Merge K sorted linked lists | Set 1
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Given K sorted linked lists of size N each, the task is to merge them all maintaining their sorted order.
Examples:
Input: K = 3, N = 4
list1 = 1->3->5->7->NULL
list2 = 2->4->6->8->NULL
list3 = 0->9->10->11->NULL
Output: 0->1->2->3->4->5->6->7->8->9->10->11
Merged lists in a sorted order where every element is greater than the previous element.Input: K = 3, N = 3
list1 = 1->3->7->NULL
list2 = 2->4->8->NULL
list3 = 9->10->11->NULL
Output: 1->2->3->4->7->8->9->10->11
Merged lists in a sorted order where every element is greater than the previous element.
Recommended Practice
Naive Approach:
A Simple Solution is to initialize the result as the first list. Now traverse all lists starting from the second list. Insert every node of the currently traversed list into the result in a sorted way.
Below is the implementation of the above approach:
C++
// C++ program to merge k sorted// arrays of size n each#include <bits/stdc++.h>using namespace std; // A Linked List nodestruct Node { int data; Node* next;}; /* Function to print nodes in a given linked list */void printList(Node* node){ while (node != NULL) { printf("%d ", node->data); node = node->next; }} // The main function that// takes an array of lists// arr[0..last] and generates// the sorted outputNode* mergeKLists(Node* arr[], int last){ // Traverse form second list to last for (int i = 1; i <= last; i++) { while (true) { // head of both the lists, // 0 and ith list. Node *head_0 = arr[0], *head_i = arr[i]; // Break if list ended if (head_i == NULL) break; // Smaller than first element if (head_0->data >= head_i->data) { arr[i] = head_i->next; head_i->next = head_0; arr[0] = head_i; } else // Traverse the first list while (head_0->next != NULL) { // Smaller than next element if (head_0->next->data >= head_i->data) { arr[i] = head_i->next; head_i->next = head_0->next; head_0->next = head_i; break; } // go to next node head_0 = head_0->next; // if last node if (head_0->next == NULL) { arr[i] = head_i->next; head_i->next = NULL; head_0->next = head_i; head_0->next->next = NULL; break; } } } } return arr[0];} // Utility function to create a new node.Node* newNode(int data){ struct Node* temp = new Node; temp->data = data; temp->next = NULL; return temp;} // Driver program to test// above functionsint main(){ // Number of linked lists int k = 3; // Number of elements in each list int n = 4; // an array of pointers storing the // head nodes of the linked lists Node* arr[k]; arr[0] = newNode(1); arr[0]->next = newNode(3); arr[0]->next->next = newNode(5); arr[0]->next->next->next = newNode(7); arr[1] = newNode(2); arr[1]->next = newNode(4); arr[1]->next->next = newNode(6); arr[1]->next->next->next = newNode(8); arr[2] = newNode(0); arr[2]->next = newNode(9); arr[2]->next->next = newNode(10); arr[2]->next->next->next = newNode(11); // Merge all lists Node* head = mergeKLists(arr, k - 1); printList(head); return 0;} |
Java
// Java program to merge k sorted// arrays of size n eachimport java.io.*; // A Linked List nodeclass Node { int data; Node next; // Utility function to create a new node. Node(int key) { data = key; next = null; }}class GFG { static Node head; static Node temp; /* Function to print nodes in a given linked list */ static void printList(Node node) { while (node != null) { System.out.print(node.data + " "); node = node.next; } System.out.println(); } // The main function that // takes an array of lists // arr[0..last] and generates // the sorted output static Node mergeKLists(Node arr[], int last) { // Traverse form second list to last for (int i = 1; i <= last; i++) { while (true) { // head of both the lists, // 0 and ith list. Node head_0 = arr[0]; Node head_i = arr[i]; // Break if list ended if (head_i == null) break; // Smaller than first element if (head_0.data >= head_i.data) { arr[i] = head_i.next; head_i.next = head_0; arr[0] = head_i; } else { // Traverse the first list while (head_0.next != null) { // Smaller than next element