EASY|| FULLY DETAILED EXPLANATION || MAKE CONNECTIONS
Intuition\n Describe your first thoughts on how to solve this problem. \nThe Intuition is as simple as it says . Just do as directed , there is no catch in the 
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe Intuition is as simple as it says . Just do as directed , there is no catch in the question . \n# GET MY LINKEDIN IN THE COMMENTS LETS CONNECT TOGETHER AND MASTER DSA.\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n\n**Steps:**\n\n1. **Initialization:**\n   - `temp`: A temporary pointer to traverse `list1`.\n   - `jointStart`: A pointer to store the node before the insertion point (`a-1`).\n   - `jointEnd`: A pointer to store the node after the insertion point (`b+1`).\n   - `index`: A counter to track the current position in `list1`.\n\n2. **Finding Insertion Points:**\n   - The loop iterates through `list1` using `temp`:\n     - If `index == a - 1`: This indicates the node before the insertion point. The code sets `jointStart` to the current node `temp`.\n     - If `index == b + 1`: This indicates the node after the insertion point. The code sets `jointEnd` to the current node `temp`.\n     - The loop increments `index` for each iteration.\n\n3. **Connecting `list1` and `list2`:**\n   - After finding the insertion points:\n     - `jointStart.next = list2`: Connects the node before the insertion point (`jointStart`) to the head of `list2`. This effectively inserts `list2` at the beginning of the desired location.\n\n4. **Finding the Tail of `list2`:**\n   - `temp2`: A temporary pointer to traverse `list2`.\n   - `tail`: A pointer to store the last node of `list2`.\n   - The loop iterates through `list2` using `temp2`:\n     - It finds the last node (`tail`) by checking for `temp2.next == null`.\n\n5. **Connecting the Tail of `list2` to `list1`:**\n   - `tail.next = jointEnd`: Connects the last node of `list2` (`tail`) to the node after the insertion point (`jointEnd`). This completes the insertion of `list2` between nodes `a` and `b` of `list1`.\n\n6. **Returning the Modified List:**\n   - The function returns `list1`, which has now been modified with `list2` inserted between the specified nodes.\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\nO(N)\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\nO(1)\n# Code\n```\n/**\n * Definition for singly-linked list.\n * public class ListNode {\n *     int val;\n *     ListNode next;\n *     ListNode() {}\n *     ListNode(int val) { this.val = val; }\n *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }\n * }\n */\nclass Solution {\n    public ListNode mergeInBetween(ListNode list1, int a, int b, ListNode list2) {\n        ListNode temp =list1;\n        ListNode jointStart=null;\n        ListNode jointEnd=null;\n        int index=0;\n        while(temp!=null){\n            if(index==a-1){\n                jointStart=temp;\n            }\n            else if(index==b+1){\n                jointEnd=temp;\n            }\n            temp=temp.next;\n            index++;\n        }\n        jointStart.next=list2;\n        //lets find tail of list 2;\n        ListNode temp2 =list2;\n        ListNode tail=null;\n        while(temp2.next!=null){\n            temp2=temp2.next;\n            tail=temp2;\n            \n        }\n        tail.next=jointEnd;\n        return list1;\n    }\n}\n```
['Linked List', 'Java']