev[1],next[1]);
}else{
resultList.add(prev);
prev = next;
}
}
resultList.add(prev); // To remember !!!
int[][] result = new int[resultList.size()][2];
for(int i = 0 ; i < resultList.size() ; i++){
result[i] = resultList.get(i);
}
return result;
}
}
class Solution {
public int[][] merge(int[][] intervals) {
if(intervals == null || intervals.length <= 1) return intervals;
PriorityQueue<int[]> minHeap = new PriorityQueue<>((a, b) -> (a[0] - b[0]));//compare and sort by start time
for(int[]interval : intervals){
minHeap.offer(interval);
}
List<int[]> merged = new ArrayList<>();
int[] prev = minHeap.poll();
while(minHeap.size() > 0){
int[] interval = minHeap.poll();
if(interval[0] <= prev[1]){ // condition 1,3,4 is merge needed
// change to prev[0] is redundant as prev[0] <= interval[0]
prev[1] = Math.max(prev[1] , interval[1]);
}else{
merged.add(prev);
prev = interval;
}
}
merged.add(prev); // To remember !!
int[][] result = new int[merged.size()][2];
for(int i = 0 ; i < merged.size() ; i++){
result[i] = merged.get(i);
}
return result;
}
}
/*
After adding all in minHeap with startTime sorted
Condition 1 :
|------------|
|------------|
Condition 2
|------------|
|------------|
Condition 3
|------------|
|------------------|
Condition 4
|------------|
|--------|
*/
//Alternate iterative
class Solution2 {
public int[][] merge(int[][] intervals) {
if(intervals == null || intervals.length == 0 || intervals[0].length <= 1){
return intervals;
}
PriorityQueue<int[]> minHeap = new PriorityQueue<>((a,b) -> (a[0]-b[0]));
for(int [] i : intervals){
minHeap.offer(i);
}
List<int[]> resultList = new ArrayList<>();
int[] prev = minHeap.poll();
while(minHeap.size() > 0){
int [] next = minHeap.poll();
if(prev[1] >= next[0]){
//if overlap then merge
next[0] = Math.min(prev[0],next[0]);
next[1] = Math.max(prev[1],next[1]);
}else{
resultList.add(prev);
}
prev = next;
}
resultList.add(prev);
int result[][] = new int [resultList.size()][2];
for(int i = 0 ; i <resultList.size() ; i++){
result[i] = resultList.get(i);
}
return result;
}
}
}