if (head_0.next.data >= head_i.data) { arr[i] = head_i.next; head_i.next = head_0.next; head_0.next = head_i; break; } // go to next node head_0 = head_0.next; // if last node if (head_0.next == null) { arr[i] = head_i.next; head_i.next = null; head_0.next = head_i; head_0.next.next = null; break; } } } } } return arr[0]; } // Driver program to test // above functions public static void main(String[] args) { // Number of linked lists int k = 3; // Number of elements in each list int n = 4; // an array of pointers storing the // head nodes of the linked lists Node[] arr = new Node[k]; arr[0] = new Node(1); arr[0].next = new Node(3); arr[0].next.next = new Node(5); arr[0].next.next.next = new Node(7); arr[1] = new Node(2); arr[1].next = new Node(4); arr[1].next.next = new Node(6); arr[1].next.next.next = new Node(8); arr[2] = new Node(0); arr[2].next = new Node(9); arr[2].next.next = new Node(10); arr[2].next.next.next = new Node(11); // Merge all lists head = mergeKLists(arr, k - 1); printList(head); }} // This code is contributed by avanitrachhadiya2155 |
Python3
# Python3 program to merge k# sorted arrays of size n each # A Linked List node class Node: def __init__(self, x): self.data = x self.next = None # Function to print nodes in# a given linked list def printList(node): while (node != None): print(node.data, end=" ") node = node.next # The main function that# takes an array of lists# arr[0..last] and generates# the sorted output def mergeKLists(arr, last): # Traverse form second # list to last for i in range(1, last + 1): while (True): # head of both the lists, # 0 and ith list. head_0 = arr[0] head_i = arr[i] # Break if list ended if (head_i == None): break # Smaller than first # element if (head_0.data >= head_i.data): arr[i] = head_i.next head_i.next = head_0 arr[0] = head_i else: # Traverse the first list while (head_0.next != None): # Smaller than next # element if (head_0.next.data >= head_i.data): arr[i] = head_i.next head_i.next = head_0.next head_0.next = head_i break # go to next node head_0 = head_0.next # if last node if (head_0.next == None): arr[i] = head_i.next head_i.next = None head_0.next = head_i head_0.next.next = None break return arr[0] # Driver codeif __name__ == '__main__': # Number of linked # lists k = 3 # Number of elements # in each list n = 4 # an array of pointers # storing the head nodes # of the linked lists arr = [None for i in range(k)] arr[0] = Node(1) arr[0].next = Node(3) arr[0].next.next = Node(5) arr[0].next.next.next = Node(7) arr[1] = Node(2) arr[1].next = Node(4) arr[1].next.next = Node(6) arr[1].next.next.next = Node(8) arr[2] = Node(0) arr[2].next = Node(9) arr[2].next.next = Node(10) arr[2].next.next.next = Node(11) # Merge all lists head = mergeKLists(arr, k - 1) printList(head) # This code is contributed by Mohit Kumar 29 |
C#
// C# program to merge k sorted// arrays of size n eachusing System; // A Linked List nodepublic class Node { public int data; public Node next; // Utility function to create a new node. public Node(int key) { data = key; next = null; }} public class GFG { static Node head; /* Function to print nodes in a given linked list */ static void printList(Node node) { while (node != null) { Console.Write(node.data + " "); node = node.next; } Console.WriteLine(); } // The main function that // takes an array of lists // arr[0..last] and generates // the sorted output static Node mergeKLists(Node[] arr, int last) { // Traverse form second list to last for (int i = 1; i <= last; i++) { while (true) { // head of both the lists, // 0 and ith list. Node head_0 = arr[0]; Node head_i = arr[i]; // Break if list ended if (head_i == null) break; // Smaller than first element if (head_0.data >= head_i.data) { arr[i] = head_i.next; head_i.next = head_0; arr[0] = head_i; } else { // Traverse the first list while (head_0.next != null) { // Smaller than next element if (head_0.next.data >= head_i.data) { arr[i] = head_i.next; head_i.next = head_0.next; head_0.next = head_i; break; } // go to next node head_0 = head_0.next; // if