Given two linked lists head1 and head2, the task is to find occurrences of head2 in head1.
Examples:
Input: Head1 = 1->2->3->2->4->5->2->4, Head2 = 2->4
Output: Occurrences of head2 in head1: 2Input: Head1 = 3->4->1->5->2, Head2 = 3->4
Output: Occurrences of Head2 in Head1: 1
Approach 1: To solve the problem using Sliding Window Approach:
The idea is to use sliding window approach. We will traverse the first linked list head1, and at each node, we will check if the current node is equal to the first node of head2. If it is, then we will start a new traversal of both linked lists from this node, and check if all nodes in head2 match the nodes in head1 starting from this node. If they match, we will count it as an occurrence.
Below are the steps for the above approach:
- Initialize a variable count = zero.
- Traverse the first linked list head1 using a while loop that runs until head1 is NULL. At each node in head1,
- Initialize two pointers p1 and p2 to the current node in head1 and the head of head2, respectively.
- Run a nested while loop that traverses both linked lists starting from p1 and p2, and checks if the values of the nodes are equal. If they are not, the loop breaks and we move on to the next node in head1. If they are equal and we reach the end of head2, it means that we have found a match, and we increment the counter variable count.
- Update p1 to point to the next node in head1, and update p2 to point to the head of head2.
- After traversing the entire first linked list, return the final value of the counter variable count.
Below is the code for the above approach:
C++
// C++ code for the above approach:#include <bits/stdc++.h>using namespace std;struct Node { int val; Node* next; Node(int x) : val(x), next(NULL) { }};// Function to count occurencesint countOccurrences(Node* head1, Node* head2){ int count = 0; while (head1 != NULL) { Node* p1 = head1; Node* p2 = head2; while (p1 != NULL && p2 != NULL && p1->val == p2->val) { p1 = p1->next; p2 = p2->next; } if (p2 == NULL) { count++; } head1 = head1->next; } return count;}// Drivers codeint main(){ // Initialize the first linked list Node* head1 = new Node(1); head1->next = new Node(2); head1->next->next = new Node(3); head1->next->next->next = new Node(2); head1->next->next->next->next = new Node(4); head1->next->next->next->next->next = new Node(5); head1->next->next->next->next->next->next = new Node(2); head1->next->next->next->next->next->next->next = new Node(4); // Initialize the second linked list Node* head2 = new Node(2); head2->next = new Node(4); // Count the occurrences of head2 in head1 int count = countOccurrences(head1, head2); // Output the result cout << "Occurrences of head2 in head1: " << count << endl; return 0;} |
Java
// Java code implementation class Node { int val; Node next; public Node(int x) { val = x; next = null; }}public class Main { // Function to count occurrences static int countOccurrences(Node head1, Node head2) { int count = 0; while (head1 != null) { Node p1 = head1; Node p2 = head2; while (p1 != null && p2 != null && p1.val == p2.val) { p1 = p1.next; p2 = p2.next; } if (p2 == null) { count++; } head1 = head1.next; } return count; } // Drivers code public static void main(String[] args) { // Initialize the first linked list Node head1 = new Node(1); head1.next = new Node(2); head1.next.next = new Node(3); head1.next.next.next = new Node(2); head1.next.next.next.next = new Node(4); head1.next.next.next.next.next = new Node(5); head1.next.next.next.next.next.next = new Node(2); head1.next.next.next.next.next.next.next = new Node(4); // Initialize the second linked list Node head2 = new Node(2); head2.next = new Node(4); // Count the occurrences of head2 in head1 int count = countOccurrences(head1, head2); // Output the result System.out.println("Occurrences of head2 in head1: " + count); }} |
Python3
