Given a linked list of 0s, 1s and 2s, The task is to sort and print it.
Examples:
Input: 1 -> 1 -> 2 -> 0 -> 2 -> 0 -> 1 -> NULL
Output: 0 -> 0 -> 1 -> 1 -> 1 -> 2 -> 2 -> NULLInput: 1 -> 1 -> 2 -> 1 -> 0 -> NULL
Output: 0 -> 1 -> 1 -> 1 -> 2 -> NULL
Source: Microsoft Interview | Set 1
Sort Linked List of 0s, 1s and 2s using frequency counting:
Follow the below steps to implement the idea:
- Traverse the list and count the number of 0s, 1s, and 2s. Let the counts be n1, n2, and n3 respectively.
- Traverse the list again, fill the first n1 nodes with 0, then n2 nodes with 1, and finally n3 nodes with 2.
Below image is a dry run of the above approach:
Below is the implementation of the above approach.
C++
// C++ Program to sort a linked list 0s, 1s or 2s #include <bits/stdc++.h>using namespace std;/* Link list node */class Node { public: int data; Node* next; }; // Function to sort a linked list of 0s, 1s and 2s void sortList(Node *head) { int count[3] = {0, 0, 0}; // Initialize count of '0', '1' and '2' as 0 Node *ptr = head; /* count total number of '0', '1' and '2' * count[0] will store total number of '0's * count[1] will store total number of '1's * count[2] will store total number of '2's */ while (ptr != NULL) { count[ptr->data] += 1; ptr = ptr->next; } int i = 0; ptr = head; /* Let say count[0] = n1, count[1] = n2 and count[2] = n3 * now start traversing list from head node, * 1) fill the list with 0, till n1 > 0 * 2) fill the list with 1, till n2 > 0 * 3) fill the list with 2, till n3 > 0 */ while (ptr != NULL) { if (count[i] == 0) ++i; else { ptr->data = i; --count[i]; ptr = ptr->next; } } } /* Function to push a node */void push (Node** head_ref, int new_data) { /* allocate node */ Node* new_node = new Node(); /* put in the data */ new_node->data = new_data; /* link the old list of the new node */ new_node->next = (*head_ref); /* move the head to point to the new node */ (*head_ref) = new_node; } /* Function to print linked list */void printList(Node *node) { while (node != NULL) { cout << node->data << " "; node = node->next; } cout << endl; } /* Driver code*/int main(void) { Node *head = NULL; push(&head, 0); push(&head, 1); push(&head, 0); push(&head, 2); push(&head, 1); push(&head, 1); push(&head, 2); push(&head, 1); push(&head, 2); cout << "Linked List before Sorting\n"; printList(head); sortList(head); cout << "Linked List after Sorting\n"; printList(head); return 0; } // This code is contributed by rathbhupendra |
C
// C Program to sort a linked list 0s, 1s or 2s#include<stdio.h>#include<stdlib.h>/* Link list node */struct Node{ int data; struct Node* next;};// Function to sort a linked list of 0s, 1s and 2svoid sortList(struct Node *head){ int count[3] = {0, 0, 0}; // Initialize count of '0', '1' and '2' as 0 struct Node *ptr = head; /* count total number of '0', '1' and '2' * count[0] will store total number of '0's * count[1] will store total number of '1's * count[2] will store total number of '2's */ while (ptr != NULL) { count[ptr->data] += 1; ptr = ptr->next; } int i = 0; ptr = head; /* Let say count[0] = n1, count[1] = n2 and count[2] = n3 * now start traversing list from head node, * 1) fill the list with 0, till n1 > 0 * 2) fill the list with 1, till n2 > 0 * 3) fill the list with 2, till n3 > 0 */ while (ptr != NULL) { if (count[i] == 0) ++i; else { ptr->data = i; --count[i]; ptr = ptr->next; } }}/* Function to push a node */void push (struct Node** head_ref, int new_data){ /* allocate node */ struct Node* new_node = (struct Node*) malloc(sizeof(struct Node)); /* put in the data */ new_node->data = new_data; /* link the old list of the new node */ new_node->next = (*head_ref); /* move the head to point to the new node */ (*head_ref) = new_node;}/* Function to print linked list */void printList(struct Node *node){ while (node != NULL) { printf("%d ", node->data); node = node->next; } printf("\n");}/* Driver program to test above function*/int main(void){ struct Node *head = NULL; push(&head, 0); push(&head, 1); push(&head, 0); push(&head, 2); push(&head, 1); push(&head, 1); push(&head, 2); push(&head, 1); push(&head, 2); printf("Linked List before Sorting \n"); printList(head); sortList(head); printf("Linked List after Sorting \n"); printList(head); return 0;} |
