diff --git a/.github/workflows/main.yml b/.github/workflows/main.yml
index 75ab8d3..3b03292 100644
--- a/.github/workflows/main.yml
+++ b/.github/workflows/main.yml
@@ -19,4 +19,7 @@ jobs:
           GITHUB_TOKEN: ${{ secrets.GITHUB_TOKEN }} # The GITHUB_TOKEN is there by default so you just need to keep it like it is and not necessarily need to add it as secret as it will throw an error. [More Details](https://docs.github.com/en/actions/security-guides/automatic-token-authentication#about-the-github_token-secret)
           OPENAI_API_KEY: ${{ secrets.OPENAI_API_KEY }}
           OPENAI_API_MODEL: "gpt-4" # Optional: defaults to "gpt-4"
-          exclude: "**/*.json, **/*.md" # Optional: exclude patterns separated by commas
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+          exclude: "**/*.json, **/*.md" # Optional: exclude patterns separated by commas
+
+
+          
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diff --git a/src/code.js b/src/code.js
new file mode 100644
index 0000000..bac1db0
--- /dev/null
+++ b/src/code.js
@@ -0,0 +1,104 @@
+/**The task is to insert a data in a linked list such that it will always remain sorted.
+ * 
+ * Ex: say ll = 1->2->9 and data=8 
+ * So 8 should be inserted between 2 & 9 such that the linked list becomes
+ * 
+ * ll = 1->2->8->9
+ * 
+ * ex: ll = 10->20->30->40, data = 25
+ * ll = 10->20->25->30->40
+ * 
+ * ex: ll = 10->15, data=5
+ * ll = 5->10->15
+ * 
+ * ex: ll = 10->20, data=30
+ * ll = 10->20->30
+ * 
+ */
+
+class Node {
+    constructor(data, next = null) {
+        this.data = data;
+        this.next = next;
+    }
+}
+class LinkedList {
+    constructor() {
+        this.head = null;
+        this.size = 0;
+    }
+
+    /**0(n),0(1)
+     * 
+     * Case1: When the head is null, then simply whatever is the data it has to be inserted as the hed of the ll.
+     * this.head === null then this.head=node;
+     * 
+     * Case2: Say ll = 10->15, data is 5
+     * Lets maintain a current and prev pointer such that current is this.head and prev is null
+     * Now we traverse such that current!==null
+     * 
+     * If at any point we see current.data>data we break, else we keep moving ahead prev= current & current= current.next.
+     * 
+     * In above case the prev will be null and current will be head.
+     * Simply its a case of adding a node at beginning of linked list.
+     * 
+     * node.next = this.head & this.head=node.
+     * 
+     * Case3: Say ll = 10->15 data is 20
+     * 
+     * Again by the concept of current and prev pointer,
+     * The current pointer will be null and prev pointer will be at the last node.
+     * Its simple case insertion at end.
+     * 
+     * this.prev.next = node.
+     * 
+     * Case4: ll= 10->15, data=12
+     * 
+     * Now the prev will not be null and will be at the node after which this new data node needs to be inserted and the current 
+     * will be at the node before which this new node has to be inserted, so its a simple case of insertion of node between two nodes.
+     * 
+     * node.next = current;
+     * perv.next = node;
+     */
+    insertSorted(data) {
+        let node = new Node(data);
+        if (this.head === null) {
+            this.head = node;
+        } else {
+            let current = this.head;
+            let prev = null
+            while (current !== null) {
+                if (current.data < data) {
+                    prev = current;
+                    current = current.next;
+                } else {
+                    break;
+                }
+            }
+            if (prev === null && current !== null) {
+                node.next = this.head;
+                this.head = node;
+            } else if (prev !== null && current === null) {
+                prev.next = node;
+            } else {
+                node.next = current;
+                prev.next = node;
+            }
+        }
+    }
+    print() {
+        if (this.head === null) {
+            return;
+        }
+        let current = this.head;
+        while (current !== null) {
+            console.log(current.data);
+            current = current.next;
+        }
+    }
+}
+let ll = new LinkedList();
+ll.insertSorted(10);
+// ll.insertSorted(20);
+// ll.insertSorted(30);
+ll.print();
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