Linked-List-Component Solution added

Author: sumitdethe27

Linked List/Linked_List_Component.java

@@ -0,0 +1,72 @@
+
+
+
+/**PROBLEM
+ * You are given the head of a linked list containing unique integer values and an integer array nums
+ *that is a subset of the linked list values.
+
+*Return the number of connected components in nums where two values are connected if
+ *they appear consecutively in the linked list.
+*/
+
+// SAMPLE I/O
+// Input: head = [0,1,2,3], nums = [0,1,3]
+// Output: 2
+// Explanation: 0 and 1 are connected, so [0, 1] and [3] are the two connected components.
+
+// Approach
+/**
+ * Creating a HashMap to store all the values of nums[]
+ * Iterating list and if current node(head) we check if the hashmap contains the value]
+ * if yes the we increment the ans by one and setting the flag to false 
+ * 
+ * 
+ * 
+ * Time Complexity: O(N)
+ * Space Complexity : O(N)
+ */
+
+
+/**
+ * Definition for singly-linked list.
+ * public class ListNode {
+ *     int val;
+ *     ListNode next;
+ *     ListNode() {}
+ *     ListNode(int val) { this.val = val; }
+ *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
+ * }
+ */
+class Solution {
+    public int numComponents(ListNode head, int[] nums) {
+        // Create a HashMap to store the values from nums as keys and their indices as values
+        HashMap<Integer, Integer> hm = new HashMap<>();
+        for (int i = 0; i < nums.length; i++) {
+            hm.put(nums[i], i);
+        }
+        
+        boolean flag = true; // Flag to track if a connected component is found
+        int ans = 0; // Variable to store the number of connected components
+        
+        // Traverse the linked list
+        while (head != null) {
+            // Check if the current node's value is present in the HashMap
+            while (head != null && hm.containsKey(head.val)) {
+                head = head.next; // Move to the next node
+                
+                // If this is the start of a new connected component, increment the answer
+                if (flag) {
+                    ans += 1;
+                    flag = false;
+                }
+            }
+            flag = true; // Reset the flag
+            
+            if (head != null) {
+                head = head.next; // Move to the next node
+            }
+        }
+        
+        return ans;
+    }
+}