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23-N-way-merge

Sep 22, 202418 min read

Case 1: each arrays has the same length

Given K sorted arrays of size N each, merge them and print the sorted output.

Input: K = 3, N = 4, 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 
Explanation: The output array is a sorted array that contains all the elements of the input matrix. 

Input: k = 4, n = 4, arr = { {1, 5, 6, 8}, {2, 4, 10, 12}, {3, 7, 9, 11}, {13, 14, 15, 16}} 
Output: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 
Explanation: The output array is a sorted array that contains all the elements of the input matrix.

Approach 1: naive / BF

Create an output array of size (N * K) and then copy all the elements into the output array followed by sorting:

  • Create an output array of size N * K. 
  • Traverse the matrix from start to end and insert all the elements in the output array.
  • Sort and print the output array.

Implementation

// C++ program to merge K sorted arrays of size n each. 
 
#include <bits/stdc++.h> 
using namespace std; 
#define N 4 
 
// A utility function to print array elements 
void printArray(int arr[], int size) 
{ 
	for (int i = 0; i < size; i++) 
		cout << arr[i] << " "; 
} 
 
// This function takes an array of arrays as an argument and 
// All arrays are assumed to be sorted. It merges them 
// together and prints the final sorted output. 
void mergeKArrays(int arr[][N], int a, int output[]) 
{ 
	int c = 0; 
 
	// traverse the matrix 
	for (int i = 0; i < a; i++) { 
		for (int j = 0; j < N; j++) 
			output = arr[i][j]; 
	} 
 
	// sort the array 
	sort(output, output + N * a); 
} 
 
// Driver's code 
int main() 
{ 
	// Change N at the top to change number of elements 
	// in an array 
	int arr[][N] = { { 2, 6, 12, 34 }, 
					{ 1, 9, 20, 1000 }, 
					{ 23, 34, 90, 2000 } }; 
	int K = sizeof(arr) / sizeof(arr[0]); 
 
	int output[N * K]; 
 
	// Function call 
	mergeKArrays(arr, 3, output); 
 
	cout << "Merged array is " << endl; 
	printArray(output, N * K); 
 
	return 0; 
}
 
# Python3 program to merge k sorted arrays of size n each. 
 
# This function takes an array of arrays as an argument 
# and 
# All arrays are assumed to be sorted. It merges them 
# together and prints the final sorted output. 
 
 
def mergeKArrays(arr, a, output): 
	c = 0
 
	# traverse the matrix 
	for i in range(a): 
		for j in range(4): 
			output = arr[i][j] 
			c += 1
 
	# sort the array 
	output.sort() 
 
# A utility function to print array elements 
 
 
def printArray(arr, size): 
	for i in range(size): 
		print(arr[i], end=" ") 
 
 
# Driver's code 
if __name__ == '__main__': 
	arr = [[2, 6, 12, 34], [1, 9, 20, 1000], [23, 34, 90, 2000]] 
	K = 4
	N = 3
	output = [0 for i in range(N * K)] 
 
	# Function call 
	mergeKArrays(arr, N, output) 
 
	print("Merged array is ") 
	printArray(output, N * K) 
 
# This code is contributed by umadevi9616

Complexity

  • Time Complexity: O(N * K * log (N*K)), Since the resulting array is of size N * K.
  • Space Complexity: O(N * K), The output array is of size N * K.

Approach 2: devide and conquer

The process begins with merging arrays into groups of two. After the first merge, there will be K/2 arrays remaining. Again merge arrays in groups, now K/4 arrays will be remaining. This is similar to merge sort. Divide K arrays into two halves containing an equal number of arrays until there are two arrays in a group. This is followed by merging the arrays in a bottom-up manner.

Follow the given steps to solve the problem:

  • Create a recursive function that takes K arrays and returns the output array.
  • In the recursive function, if the value of K is 1 then return the array else if the value of K is 2 then merge the two arrays in linear time and return the array.
  • If the value of K is greater than 2 then divide the group of k elements into two equal halves and recursively call the function, i.e 0 to K/2 array in one recursive function and K/2 to K array in another recursive function.
  • Print the output array.

Implementation

// C++ program to merge K sorted arrays of size n each. 
 