last node if (head_0.next == null) { arr[i] = head_i.next; head_i.next = null; head_0.next = head_i; head_0.next.next = null; break; } } } } } return arr[0]; } static public void Main() { // Number of linked lists int k = 3; // an array of pointers storing the // head nodes of the linked lists Node[] arr = new Node[k]; arr[0] = new Node(1); arr[0].next = new Node(3); arr[0].next.next = new Node(5); arr[0].next.next.next = new Node(7); arr[1] = new Node(2); arr[1].next = new Node(4); arr[1].next.next = new Node(6); arr[1].next.next.next = new Node(8); arr[2] = new Node(0); arr[2].next = new Node(9); arr[2].next.next = new Node(10); arr[2].next.next.next = new Node(11); // Merge all lists head = mergeKLists(arr, k - 1); printList(head); }} // This code is contributed by rag2127 |
Javascript
<script>// Javascript program to merge k sorted// arrays of size n each // A Linked List node class Node { // Utility function to create a new node. constructor(key) { this.data=key; this.next=null; } } let head; let temp; /* Function to print nodes ina given linked list */function printList(node){ while(node != null) { document.write(node.data + " "); node = node.next; } document.write("<br>"); } // The main function that// takes an array of lists// arr[0..last] and generates// the sorted outputfunction mergeKLists(arr,last){ // Traverse form second list to last for (let i = 1; i <= last; i++) { while(true) { // head of both the lists, // 0 and ith list. let head_0 = arr[0]; let head_i = arr[i]; // Break if list ended if (head_i == null) break; // Smaller than first element if(head_0.data >= head_i.data) { arr[i] = head_i.next; head_i.next = head_0; arr[0] = head_i; } else { // Traverse the first list while (head_0.next != null) { // Smaller than next element if (head_0.next.data >= head_i.data) { arr[i] = head_i.next; head_i.next = head_0.next; head_0.next = head_i; break; } // go to next node head_0 = head_0.next; // if last node if (head_0.next == null) { arr[i] = head_i.next; head_i.next = null; head_0.next = head_i; head_0.next.next = null; break; } } } } } return arr[0];} // Driver program to test// above functions // Number of linked listslet k = 3;// Number of elements in each listlet n = 4;// an array of pointers storing the// head nodes of the linked listslet arr = new Array(k);arr[0] = new Node(1);arr[0].next = new Node(3);arr[0].next.next = new Node(5);arr[0].next.next.next = new Node(7); arr[1] = new Node(2);arr[1].next = new Node(4);arr[1].next.next = new Node(6);arr[1].next.next.next = new Node(8); arr[2] = new Node(0);arr[2].next = new Node(9);arr[2].next.next = new Node(10);arr[2].next.next.next = new Node(11); // Merge all listshead = mergeKLists(arr, k - 1);printList(head); // This code is contributed by unknown2108 </script> |
Output
0 1 2 3 4 5 6 7 8 9 10 11
Time complexity: O(NK-1), Traversing N times on each of the K lists.
Auxiliary Space: O(1).
Merge K sorted linked lists using Min Heap:
This solution is based on the Min Heap approach. The process must start with creating a MinHeap and inserting the first element of all the K Linked Lists. Remove the root element of Minheap and put it in the output Linked List and insert the next element from the Linked List of the removed element. To get the result the step must continue until there is no element left in the MinHeap.
For a more detailed solution and code checkout, this article Merge k sorted linked lists | Set 2 (Using Min Heap).
Time Complexity: O(N*K*LogK)
Auxiliary Space: O(K)
Merge K sorted linked lists using Divide and Conquer:
The idea is to pair up a sorted list after which K/2 list will be left to be merged and repeat this till all the lists gets merged.
Follow the steps below to solve the problem:
- Pair up K lists and merge each pair in linear time using O(N) space.
- After the first cycle, K/2 lists are left each of size 2*N. After the second cycle, K/4 lists are left each of size 4*N and so on.
- Repeat the procedure until we have only one list left.
Below is the implementation of the above idea.