Java
// Java program to sort a linked list of 0, 1 and 2class LinkedList{ Node head; // head of list /* Linked list Node*/ class Node { int data; Node next; Node(int d) {data = d; next = null; } } void sortList() { // initialise count of 0 1 and 2 as 0 int count[] = {0, 0, 0}; Node ptr = head; /* count total number of '0', '1' and '2' * count[0] will store total number of '0's * count[1] will store total number of '1's * count[2] will store total number of '2's */ while (ptr != null) { count[ptr.data]++; ptr = ptr.next; } int i = 0; ptr = head; /* Let say count[0] = n1, count[1] = n2 and count[2] = n3 * now start traversing list from head node, * 1) fill the list with 0, till n1 > 0 * 2) fill the list with 1, till n2 > 0 * 3) fill the list with 2, till n3 > 0 */ while (ptr != null) { if (count[i] == 0) i++; else { ptr.data= i; --count[i]; ptr = ptr.next; } } } /* Utility functions */ /* Inserts a new Node at front of the list. */ public void push(int new_data) { /* 1 & 2: Allocate the Node & Put in the data*/ Node new_node = new Node(new_data); /* 3. Make next of new Node as head */ new_node.next = head; /* 4. Move the head to point to new Node */ head = new_node; } /* Function to print linked list */ void printList() { Node temp = head; while (temp != null) { System.out.print(temp.data+" "); temp = temp.next; } System.out.println(); } /* Driver program to test above functions */ public static void main(String args[]) { LinkedList llist = new LinkedList(); /* Constructed Linked List is 1->2->3->4->5->6->7-> 8->8->9->null */ llist.push(0); llist.push(1); llist.push(0); llist.push(2); llist.push(1); llist.push(1); llist.push(2); llist.push(1); llist.push(2); System.out.println("Linked List before sorting"); llist.printList(); llist.sortList(); System.out.println("Linked List after sorting"); llist.printList(); }} /* This code is contributed by Rajat Mishra */ |
Python3
# Python program to sort a linked list of 0, 1 and 2class LinkedList(object): def __init__(self): # head of list self.head = None # Linked list Node class Node(object): def __init__(self, d): self.data = d self.next = None def sortList(self): # initialise count of 0 1 and 2 as 0 count = [0, 0, 0] ptr = self.head # count total number of '0', '1' and '2' # * count[0] will store total number of '0's # * count[1] will store total number of '1's # * count[2] will store total number of '2's while ptr != None: count[ptr.data]+=1 ptr = ptr.next i = 0 ptr = self.head # Let say count[0] = n1, count[1] = n2 and count[2] = n3 # * now start traversing list from head node, # * 1) fill the list with 0, till n1 > 0 # * 2) fill the list with 1, till n2 > 0 # * 3) fill the list with 2, till n3 > 0 while ptr != None: if count[i] == 0: i+=1 else: ptr.data = i count[i]-=1 ptr = ptr.next # Utility functions # Inserts a new Node at front of the list. def push(self, new_data): # 1 & 2: Allocate the Node & # Put in the data new_node = self.Node(new_data) # 3. Make next of new Node as head new_node.next = self.head # 4. Move the head to point to new Node self.head = new_node # Function to print linked list def printList(self): temp = self.head while temp != None: print (str(temp.data),end=" ") temp = temp.next print()# Driver program to test above functionsllist = LinkedList()llist.push(0)llist.push(1)llist.push(0)llist.push(2)llist.push(1)llist.push(1)llist.push(2)llist.push(1)llist.push(2)print ("Linked List before sorting")llist.printList()llist.sortList()print ("Linked List after sorting")llist.printList()# This code is contributed by BHAVYA JAIN |
C#
