Merge K Sorted Arrays in Java

This is a classic interview question. Another similar problem is "merge k sorted lists".

This problem can be solved by using a heap. The time complexity is O(nlog(k)), where n is the total number of elements and k is the number of arrays.

It takes O(log(k)) to insert an element to the heap and it takes O(log(k)) to delete the minimum element.

class ArrayContainer implements Comparable<ArrayContainer> {
	int[] arr;
	int index;
 
	public ArrayContainer(int[] arr, int index) {
		this.arr = arr;
		this.index = index;
	}
 
	@Override
	public int compareTo(ArrayContainer o) {
		return this.arr[this.index] - o.arr[o.index];
	}
}

public class KSortedArray {
	public static int[] mergeKSortedArray(int[][] arr) {
		//PriorityQueue is heap in Java 
		PriorityQueue<ArrayContainer> queue = new PriorityQueue<ArrayContainer>();
		int total=0;
 
		//add arrays to heap
		for (int i = 0; i < arr.length; i++) {
			queue.add(new ArrayContainer(arr[i], 0));
			total = total + arr[i].length;
		}
 
		int m=0;
		int result[] = new int[total];
 
		//while heap is not empty
		while(!queue.isEmpty()){
			ArrayContainer ac = queue.poll();
			result[m++]=ac.arr[ac.index];
 
			if(ac.index < ac.arr.length-1){
				queue.add(new ArrayContainer(ac.arr, ac.index+1));
			}
		}
 
		return result;
	}
 
	public static void main(String[] args) {
		int[] arr1 = { 1, 3, 5, 7 };
		int[] arr2 = { 2, 4, 6, 8 };
		int[] arr3 = { 0, 9, 10, 11 };
 
		int[] result = mergeKSortedArray(new int[][] { arr1, arr2, arr3 });
		System.out.println(Arrays.toString(result));
	}
}

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Shital Devalkar

Java program for Merge n number of sorted arrays.

public class MergeSotedArrays {
public static void main(String[] args) {
int[] a = { 1, 4, 5 };
int[] b = { 2, 3, 6, 7 };
int[] c = { 1, 8, 2, 10 };
int[] d = { 3, 11, 50, 60 };
int[] result = merge(d,a,c,b);

for (int idx = 0; idx < result.length; idx++) {
System.out.print(result[idx] + " ");
}
}

public static int[] merge(int[] …inputArrays) {
int[] finalResult = inputArrays[0];
for (int arrayno = 1; arrayno < inputArrays.length; arrayno++) {
int tempResult[] = new int[finalResult.length + inputArrays[arrayno].length];
int resultIndex = 0;
for (int j = 0, k = 0; (j < finalResult.length || k < inputArrays[arrayno].length);) {
if (j < finalResult.length && k < inputArrays[arrayno].length)
tempResult[resultIndex++] = (finalResult[j] <= inputArrays[arrayno][k]) ? finalResult[j++] : inputArrays[arrayno][k++];
else
tempResult[resultIndex++] = (j < finalResult.length) ? finalResult[j++] : inputArrays[arrayno][k++];
}
finalResult = tempResult;
}
return finalResult;
}
}
Bonsai

You are right.
p.andrey

I asked my self the same question… in my opinion is O(n*log(k)) = O(n)… unless there is something I do not understand. So, without going into details (according to the PriorityQueue documentation), each poll is O(log(k)), each add is O(log(k)), these are called n times, where n is the sum of the lengths of all array. In conclusion I think is O(n*log(k)) = O(n). Am I missing something?

I think there is mistake here, because if you look to: https://www.programcreek.com/2013/02/leetcode-merge-k-sorted-lists-java/ is O(n), which is correct.
rambabu kokkiligadda

Can you please explain how time complexity is O(n log n)? I assume space complexity is O(1).
Gaurang Bhatt

Removing a root from a PriorityQueue is log(n) because it has to move the rest of the elements to acount for the empty spot.
Hayro

Oh, sorry. The algo should be removing only the root (min element) each time, which is O(1)
Hayro

it is the time to remove it from the array
Sheetal

This method doesn’t work if all array lengths are not equal.
Kurt Maegerle

The problem without ArrayContainer is that, as you described, after removing from the heap you must add the “next element from the exact array which we had root from.” Knowing what the next element in the array is or what array an element came from gets pretty ugly without something like an ArrayContainer class.

You could instead enqueue an object that has a value, an index indicated what array you are talking about, and an index indicating the current index of this array. Alternatively, it could be done with maps, but it’s even worse.

Ultimately, I think ArrayContainer is the way to do this that makes the most sense.
Ankit Shah

//you can rewrite the compareTo method like this

public int compareTo(ArrayHolder o) {
return this.arr[this.index] – o.arr[o.index];
}
srrm

from the javadoc for PriorityQueue – http://docs.oracle.com/javase/7/docs/api/java/util/PriorityQueue.html

“Implementation note: this implementation provides O(log(n)) time for the enqueing and dequeing methods (offer, poll, remove() and add);”
srrm

isn’t it O(log(n)) to remove from a heap (i.e. PriorityQueue)?
Holden

For example, I know we should put first elements of all arrays into the heap first, heapify it, then remove the minimum (which is root), and then we should put next element from the exact array which we had root from, to heap again, and again heapify and do this process till the end. But I cannot follow this procedure in here:

while(!queue.isEmpty()){
ArrayContainer ac = queue.poll();
result[m++]=ac.arr[ac.index];

if(ac.index < ac.arr.length-1){
queue.add(new ArrayContainer(ac.arr, ac.index+1));
}
}
Holden

Thank you for your great post. Could you please post a simpler solution without “ArrayContainer class”? It made it difficult to follow the solution for me. Thank you

Merge K Sorted Arrays in Java

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