C++
// C++ program to merge k sorted// arrays of size n each#include <bits/stdc++.h>using namespace std; // A Linked List nodestruct Node { int data; Node* next;}; /* Function to print nodes in a given linked list */void printList(Node* node){ while (node != NULL) { printf("%d ", node->data); node = node->next; }} /* Takes two lists sorted in increasing order, and merge their nodes together to make one big sorted list. Below function takes O(n) extra space for recursive calls, */Node* SortedMerge(Node* a, Node* b){ Node* result = NULL; /* Base cases */ if (a == NULL) return (b); else if (b == NULL) return (a); /* Pick either a or b, and recur */ if (a->data <= b->data) { result = a; result->next = SortedMerge(a->next, b); } else { result = b; result->next = SortedMerge(a, b->next); } return result;} // The main function that takes an array of lists// arr[0..last] and generates the sorted outputNode* mergeKLists(Node* arr[], int last){ // repeat until only one list is left while (last != 0) { int i = 0, j = last; // (i, j) forms a pair while (i < j) { // merge List i with List j and store // merged list in List i arr[i] = SortedMerge(arr[i], arr[j]); // consider next pair i++, j--; // If all pairs are merged, update last if (i >= j) last = j; } } return arr[0];} // Utility function to create a new node.Node* newNode(int data){ struct Node* temp = new Node; temp->data = data; temp->next = NULL; return temp;} // Driver program to test above functionsint main(){ int k = 3; // Number of linked lists int n = 4; // Number of elements in each list // an array of pointers storing the head nodes // of the linked lists Node* arr[k]; arr[0] = newNode(1); arr[0]->next = newNode(3); arr[0]->next->next = newNode(5); arr[0]->next->next->next = newNode(7); arr[1] = newNode(2); arr[1]->next = newNode(4); arr[1]->next->next = newNode(6); arr[1]->next->next->next = newNode(8); arr[2] = newNode(0); arr[2]->next = newNode(9); arr[2]->next->next = newNode(10); arr[2]->next->next->next = newNode(11); // Merge all lists Node* head = mergeKLists(arr, k - 1); printList(head); return 0;} |
Java
// Java program to merge k sorted arrays of size n eachpublic class MergeKSortedLists { /* Takes two lists sorted in increasing order, and merge their nodes together to make one big sorted list. Below function takes O(Log n) extra space for recursive calls, but it can be easily modified to work with same time and O(1) extra space */ public static Node SortedMerge(Node a, Node b) { Node result = null; /* Base cases */ if (a == null) return b; else if (b == null) return a; /* Pick either a or b, and recur */ if (a.data <= b.data) { result = a; result.next = SortedMerge(a.next, b); } else { result = b; result.next = SortedMerge(a, b.next); } return result; } // The main function that takes an array of lists // arr[0..last] and generates the sorted output public static Node mergeKLists(Node arr[], int last) { // repeat until only one list is left while (last != 0) { int i = 0, j = last; // (i, j) forms a pair while (i < j) { // merge List i with List j and store // merged list in List i arr[i] = SortedMerge(arr[i], arr[j]); // consider next pair i++; j--; // If all pairs are merged, update last if (i >= j) last = j; } } return arr[0]; } /* Function to print nodes in a given linked list */ public static void printList(Node node) { while (node != null) { System.out.print(node.data + " "); node = node.next; } } public static void main(String args[]) { int k = 3; // Number of linked lists int n = 4; // Number of elements in each list // an array of pointers storing the head nodes // of the linked lists Node arr[] = new Node[k]; arr[0] = new Node(1); arr[0].next = new Node(3); arr[0].next.next = new Node(5); arr[0].next.next.next = new Node(7); arr[1] = new Node(2); arr[1].next = new Node(4); arr[1].next.next = new Node(6); arr[1].next.next.next = new Node(8); arr[2] = new Node(0); arr[2].next = new Node(9); arr[2].next.next = new Node(10); arr[2].next.next.next = new Node(11); // Merge all lists Node head = mergeKLists(arr, k - 1); printList(head); }} class Node { int data; Node next; Node(int data) { this.data = data; }}// This code is contributed by Gaurav Tiwari |
Python3