// C# program to sort a linked // list of 0, 1 and 2using System;public class LinkedList{ Node head; // head of list /* Linked list Node*/ class Node { public int data; public Node next; public Node(int d) { data = d; next = null; } } void sortList() { // initialise count of 0 1 and 2 as 0 int []count = {0, 0, 0}; Node ptr = head; /* count total number of '0', '1' and '2' * count[0] will store total number of '0's * count[1] will store total number of '1's * count[2] will store total number of '2's */ while (ptr != null) { count[ptr.data]++; ptr = ptr.next; } int i = 0; ptr = head; /* Let say count[0] = n1, count[1] = n2 and count[2] = n3 * now start traversing list from head node, * 1) fill the list with 0, till n1 > 0 * 2) fill the list with 1, till n2 > 0 * 3) fill the list with 2, till n3 > 0 */ while (ptr != null) { if (count[i] == 0) i++; else { ptr.data= i; --count[i]; ptr = ptr.next; } } } /* Utility functions */ /* Inserts a new Node at front of the list. */ public void push(int new_data) { /* 1 & 2: Allocate the Node & Put in the data*/ Node new_node = new Node(new_data); /* 3. Make next of new Node as head */ new_node.next = head; /* 4. Move the head to point to new Node */ head = new_node; } /* Function to print linked list */ void printList() { Node temp = head; while (temp != null) { Console.Write(temp.data+" "); temp = temp.next; } Console.WriteLine(); } /* Driver code */ public static void Main(String []args) { LinkedList llist = new LinkedList(); /* Constructed Linked List is 1->2->3->4-> 5->6->7->8->8->9->null */ llist.push(0); llist.push(1); llist.push(0); llist.push(2); llist.push(1); llist.push(1); llist.push(2); llist.push(1); llist.push(2); Console.WriteLine("Linked List before sorting"); llist.printList(); llist.sortList(); Console.WriteLine("Linked List after sorting"); llist.printList(); }} /* This code is contributed by 29AjayKumar */ |
Javascript
<script>// Javascript program to sort a // linked list of 0, 1 and 2var head; // head of list /* Linked list Node */ class Node { constructor(val) { this.data = val; this.next = null; } } function sortList() { // initialise count of 0 1 and 2 as 0 var count = [ 0, 0, 0 ];var ptr = head; /* count total number of '0', '1' and '2' count[0] will store total number of '0's count[1] will store total number of '1's count[2] will store total number of '2's */ while (ptr != null) { count[ptr.data]++; ptr = ptr.next; } var i = 0; ptr = head; /* Let say count[0] = n1, count[1] = n2 and count[2] = n3 now start traversing list from head node, 1) fill the list with 0, till n1 > 0 2) fill the list with 1, till n2 > 0 3) fill the list with 2, till n3 > 0 */ while (ptr != null) { if (count[i] == 0) i++; else { ptr.data = i; --count[i]; ptr = ptr.next; } } } /* Utility functions */ /* Inserts a new Node at front of the list. */ function push(new_data) { /* * 1 & 2: Allocate the Node & Put in the data */var new_node = new Node(new_data); /* 3. Make next of new Node as head */ new_node.next = head; /* 4. Move the head to point to new Node */ head = new_node; } /* Function to print linked list */ function printList() { var temp = head; while (temp != null) { document.write(temp.data + " "); temp = temp.next; } document.write("<br/>"); } /* Driver program to test above functions */ /* Constructed Linked List is 1->2->3->4->5->6->7-> 8->8->9->null */ push(0); push(1); push(0); push(2); push(1); push(1); push(2); push(1); push(2); document.write("Linked List before sorting<br/>"); printList(); sortList(); document.write("Linked List after sorting<br/>"); printList();// This code is contributed by todaysgaurav</script> |
Output
Linked List before Sorting 2 1 2 1 1 2 0 1 0 Linked List after Sorting 0 0 1 1 1 1 2 2 2
Time Complexity: O(n) where n is the number of nodes in the linked list.
