Merge K Sorted Arrays Problem Statement
Given k arrays sorted in the same order - either non-increasing or non-decreasing - merge them all into one array sorted in the same order.
Example
{
"arr": [
[1, 3, 5, 7],
[2, 4, 6, 8],
[0, 9, 10, 11]
]
}
Output:
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11]
Notes
Constraints:
- 1 <=
k <= 500
- 2 <= size of an array <= 500
- All
k arrays are of the same size
- -106 <= values in arrays <= 106
We have provided one solution for this problem.
Merge K Sorted Arrays Solution: Optimal
First step is to check the direction of sorting in the input data. Let us solve it for increasingly sorted input.
A naive approach would be to add all elements to one collection and then sort them out. We can build on our solution following the idea of the naive solution. At any given point of time, the smallest element would be from the pool of candidate smallest elements formed by adding the elements at start of all arrays. When we remove the smallest element from the pool, we will add the next element from that array.
We would use a min priority queue to carry out these operations. For input sorted in the decreasing order - a max priority queue.
Time Complexity
O(n * log(k)) where n is the total number of elements in all k arrays.
Auxiliary Space Used
O(k).
Space Complexity
O(k + n) where n is the total number of elements in all k arrays.
Code For Merge K Sorted Arrays Solution: Optimal
/*
* Asymptotic complexity in terms of total number of elements in all arrays \`n\`
* and number of arrays \`k\`:
* Time: O(n * log(k)).
* Auxiliary space: O(k).
* Total space: O(n + k).
*/
static ArrayList<Integer> merge_arrays(ArrayList<ArrayList<Integer>> arr) {
int k = arr.size();
int N = arr.get(0).size();
// Get appropriate priority queue
PriorityQueue<Node> priorityQueue = getPriorityQueue(arr);
for (int i = 0; i < k; i++) {
priorityQueue.add(new Node(arr.get(i).get(0), i, 0));
}
ArrayList<Integer> ans = new ArrayList<Integer>();
while (ans.size() < N * k) {
Node rem = priorityQueue.poll();
ans.add(rem.value);
// Add the next element from the same row from which element is removed, if available
if (rem.column + 1 < N) {
priorityQueue.add(new Node(arr.get(rem.row).get(rem.column + 1), rem.row,
rem.column + 1));
}
}
return ans;
}
static private PriorityQueue<Node> getPriorityQueue(ArrayList<ArrayList<Integer>> arr) {
boolean isIncreasing = false, isDecreasing = false;
// We will check if the input is sorted in increasing manner or
// decreasing manner
for (int i = 0; i < arr.size(); i++) {
if (arr.get(i).get(0) < arr.get(i).get(arr.get(i).size() - 1)) {
isIncreasing = true;
}
if (arr.get(i).get(0) > arr.get(i).get(arr.get(i).size() - 1)) {
isDecreasing = true;
}
}
if (isIncreasing) {
// Make a min priority queue
return new PriorityQueue<>();
}
if (isDecreasing) {
// Make a max priority queue
return new PriorityQueue<>(arr.size(), Collections.reverseOrder());
}
// Some error in test case; Code should never reach here;
return null;
}
static class Node implements Comparable<Node> {
int value;
int row;
int column;
Node(int xx, int yy, int zz) {
value = xx;
row = yy;
column = zz;
}
@Override
public int compareTo(Node o) {
return Long.compare(this.value, o.value);
}
}
We hope that these solutions to merge k sorted arrays problem have helped you level up your coding skills. You can expect problems like these at top tech companies like Amazon and Google.
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