# Python3 program to merge k sorted# arrays of size n each # A Linked List nodeclass Node: def __init__(self): self.data = 0 self.next = None # Function to print nodes in a# given linked listdef printList(node): while (node != None): print(node.data, end=' ') node = node.next # Takes two lists sorted in increasing order,# and merge their nodes together to make one# big sorted list. Below function takes# O(Log n) extra space for recursive calls,# but it can be easily modified to work with# same time and O(1) extra spacedef SortedMerge(a, b): result = None # Base cases if (a == None): return(b) elif (b == None): return(a) # Pick either a or b, and recur if (a.data <= b.data): result = a result.next = SortedMerge(a.next, b) else: result = b result.next = SortedMerge(a, b.next) return result # The main function that takes an array of lists# arr[0..last] and generates the sorted outputdef mergeKLists(arr, last): # Repeat until only one list is left while (last != 0): i = 0 j = last # (i, j) forms a pair while (i < j): # Merge List i with List j and store # merged list in List i arr[i] = SortedMerge(arr[i], arr[j]) # Consider next pair i += 1 j -= 1 # If all pairs are merged, update last if (i >= j): last = j return arr[0] # Utility function to create a new node.def newNode(data): temp = Node() temp.data = data temp.next = None return temp # Driver codeif __name__ == '__main__': # Number of linked lists k = 3 # Number of elements in each list n = 4 # An array of pointers storing the # head nodes of the linked lists arr = [0 for i in range(k)] arr[0] = newNode(1) arr[0].next = newNode(3) arr[0].next.next = newNode(5) arr[0].next.next.next = newNode(7) arr[1] = newNode(2) arr[1].next = newNode(4) arr[1].next.next = newNode(6) arr[1].next.next.next = newNode(8) arr[2] = newNode(0) arr[2].next = newNode(9) arr[2].next.next = newNode(10) arr[2].next.next.next = newNode(11) # Merge all lists head = mergeKLists(arr, k - 1) printList(head) # This code is contributed by rutvik_56 |
C#
// C# program to merge k sorted arrays of size n eachusing System; public class MergeKSortedLists { /* Takes two lists sorted in increasing order, and merge their nodes together to make one big sorted list. Below function takes O(Log n) extra space for recursive calls, but it can be easily modified to work with same time and O(1) extra space */ public static Node SortedMerge(Node a, Node b) { Node result = null; /* Base cases */ if (a == null) return b; else if (b == null) return a; /* Pick either a or b, and recur */ if (a.data <= b.data) { result = a; result.next = SortedMerge(a.next, b); } else { result = b; result.next = SortedMerge(a, b.next); } return result; } // The main function that takes // an array of lists arr[0..last] // and generates the sorted output public static Node mergeKLists(Node[] arr, int last) { // repeat until only one list is left while (last != 0) { int i = 0, j = last; // (i, j) forms a pair while (i < j) { // merge List i with List j and store // merged list in List i arr[i] = SortedMerge(arr[i], arr[j]); // consider next pair i++; j--; // If all pairs are merged, update last if (i >= j) last = j; } } return arr[0]; } /* Function to print nodes in a given linked list */ public static void printList(Node node) { while (node != null) { Console.Write(node.data + " "); node = node.next; } } public static void Main() { int k = 3; // Number of linked lists // int n = 4; // Number of elements in each list // An array of pointers storing the head nodes // of the linked lists Node[] arr = new Node[k]; arr[0] = new Node(1); arr[0].next = new Node(3); arr[0].next.next = new Node(5); arr[0].next.next.next = new Node(7); arr[1] = new Node(2); arr[1].next = new Node(4); arr[1].next.next = new Node(6); arr[1].next.next.next = new Node(8); arr[2] = new Node(0); arr[2].next = new Node(9); arr[2].next.next = new Node(10); arr[2].next.next.next = new Node(11); // Merge all lists Node head = mergeKLists(arr, k - 1); printList(head); }} public class Node { public int data; public Node next; public Node(int data) { this.data = data; }} /* This code contributed by PrinciRaj1992 */ |
Javascript