# Python code for the above approachclass Node: def __init__(self, x): self.val = x self.next = None# Function to count occurrencesdef countOccurrences(head1, head2): count = 0 while head1 != None: p1 = head1 p2 = head2 while p1 != None and p2 != None and p1.val == p2.val: p1 = p1.next p2 = p2.next if p2 == None: count += 1 head1 = head1.next return count# Driver codeif __name__ == '__main__': # Initialize the first linked list head1 = Node(1) head1.next = Node(2) head1.next.next = Node(3) head1.next.next.next = Node(2) head1.next.next.next.next = Node(4) head1.next.next.next.next.next = Node(5) head1.next.next.next.next.next.next = Node(2) head1.next.next.next.next.next.next.next = Node(4) # Initialize the second linked list head2 = Node(2) head2.next = Node(4) # Count the occurrences of head2 in head1 count = countOccurrences(head1, head2) # Output the result print("Occurrences of head2 in head1:", count) |
C#
using System;class Node{ public int val; public Node next; public Node(int x) { val = x; next = null; }}class Program{ // Function to count occurrences static int CountOccurrences(Node head1, Node head2) { int count = 0; while (head1 != null) { Node p1 = head1; Node p2 = head2; while (p1 != null && p2 != null && p1.val == p2.val) { p1 = p1.next; p2 = p2.next; } if (p2 == null) { count++; } head1 = head1.next; } return count; } // Driver code static void Main() { // Initialize the first linked list Node head1 = new Node(1); head1.next = new Node(2); head1.next.next = new Node(3); head1.next.next.next = new Node(2); head1.next.next.next.next = new Node(4); head1.next.next.next.next.next = new Node(5); head1.next.next.next.next.next.next = new Node(2); head1.next.next.next.next.next.next.next = new Node(4); // Initialize the second linked list Node head2 = new Node(2); head2.next = new Node(4); // Count the occurrences of head2 in head1 int count = CountOccurrences(head1, head2); // Output the result Console.WriteLine("Occurrences of head2 in head1: " + count); }} |
Javascript
// Javascript code for the above approach:class Node { constructor(val) { this.val = val; this.next = null; }}// Function to count occurrencesfunction countOccurrences(head1, head2) { let count = 0; while (head1 !== null) { let p1 = head1; let p2 = head2; while (p1 !== null && p2 !== null && p1.val === p2.val) { p1 = p1.next; p2 = p2.next; } if (p2 === null) { count++; } head1 = head1.next; } return count;}// Drivers code// Initialize the first linked listconst head1 = new Node(1);head1.next = new Node(2);head1.next.next = new Node(3);head1.next.next.next = new Node(2);head1.next.next.next.next = new Node(4);head1.next.next.next.next.next = new Node(5);head1.next.next.next.next.next.next = new Node(2);head1.next.next.next.next.next.next.next = new Node(4);// Initialize the second linked listconst head2 = new Node(2);head2.next = new Node(4);// Count the occurrences of head2 in head1const count = countOccurrences(head1, head2);// Output the resultconsole.log("Occurrences of head2 in head1:", count); // this code is contributed by uttamdp_10 |
Output
Occurrences of head2 in head1: 2
Time Complexity: O(m*n), where m and n are the lengths of the two linked lists
Auxiliary Space: O(1)
Approach 2: To solve the problem using Knuth-Morris-Pratt (KMP) algorithm
The basic idea of using a string matching algorithm is to treat the linked lists as strings and search for the second linked list within the first linked list as a substring.
Below are the steps for the above approach:
- Convert the linked lists to strings by concatenating their values in order.
- Use the KMP algorithm to search for the second string within the first string. The KMP algorithm searches for the occurrence of a pattern within a text by using a preprocessed table that is based on the pattern.
- Initialize two pointers p1 and p2 to the head of the first linked list and the second linked list, respectively.
- Traverse the first linked list using p1 and update p2 based on the preprocessed table. If we reach the end of the second linked list, it means that we have found a match, and we can increment the count.
- After traversing the entire first linked list, return the final value of the count.