#include <bits/stdc++.h> 
using namespace std; 
#define N 4 
 
// Merge arr1[0..N1-1] and arr2[0..N2-1] into 
// arr3[0..N1+N2-1] 
void mergeArrays(int arr1[], int arr2[], int N1, int N2, 
				int arr3[]) 
{ 
	int i = 0, j = 0, k = 0; 
 
	// Traverse both array 
	while (i < N1 && j < N2) { 
		// Check if current element of first 
		// array is smaller than current element 
		// of second array. If yes, store first 
		// array element and increment first array 
		// index. Otherwise do same with second array 
		if (arr1[i] < arr2[j]) 
			arr3[k++] = arr1[i++]; 
		else
			arr3[k++] = arr2[j++]; 
	} 
 
	// Store remaining elements of first array 
	while (i < N1) 
		arr3[k++] = arr1[i++]; 
 
	// Store remaining elements of second array 
	while (j < N2) 
		arr3[k++] = arr2[j++]; 
} 
 
// A utility function to print array elements 
void printArray(int arr[], int size) 
{ 
	for (int i = 0; i < size; i++) 
		cout << arr[i] << " "; 
} 
 
// This function takes an array of arrays as an argument and 
// All arrays are assumed to be sorted. It merges them 
// together and prints the final sorted output. 
void mergeKArrays(int arr[][N], int i, int j, int output[]) 
{ 
	// If one array is in range 
	if (i == j) { 
		for (int p = 0; p < N; p++) 
			output[p] = arr[i][p]; 
		return; 
	} 
 
	// if only two arrays are left them merge them 
	if (j - i == 1) { 
		mergeArrays(arr[i], arr[j], N, N, output); 
		return; 
	} 
 
	// Output arrays 
	int out1[N * (((i + j) / 2) - i + 1)], 
		out2[N * (j - ((i + j) / 2))]; 
 
	// Divide the array into halves 
	mergeKArrays(arr, i, (i + j) / 2, out1); 
	mergeKArrays(arr, (i + j) / 2 + 1, j, out2); 
 
	// Merge the output array 
	mergeArrays(out1, out2, N * (((i + j) / 2) - i + 1), 
				N * (j - ((i + j) / 2)), output); 
} 
 
// Driver's code 
int main() 
{ 
	// Change N at the top to change number of elements 
	// in an array 
	int arr[][N] = { { 2, 6, 12, 34 }, 
					{ 1, 9, 20, 1000 }, 
					{ 23, 34, 90, 2000 } }; 
	int K = sizeof(arr) / sizeof(arr[0]); 
	int output[N * K]; 
	mergeKArrays(arr, 0, 2, output); 
 
	// Function call 
	cout << "Merged array is " << endl; 
	printArray(output, N * K); 
 
	return 0; 
}

Complexity

  • Time Complexity: O(N * K * log K).  There are log K levels as in each level the K arrays are divided in half and at each level, the K arrays are traversed.
  • Auxiliary Space: O(N * K * log K). In each level O(N * K) space is required.

Approach 3: Min-heap

The idea is to use Min-Heap. This MinHeap based solution has the same time complexity which is O(N*K*logK). But for a different and particular sized array, this solution works much better. The process must start with creating a MinHeap and inserting the first element of all the k arrays. Remove the root element of Minheap and put it in the output array and insert the next element from the array of removed element. To get the result the step must continue until there is no element left in the MinHeap.

Follow the given steps to solve the problem:

  • Create a min Heap and insert the first element of all the K arrays.
  • Run a loop until the size of MinHeap is greater than zero.
    • Remove the top element of the MinHeap and print the element.
    • Now insert the next element from the same array in which the removed element belonged.
    • If the array doesn’t have any more elements, then replace root with infinite. After replacing the root, heapify the tree.
  • Return the output array

Implementation

// C++ program to merge K sorted 
// arrays of size N each. 
#include <bits/stdc++.h> 
using namespace std; 
 
#define N 4 
 
// A min-heap node 
struct MinHeapNode { 
	// The element to be stored 
	int element; 
 
	// index of the array from which the element is taken 
	int i; 
 
	// index of the next element to be picked from the array 
	int j; 
}; 
 
// Prototype of a utility function to swap two min-heap 
// nodes 
void swap(MinHeapNode* x, MinHeapNode* y); 
 