Auxiliary Space: O(1)
Sort a linked list of 0s, 1s and 2s by changing links
Using a stable sort algorithm:
C++
//C++ code to implement the above approach#include <iostream>// Define a structure for a node in a linked liststruct Node { int data; Node *next;};// Function to insert a new node at the end of the linked listvoid push(Node **head, int data) { Node *newNode = new Node(); newNode->data = data; newNode->next = nullptr; if (*head == nullptr) { *head = newNode; return; } Node *last = *head; while (last->next != nullptr) { last = last->next; } last->next = newNode;}// Function to print the linked listvoid printList(Node *head) { Node *current = head; while (current != nullptr) { std::cout << current->data << " "; current = current->next; } std::cout << std::endl;}// Function to sort the linked list containing 0's, 1's, and 2'svoid sortList(Node *head) { int count[3] = {0, 0, 0}; // Count the number of 0's, 1's, and 2's in the linked list Node *current = head; while (current != nullptr) { count[current->data]++; current = current->next; } // Overwrite the linked list with the sorted elements current = head; int i = 0; while (current != nullptr) { if (count[i] == 0) { i++; } else { current->data = i; count[i]--; current = current->next; } }}int main() { Node *head = nullptr; // Insert some elements into the linked list push(&head, 0); push(&head, 1); push(&head, 0); push(&head, 2); push(&head, 1); push(&head, 1); push(&head, 2); push(&head, 1); push(&head, 2); std::cout << "Linked List before Sorting: "<<std::endl; printList(head); sortList(head); std::cout << "Linked List after Sorting: "<<std::endl; printList(head); return 0;}//This code is contributed by Veerendra Singh Rajpoot |
Java
// Java code to implement the above approachimport java.util.*;public class Main { // Define a class for a node in a linked list static class Node { int data; Node next; // Constructor to initialize the node with data Node(int data) { this.data = data; this.next = null; } } // Function to insert a new node at the end of the // linked list static Node push(Node head, int data) { Node newNode = new Node(data); if (head == null) { head = newNode; return head; } Node last = head; while (last.next != null) { last = last.next; } last.next = newNode; return head; } // Function to print the linked list static void printList(Node head) { Node current = head; while (current != null) { System.out.print(current.data + " "); current = current.next; } System.out.println(); } // Function to sort the linked list containing 0's, 1's, // and 2's static void sortList(Node head) { int[] count = { 0, 0, 0 }; // Count the number of 0's, 1's, and 2's in the // linked list Node current = head; while (current != null) { count[current.data]++; current = current.next; } // Overwrite the linked list with the sorted // elements current = head; int i = 0; while (current != null) { if (count[i] == 0) { i++; } else { current.data = i; count[i]--; current = current.next; } } } public static void main(String[] args) { Node head = null; // Insert some elements into the linked list head = push(head, 0); head = push(head, 1); head = push(head, 0); head = push(head, 2); head = push(head, 1); head = push(head, 1); head = push(head, 2); head = push(head, 1); head = push(head, 2); System.out.println("Linked List before Sorting: "); printList(head); sortList(head); System.out.println("Linked List after Sorting: "); printList(head); }} |
Python3
# Define a class for a node in a linked listclass Node: def __init__(self, data): self.data = data self.next = None# Function to insert a new node at the end of the linked listdef push(head, data): new_node = Node(data) if head is None: head = new_node return head last = head while last.next is not None: last = last.next last.next = new_node return head# Function to print the linked listdef printList(head): current = head while current is not None: print(current.data, end=' ') current = current.next print()# Function to sort the linked list containing 0's, 1's, and 2'sdef sortList(head): count = [0, 0, 0] # Count the number of 0's, 1's, and 2's in the linked list current = head while current is not None: count[current.data] += 1 current = current.next # Overwrite the linked list with the sorted elements current = head i = 0 while current is not None: if count[i] == 0: i += 1 else: current.data = i count[i] -= 1 current = current.next# Main function to test the implementationif __name__ == '__main__': head = None # Insert some elements into the linked list head = push(head, 0) head = push(head, 1) head = push(head, 0) head = push(head, 2) head = push(head, 1) head = push(head, 1) head = push(head, 2) head = push(head, 1) head = push(head, 2) print("Linked List before Sorting:") printList(head) sortList(head) print("Linked List after Sorting:") printList(head) |
C#