Below is the code for the above approach:
C++
// C++ code for the above approach:#include <bits/stdc++.h>using namespace std;struct Node { int val; Node* next; Node(int x) : val(x), next(NULL) { }};// Function to concatenate the values// of a linked list into a stringstring concatenateList(Node* head){ string result = ""; Node* current = head; while (current != NULL) { result += to_string(current->val); current = current->next; } return result;}// Function to compute the prefix table// for the KMP algorithmvoid computePrefixTable(string pattern, int* prefixTable){ int m = pattern.length(); int j = 0; prefixTable[0] = 0; for (int i = 1; i < m; i++) { while (j > 0 && pattern[j] != pattern[i]) { j = prefixTable[j - 1]; } if (pattern[j] == pattern[i]) { j++; } prefixTable[i] = j; }}// Function to count the occurrences of a// second linked list within a// first linked listint countOccurrences(Node* head1, Node* head2){ string text = concatenateList(head1); string pattern = concatenateList(head2); int n = text.length(); int m = pattern.length(); if (m == 0) { return 0; } int prefixTable[m]; computePrefixTable(pattern, prefixTable); int count = 0; int i = 0; int j = 0; while (i < n) { if (text[i] == pattern[j]) { i++; j++; if (j == m) { count++; j = prefixTable[j - 1]; } } else if (j > 0) { j = prefixTable[j - 1]; } else { i++; } } return count;}// Drivers codeint main(){ // Initialize the first linked list Node* head1 = new Node(1); head1->next = new Node(2); head1->next->next = new Node(3); head1->next->next->next = new Node(2); head1->next->next->next->next = new Node(4); head1->next->next->next->next->next = new Node(5); head1->next->next->next->next->next->next = new Node(2); head1->next->next->next->next->next->next->next = new Node(4); // Initialize the second linked list Node* head2 = new Node(2); head2->next = new Node(4); // Function Call int result = countOccurrences(head1, head2); cout << "Occurrences of head2 in head1: " << result << endl; return 0;} |
Java
import java.io.*;import java.util.*;class Node { int val; Node next; Node(int x) { val = x; next = null; }}public class GFG { // Function to concatenate the values // of a linked list into string public static String concatenateList(Node head) { StringBuilder result = new StringBuilder(); Node current = head; while (current != null) { result.append(current.val); current = current.next; } return result.toString(); } // Function to compute the prefix table // for the KMP algorithm public static void computePrefixTable(String pattern, int[] prefixTable) { int m = pattern.length(); int j = 0; prefixTable[0] = 0; for (int i = 1; i < m; i++) { while (j > 0 && pattern.charAt(j) != pattern.charAt(i)) { j = prefixTable[j - 1]; } if (pattern.charAt(j) == pattern.charAt(i)) { j++; } prefixTable[i] = j; } } // Function to count the occurrences of // second linked list within a // first linked list public static int countOccurrences(Node head1, Node head2) { String text = concatenateList(head1); String pattern = concatenateList(head2); int n = text.length(); int m = pattern.length(); if (m == 0) { return 0; } int[] prefixTable = new int[m]; computePrefixTable(pattern, prefixTable); int count = 0; int i = 0; int j = 0; while (i < n) { if (text.charAt(i) == pattern.charAt(j)) { i++; j++; if (j == m) { count++; j = prefixTable[j - 1]; } } else if (j > 0) { j = prefixTable[j - 1]; } else { i++; } } return count; } // Drivers code public static void main(String[] args) { // Initialize the first linked list Node head1 = new Node(1); head1.next = new Node(2); head1.next.next = new Node(3); head1.next.next.next = new Node(2); head1.next.next.next.next = new Node(4); head1.next.next.next.next.next = new Node(5); head1.next.next.next.next.next.next = new Node(2); head1.next.next.next.next.next.next.next = new Node(4); // Initialize the second linked list Node head2 = new Node(2); head2.next = new Node(4); // Function Call int result = countOccurrences(head1, head2); System.out.println("Occurrences of head2 in head1: " + result); }} |
Python3
# Python3 code for the above approachclass Node: def __init__(self, x): self.val = x self.next = None# Function to concatenate the values # of a linked list into a stringdef concatenateList(head): result = "" current = head while current is not None: result += str(current.val) current = current.next return result# Function to compute the prefix table # for the KMP algorithmdef computePrefixTable(pattern, prefixTable): m = len(pattern) j = 0 prefixTable[0] = 0 for i in range(1, m): while j > 0 and pattern[j] != pattern[i]: j = prefixTable[j - 1] if pattern[j] == pattern[i]: j += 1 prefixTable[i] = j# Function to count the occurrences of a # second linked list within a # first linked listdef countOccurrences(head1, head2): text = concatenateList(head1) pattern = concatenateList(head2) n = len(text) m = len(pattern) if m == 0: return 0 prefixTable = [0] * m computePrefixTable(pattern, prefixTable) count = 0 i = 0 j = 0 while i < n: if text[i] == pattern[j]: i += 1 j += 1 if j == m: count += 1 j = prefixTable[j - 1] elif j > 0: j = prefixTable[j - 1] else: i += 1 return count# Drivers codeif __name__ == "__main__": # Initialize the first linked list head1 = Node(1) head1.next = Node(2) head1.next.next = Node(3) head1.next.next.next = Node(2) head1.next.next.next.next = Node(4) head1.next.next.next.next.next = Node(5) head1.next.next.next.next.next.next = Node(2) head1.next.next.next.next.next.next.next = Node(4) # Initialize the second linked list head2 = Node(2) head2.next = Node(4) # Function Call result = countOccurrences(head1, head2) print("Occurrences of head2 in head1:", result) |
C#