// A class for Min Heap 
class MinHeap { 
 
	// pointer to array of elements in heap 
	MinHeapNode* harr; 
 
	// size of min heap 
	int heap_size; 
 
public: 
	// Constructor: creates a min heap of given size 
	MinHeap(MinHeapNode a[], int size); 
 
	// to heapify a subtree with root at given index 
	void MinHeapify(int); 
 
	// to get index of left child of node at index i 
	int left(int i) { return (2 * i + 1); } 
 
	// to get index of right child of node at index i 
	int right(int i) { return (2 * i + 2); } 
 
	// to get the root 
	MinHeapNode getMin() { return harr[0]; } 
 
	// to replace root with new node x and heapify() new 
	// root 
	void replaceMin(MinHeapNode x) 
	{ 
		harr[0] = x; 
		MinHeapify(0); 
	} 
}; 
 
// This function takes an array of arrays as an argument and 
// All arrays are assumed to be sorted. It merges them 
// together and prints the final sorted output. 
int* mergeKArrays(int arr[][N], int K) 
{ 
 
	// To store output array 
	int* output = new int[N * K]; 
 
	// Create a min heap with k heap nodes. 
	// Every heap node has first element of an array 
	MinHeapNode* harr = new MinHeapNode[K]; 
	for (int i = 0; i < K; i++) { 
 
		// Store the first element 
		harr[i].element = arr[i][0]; 
 
		// index of array 
		harr[i].i = i; 
 
		// Index of next element to be stored from the array 
		harr[i].j = 1; 
	} 
 
	// Create the heap 
	MinHeap hp(harr, K); 
 
	// Now one by one get the minimum element from min 
	// heap and replace it with next element of its array 
	for (int count = 0; count < N * K; count++) { 
		// Get the minimum element and store it in output 
		MinHeapNode root = hp.getMin(); 
		output[count] = root.element; 
 
		// Find the next element that will replace current 
		// root of heap. The next element belongs to same 
		// array as the current root. 
		if (root.j < N) { 
			root.element = arr[root.i][root.j]; 
			root.j += 1; 
		} 
		// If root was the last element of its array 
		// INT_MAX is for infinite 
		else
			root.element = INT_MAX; 
 
		// Replace root with next element of array 
		hp.replaceMin(root); 
	} 
 
	return output; 
} 
 
// FOLLOWING ARE IMPLEMENTATIONS OF 
// STANDARD MIN HEAP METHODS FROM CORMEN BOOK 
// Constructor: Builds a heap from a given 
// array a[] of given size 
MinHeap::MinHeap(MinHeapNode a[], int size) 
{ 
	heap_size = size; 
	harr = a; // store address of array 
	int i = (heap_size - 1) / 2; 
	while (i >= 0) { 
		MinHeapify(i); 
		i--; 
	} 
} 
 
// A recursive method to heapify a 
// subtree with root at given index. 
// This method assumes that the subtrees 
// are already heapified 
void MinHeap::MinHeapify(int i) 
{ 
	int l = left(i); 
	int r = right(i); 
	int smallest = i; 
 
	if (l < heap_size && harr[l].element < harr[i].element) 
		smallest = l; 
 
	if (r < heap_size 
		&& harr[r].element < harr[smallest].element) 
		smallest = r; 
 
	if (smallest != i) { 
		swap(&harr[i], &harr[smallest]); 
		MinHeapify(smallest); 
	} 
} 
 
// A utility function to swap two elements 
void swap(MinHeapNode* x, MinHeapNode* y) 
{ 
	MinHeapNode temp = *x; 
	*x = *y; 
	*y = temp; 
} 
 
// A utility function to print array elements 
void printArray(int arr[], int size) 
{ 
	for (int i = 0; i < size; i++) 
		cout << arr[i] << " "; 
} 
 
// Driver's code 
int main() 
{ 
	// Change N at the top to change number of elements 
	// in an array 
	int arr[][N] = { { 2, 6, 12, 34 }, 
					{ 1, 9, 20, 1000 }, 
					{ 23, 34, 90, 2000 } }; 
	int K = sizeof(arr) / sizeof(arr[0]); 
 