using System;class Gfg{// Define a structure for a node in a linked listpublic class Node { public int data; public Node next;}// Function to insert a new node at the end of the linked listpublic static void Push(ref Node head, int data) { Node newNode = new Node(); newNode.data = data; newNode.next = null; if (head == null) { head = newNode; return; } Node last = head; while (last.next != null) { last = last.next; } last.next = newNode;}// Function to print the linked listpublic static void PrintList(Node head) { Node current = head; while (current != null) { Console.Write(current.data + " "); current = current.next; } Console.WriteLine();}// Function to sort the linked list containing 0's, 1's, and 2'spublic static void SortList(Node head) { int[] count = {0, 0, 0}; // Count the number of 0's, 1's, and 2's in the linked list Node current = head; while (current != null) { count[current.data]++; current = current.next; } // Overwrite the linked list with the sorted elements current = head; int i = 0; while (current != null) { if (count[i] == 0) { i++; } else { current.data = i; count[i]--; current = current.next; } }}public static void Main() { Node head = null; // Insert some elements into the linked list Push(ref head, 0); Push(ref head, 1); Push(ref head, 0); Push(ref head, 2); Push(ref head, 1); Push(ref head, 1); Push(ref head, 2); Push(ref head, 1); Push(ref head, 2); Console.WriteLine("Linked List before Sorting:"); PrintList(head); SortList(head); Console.WriteLine("Linked List after Sorting:"); PrintList(head);}} |
Javascript
// Define a class for a node in a linked listclass Node {constructor(data) {this.data = data;this.next = null;}}// Function to insert a new node at the end of the linked listfunction push(head, data) {const new_node = new Node(data);if (head === null) {head = new_node;return head;}let last = head;while (last.next !== null) {last = last.next;}last.next = new_node;return head;}// Function to print the linked listfunction printList(head) {let current = head;temp="";while (current !== null) { temp = temp + current.data+ ' ';current = current.next;}console.log(temp);}// Function to sort the linked list containing 0's, 1's, and 2'sfunction sortList(head) {const count = [0, 0, 0];// Count the number of 0's, 1's, and 2's in the linked listlet current = head;while (current !== null) {count[current.data] += 1;current = current.next;}// Overwrite the linked list with the sorted elementscurrent = head;let i = 0;while (current !== null) {if (count[i] === 0) {i += 1;} else {current.data = i;count[i] -= 1;current = current.next;}}}// Main function to test the implementationlet head = null;// Insert some elements into the linked listhead = push(head, 0);head = push(head, 1);head = push(head, 0);head = push(head, 2);head = push(head, 1);head = push(head, 1);head = push(head, 2);head = push(head, 1);head = push(head, 2);console.log("Linked List before Sorting:");printList(head);sortList(head);console.log("Linked List after Sorting:");printList(head); |
Output
Linked List before Sorting: 0 1 0 2 1 1 2 1 2 Linked List after Sorting: 0 0 1 1 1 1 2 2 2
Explanation of the above approach:
- The Node structure: This structure is used to define a node in a linked list. It contains an integer data to store the data of the node and a pointer next to the next node in the linked list.
- The push function: This function is used to insert a new node at the end of the linked list. It takes two parameters: a pointer to the head of the linked list and an integer data to be stored in the new node. The function first creates a new node with the given data and sets its next pointer to nullptr. If the linked list is empty, the function sets the head of the linked list to the new node. Otherwise, it iterates through the linked list to find the last node and inserts the new node after it.
- The printList function: This function is used to print the linked list. It takes one parameter, a pointer to the head of the linked list, and iterates through the list while printing the data of each node.
- The sortList function: This function is used to sort the linked list. It takes one parameter, a pointer to the head of the linked list. The function first initializes an array count of size 3 to store the count of 0’s, 1’s, and 2’s in the linked list. It then iterates through the linked list and increments the count of each element as it appears. Finally, it overwrites the linked list with the sorted elements. The function iterates through the linked list again and whenever it encounters a node with a value of 0, it sets its data to 0 and decrements the count of 0’s. Similarly, whenever it encounters a node with a value of 1, it sets its data to 1 and decrements the count of 1’s. And whenever it encounters a node with a value of 2, it sets its data to 2 and decrements the count of 2’s.
- The main function: This function is the entry point of the program. It first creates an empty linked list and inserts some elements into it. It then calls the sortList function to sort the linked list. Finally, it prints the original and sorted linked lists to verify that the sorting is correct.
Note: This is a simple implementation of the “Using a stable sort algorithm” approach for sorting a linked list containing only 0’s, 1’s, and 2’s. It may not be the most efficient implementation for large linked lists, but it should give you a good understanding of the basic idea
Time complexity: O(n),
The time complexity of this implementation is O(n), where n is the number of elements in the linked list. This is because the program iterates through the linked list twice: once to count the number of 0’s, 1’s, and 2’s, and once to overwrite the linked list with the sorted elements. The count operation takes O(n) time and the overwrite operation takes O(n) time, so the total time complexity is O(n) + O(n) = O(n).
Space complexity: O(1),
The space complexity of this implementation is O(1), because it uses only a constant amount of extra memory to store the count array of size 3. The linked list itself is not used to store any additional information during the sorting, so its space complexity remains the same.
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