<script> // JavaScript program to merge k sorted// arrays of size n eachclass Node { constructor(val) { this.data = val; this.next = null; }} /* Takes two lists sorted in increasing order, and merge their nodes together to make one big sorted list. Below function takes O(Log n) extra space for * recursive calls, but it can be easily modified to work with same time and * O(1) extra space */ function SortedMerge(a, b) {var result = null; /* Base cases */ if (a == null) return b; else if (b == null) return a; /* Pick either a or b, and recur */ if (a.data <= b.data) { result = a; result.next = SortedMerge(a.next, b); } else { result = b; result.next = SortedMerge(a, b.next); } return result; } // The main function that takes an array of lists // arr[0..last] and generates the sorted output function mergeKLists(arr , last) { // repeat until only one list is left while (last != 0) { var i = 0, j = last; // (i, j) forms a pair while (i < j) { // merge List i with List j and store // merged list in List i arr[i] = SortedMerge(arr[i], arr[j]); // consider next pair i++; j--; // If all pairs are merged, update last if (i >= j) last = j; } } return arr[0]; } /* Function to print nodes in a given linked list */ function printList(node) { while (node != null) { document.write(node.data + " "); node = node.next; } } var k = 3; // Number of linked lists var n = 4; // Number of elements in each list // an array of pointers storing the head nodes // of the linked listsvar arr = Array(k); arr[0] = new Node(1); arr[0].next = new Node(3); arr[0].next.next = new Node(5); arr[0].next.next.next = new Node(7); arr[1] = new Node(2); arr[1].next = new Node(4); arr[1].next.next = new Node(6); arr[1].next.next.next = new Node(8); arr[2] = new Node(0); arr[2].next = new Node(9); arr[2].next.next = new Node(10); arr[2].next.next.next = new Node(11); // Merge all listsvar head = mergeKLists(arr, k - 1); printList(head); // This code contributed by gauravrajput1 </script> |
Output
0 1 2 3 4 5 6 7 8 9 10 11
Time Complexity: O(N * K * log K), As outer while loop in function mergeKLists() runs log K times and every time it processes N*K elements.
Auxiliary Space: O(N * K), Because recursion is used in SortedMerge() and to merge the final 2 linked lists of size N/2, N recursive calls will be made.
Merge K sorted linked lists by Selecting min of top element:
The idea is to select the minimum of top elements iteratively store that in a new node and increment the pointer of the minimum element.
Follow the steps below to solve the problem:
- Find the node with the smallest value in all the K lists and
- Increment the current pointer to the next node of the list where the smallest node is found.
- Now make a new node and append the node to the head node of the resultant list and point the head list with this new node
- Repeat these steps till all nodes have been used.
Below is the implementation of the above approach:
C++
// C++ program to merge k sorted arrays of size n each#include <bits/stdc++.h>using namespace std; // A Linked List nodestruct Node { int data; Node* next; Node(int x) { data = x; next = NULL; }}; /* Function to print nodes in a given linked list */void printList(Node* node){ while (node != NULL) { printf("%d ", node->data); node = node->next; } cout << endl;} /*Linked list Node structure struct Node{int data;Node* next;Node(int x){ data = x; next = NULL;}};*/ // Function to merge K sorted linked list. Node* mergeKLists(Node* arr[], int K) { Node* head = NULL; while (1) { int a = 0; int z; Node* curr; int min = INT_MAX; for (int i = 0; i < K; i++) { if (arr[i] != NULL) { a++; if (arr[i]->data < min) { min = arr[i]->data; z = i; } } } if (a != 0) { arr[z] = arr[z]->next; Node* temp = new Node(min); if (head == NULL) { head = temp; curr = temp; } else { curr->next = temp; curr = temp; } } else { return head; } } } // { Driver Code Starts. // Driver program to test above functionsint main(){ int k = 3; // Number of linked lists int n = 4; // Number of elements in each list // an array of pointers storing the head nodes // of the linked lists Node* arr[k]; arr[0] = new Node(1); arr[0]->next = new Node(3); arr[0]->next->next = new Node(5); arr[0]->next->next->next = new Node(7); arr[1] = new Node(2); arr[1]->next = new Node(4); arr[1]->next->next = new Node(6); arr[1]->next->next->next = new Node(8); arr[2] = new Node(0); arr[2]->next = new Node(9); arr[2]->next->next = new Node(10); arr[2]->next->next->next = new Node(11); Node* head = mergeKLists(arr, k); printList(head); return 0;} // } Driver Code Ends |
Output
0 1 2 3 4 5 6 7 8 9 10 11
Time complexity: O(N*K2), There are N*K nodes in total and to find the smallest node it takes K times so for the N*K nodes it will take N*K*K time.
Auxiliary Space: O(N)
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Last Updated :
09 May, 2023
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