	// Function call 
	int* output = mergeKArrays(arr, K); 
 
	cout << "Merged array is " << endl; 
	printArray(output, N * K); 
 
	return 0; 
}
 
import sys 
 
# Python 3 program to merge K sorted 
# arrays of size N each. 
# A Min heap node 
class MinHeapNode : 
	element = 0
	
	# The element to be stored 
	# index of the array from 
	# which the element is taken 
	i = 0
	
	# index of the next element 
	# to be picked from array 
	j = 0
	def __init__(self, element, i, j) : 
		self.element = element 
		self.i = i 
		self.j = j 
		
# A class for Min Heap 
class MinHeap : 
	harr = None
	
	# Array of elements in heap 
	heap_size = 0
	
	# Current number of elements in min heap 
	# Constructor: Builds a heap from 
	# a given array a[] of given size 
	def __init__(self, a, size) : 
		self.heap_size = size 
		self.harr = a 
		i = int((self.heap_size - 1) / 2) 
		while (i >= 0) : 
			self.MinHeapify(i) 
			i -= 1
			
	# A recursive method to heapify a subtree 
	# with the root at given index This method 
	# assumes that the subtrees are already heapified 
	def MinHeapify(self, i) : 
		l = self.left(i) 
		r = self.right(i) 
		smallest = i 
		if (l < self.heap_size and self.harr[l].element < self.harr[i].element) : 
			smallest = l 
		if (r < self.heap_size and self.harr[r].element < self.harr[smallest].element) : 
			smallest = r 
		if (smallest != i) : 
			self.swap(self.harr, i, smallest) 
			self.MinHeapify(smallest) 
			
	# to get index of left child of node at index i 
	def left(self, i) : 
		return (2 * i + 1) 
		
	# to get index of right child of node at index i 
	def right(self, i) : 
		return (2 * i + 2) 
		
	# to get the root 
	def getMin(self) : 
		if (self.heap_size <= 0) : 
			print("Heap underflow") 
			return None
		return self.harr[0] 
		
	# to replace root with new node 
	# "root" and heapify() new root 
	def replaceMin(self, root) : 
		self.harr[0] = root 
		self.MinHeapify(0) 
		
	# A utility function to swap two min heap nodes 
	def swap(self, arr, i, j) : 
		temp = arr[i] 
		arr[i] = arr[j] 
		arr[j] = temp 
		
	# A utility function to print array elements 
	@staticmethod
	def printArray( arr) : 
		for i in arr : 
			print(str(i) + " ", end ="") 
		print() 
		
	# This function takes an array of 
	# arrays as an argument and All 
	# arrays are assumed to be sorted. 
	# It merges them together and 
	# prints the final sorted output. 
	@staticmethod
	def mergeKSortedArrays( arr, K) : 
		hArr = [None] * (K) 
		resultSize = 0
		i = 0
		while (i < len(arr)) : 
			node = MinHeapNode(arr[i][0], i, 1) 
			hArr[i] = node 
			resultSize += len(arr[i]) 
			i += 1
			
		# Create a min heap with k heap nodes. Every heap 
		# node has first element of an array 
		mh = MinHeap(hArr, K) 
		result = [0] * (resultSize) 
		
		# To store output array 
		# Now one by one get the minimum element from min 
		# heap and replace it with next element of its 
		# array 
		i = 0
		while (i < resultSize) : 
			
			# Get the minimum element and store it in 
			# result 
			root = mh.getMin() 
			result[i] = root.element 
			
			# Find the next element that will replace 
			# current root of heap. The next element 
			# belongs to same array as the current root. 
			if (root.j < len(arr[root.i])) : 
				root.element = arr[root.i][root.j] 
				root.j += 1
			else : 
				root.element = sys.maxsize 
				
			# Replace root with next element of array 
			mh.replaceMin(root) 
			i += 1
		MinHeap.printArray(result) 
		
	# Driver's code 
	@staticmethod
	def main( args) : 
		arr = [[2, 6, 12, 34], [1, 9, 20, 1000], [23, 34, 90, 2000]] 
		print("Merged array is :") 
		# Function call 
		MinHeap.mergeKSortedArrays(arr, len(arr)) 
	
if __name__=="__main__": 
	MinHeap.main([]) 
	
	# This code is contributed by aadityaburujwale.

Complexity

  • Time Complexity: O(N * K * log K), Insertion and deletion in a Min Heap requires log K time.
  • Auxiliary Space: O(K), If Output is not stored then the only space required is the Min-Heap of K elements.