// C# code implementation of above approach using System;public class Node { public int val; public Node next; public Node(int x) { val = x; next = null; }}public class LinkedListOccurrence { // Function to concatenate the values // of a linked list into a string public static string ConcatenateList(Node head) { string result = ""; Node current = head; while (current != null) { result += current.val.ToString(); current = current.next; } return result; } // Function to compute the prefix table // for the KMP algorithm public static void ComputePrefixTable(string pattern, int[] prefixTable) { int m = pattern.Length; int j = 0; prefixTable[0] = 0; for (int i = 1; i < m; i++) { while (j > 0 && pattern[j] != pattern[i]) { j = prefixTable[j - 1]; } if (pattern[j] == pattern[i]) { j++; } prefixTable[i] = j; } } // Function to count the occurrences of a // second linked list within a // first linked list public static int CountOccurrences(Node head1, Node head2) { string text = ConcatenateList(head1); string pattern = ConcatenateList(head2); int n = text.Length; int m = pattern.Length; if (m == 0) { return 0; } int[] prefixTable = new int[m]; ComputePrefixTable(pattern, prefixTable); int count = 0; int i = 0; int j = 0; while (i < n) { if (text[i] == pattern[j]) { i++; j++; if (j == m) { count++; j = prefixTable[j - 1]; } } else if (j > 0) { j = prefixTable[j - 1]; } else { i++; } } return count; } // Drivers code public static void Main() { // Initialize the first linked list Node head1 = new Node(1); head1.next = new Node(2); head1.next.next = new Node(3); head1.next.next.next = new Node(2); head1.next.next.next.next = new Node(4); head1.next.next.next.next.next = new Node(5); head1.next.next.next.next.next.next = new Node(2); head1.next.next.next.next.next.next.next = new Node(4); // Initialize the second linked list Node head2 = new Node(2); head2.next = new Node(4); // Function Call int result = CountOccurrences(head1, head2); Console.WriteLine("Occurrences of head2 in head1: " + result); }} |
Javascript
// Javascript code for above approach :class Node { constructor(val) { this.val = val; this.next = null; }}// Function to concatenate the values// of a linked list into a stringfunction concatenateList(head) { let result = ""; let current = head; while (current !== null) { result += current.val.toString(); current = current.next; } return result;}// Function to compute the prefix table// for the KMP algorithmfunction computePrefixTable(pattern) { const m = pattern.length; const prefixTable = new Array(m).fill(0); let j = 0; for (let i = 1; i < m; i++) { while (j > 0 && pattern[j] !== pattern[i]) { j = prefixTable[j - 1]; } if (pattern[j] === pattern[i]) { j++; } prefixTable[i] = j; } return prefixTable;}// Function to count the occurrences of a// second linked list within a// first linked listfunction countOccurrences(head1, head2) { const text = concatenateList(head1); const pattern = concatenateList(head2); const n = text.length; const m = pattern.length; if (m === 0) { return 0; } const prefixTable = computePrefixTable(pattern); let count = 0; let i = 0; let j = 0; while (i < n) { if (text[i] === pattern[j]) { i++; j++; if (j === m) { count++; j = prefixTable[j - 1]; } } else if (j > 0) { j = prefixTable[j - 1]; } else { i++; } } return count;}// Driver's code// Initialize the first linked listconst head1 = new Node(1);head1.next = new Node(2);head1.next.next = new Node(3);head1.next.next.next = new Node(2);head1.next.next.next.next = new Node(4);head1.next.next.next.next.next = new Node(5);head1.next.next.next.next.next.next = new Node(2);head1.next.next.next.next.next.next.next = new Node(4);// Initialize the second linked listconst head2 = new Node(2);head2.next = new Node(4);// Function Callconst result = countOccurrences(head1, head2);console.log("Occurrences of head2 in head1:", result);// this code is contributed by uttamdp_10 |
Output
Occurrences of head2 in head1: 2
Time Complexity: O(m+n), where m and n are the lengths of the two linked lists.
Auxiliary Space: O(m), where m is the length of the pattern (the second linked list).
Related articles:
- Introduction to Linked List – Data Structure and Algorithm Tutorials
- Introduction to Strings – Data Structure and Algorithm Tutorials
- KMP Algorithm for Pattern Searching
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