Case 2: each arrays has random length

Given k sorted arrays of possibly different sizes, merge them and print the sorted output.

Examples: 

Input: k = 3   
      arr[][] = { {1, 3},  
                  {2, 4, 6},  
                  {0, 9, 10, 11}} ;  
Output: 0 1 2 3 4 6 9 10 11   
Input: k = 2  
      arr[][] = { {1, 3, 20},  
                  {2, 4, 6}} ;  
Output: 1 2 3 4 6 20

We have discussed a solution that works for all arrays of the same size in Merge k sorted arrays | Set 1.
A simple solution is to create an output array and one by one copy all k arrays to it. Finally, sort the output array. This approach takes O(N Log N) time where N is the count of all elements.

An efficient solution is to use a heap data structure. The time complexity of the heap-based solution is O(N Log k).

  1. Create an output array. 
  2. Create a min-heap of size k and insert 1st element in all the arrays into the heap 
  3. Repeat the following steps while the priority queue is not empty. 
    1. Remove the minimum element from the heap (minimum is always at the root) and store it in the output array. 
    2. Insert the next element from the array from which the element is extracted. If the array doesn’t have any more elements, then do nothing.

Below is the implementation of the above approach:

Implementation

// C++ program to merge k sorted arrays
// of size n each.
#include <bits/stdc++.h>
using namespace std;
 
// A pair of pairs, first element is going to
// store value, second element index of array
// and third element index in the array.
typedef pair<int, pair<int, int> > ppi;
 
// This function takes an array of arrays as an
// argument and all arrays are assumed to be
// sorted. It merges them together and prints
// the final sorted output.
vector<int> mergeKArrays(vector<vector<int> > arr)
{
	vector<int> output;
 
	// Create a min heap with k heap nodes. Every
	// heap node has first element of an array
	priority_queue<ppi, vector<ppi>, greater<ppi> > pq;
 
	for (int i = 0; i < arr.size(); i++)
		pq.push({ arr[i][0], { i, 0 } });
 
	// Now one by one get the minimum element
	// from min heap and replace it with next
	// element of its array
	while (pq.empty() == false) {
		ppi curr = pq.top();
		pq.pop();
 
		// i ==> Array Number
		// j ==> Index in the array number
		int i = curr.second.first;
		int j = curr.second.second;
 
		output.push_back(curr.first);
 
		// The next element belongs to same array as
		// current.
		if (j + 1 < arr[i].size())
			pq.push({ arr[i][j + 1], { i, j + 1 } });
	}
 
	return output;
}
 
// Driver program to test above functions
int main()
{
	// Change n at the top to change number
	// of elements in an array
	vector<vector<int> > arr{ { 2, 6, 12 },
							{ 1, 9 },
							{ 23, 34, 90, 2000 } };
 
	vector<int> output = mergeKArrays(arr);
 
	cout << "Merged array is " << endl;
	for (auto x : output)
		cout << x << " ";
 
	return 0;
}
 
# Python code to implement the approach
 
import heapq
 
# Merge function merge two arrays
# of different or same length
# if n = max(n1, n2)
# time complexity of merge is (o(n log(n)))
 
# Function for meging k arrays
def mergeK(arr, k):
	res = []
	
	# Declaring min heap
	h = []
	
	# Inserting the first elements of each row
	for i in range(len(arr)):
		heapq.heappush(h, (arr[i][0], i, 0))
		
	# Loop to merge all the arrays
	while h:
	
		# ap stores the row number, 
		# vp stores the column number
		val, ap, vp = heapq.heappop(h)
		res.append(val)
		if vp+1 < len(arr[ap]):
			heapq.heappush(h, (arr[ap][vp+1], ap, vp+1))
	return res	 
		
 
# Driver code
if __name__ == '__main__':
	arr =[[2, 6, 12 ],
		[ 1, 9 ],
		[23, 34, 90, 2000 ]]
	k = 3
	l = mergeK(arr, k)
	print(*l)

Complexity

  • Time Complexity: O(N*K*logK)
  • Auxiliary Space: O(N)

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