Simple solution || Explained step by step || Java ✅|| Beats 100% ✌️
Intuition\n1. We can solve this by iterating through the first list list1 to find the nodes where we want to insert list2.\n2. Once we find those nodes, we can 
# Intuition\n1. We can solve this by iterating through the first list `list1` to find the nodes where we want to insert `list2.`\n2. Once we find those nodes, we can manipulate the pointers to detach the segment in `list1` and connect `list2` in its place.\n\n# Approach\n1. **Find Insertion and Removal Points:** *Iterate through* `list1` keeping track of positions. Store the node before position `a` and the node after position `b`.\n2. **Detach Segment:** Set the `next` pointer of the node before `a` to point to the head of `list2`.\n3. **Connect list2 Tail:** Traverse `list2` to find its last node and set its `next` pointer to the node after position `b` in `list1`.\n4. **Return Modified List:** The head of `list1` now points to the merged list, so return it.\n\n# Complexity\n- Time complexity:\n**O(m+n)**\n\n- Space complexity:\n**O(1)**\n\n# Code\n```Java []\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 current = list1;\n        for (int i = 0; i < a - 1; i++) {\n            current = current.next;\n        }\n        \n        ListNode startNode = current;\n        \n        for (int i = a; i <= b + 1; i++) {\n            current = current.next;\n        }\n        \n        startNode.next = list2;\n        \n        while (list2.next != null) {\n            list2 = list2.next;\n        }\n        \n        list2.next = current;\n        \n        return list1;\n    }\n}\n\n```\n```C++ []\n/**\n * Definition for singly-linked list.\n * struct ListNode {\n *     int val;\n *     ListNode *next;\n *     ListNode() : val(0), next(nullptr) {}\n *     ListNode(int x) : val(x), next(nullptr) {}\n *     ListNode(int x, ListNode *next) : val(x), next(next) {}\n * };\n */\nclass Solution {\npublic:\n    ListNode* mergeInBetween(ListNode* list1, int a, int b, ListNode* list2) {\n        // Traverse to the node before \'a\'\n        ListNode* current = list1;\n        for (int i = 0; i < a - 1; i++) {\n            current = current->next;\n        }\n        \n        // Mark the node before \'a\'\n        ListNode* startNode = current;\n        \n        // Traverse to the node after \'b\'\n        for (int i = a; i <= b + 1; i++) {\n            current = current->next;\n        }\n        \n        // Connect list2 at the position of \'a\' to \'b\'\n        startNode->next = list2;\n        \n        // Traverse to the end of list2\n        while (list2->next != nullptr) {\n            list2 = list2->next;\n        }\n        \n        // Connect the end of list2 to the node after \'b\'\n        list2->next = current;\n        \n        return list1; // Return the head of modified linked list\n    }\n};\n\n```\n```javascript []\n/**\n * Definition for singly-linked list.\n */\nfunction ListNode(val, next) {\n    this.val = (val === undefined ? 0 : val);\n    this.next = (next === undefined ? null : next);\n}\n\n/**\n * @param {ListNode} list1\n * @param {number} a\n * @param {number} b\n * @param {ListNode} list2\n * @return {ListNode}\n */\nvar mergeInBetween = function(list1, a, b, list2) {\n    let current = list1;\n    \n    // Traverse to the node before \'a\'\n    for (let i=0; i<a-1; i++) {\n      current=current.next;\n    }\n    \n    // Mark the node before \'a\'\n    let startNode=current;\n    \n    // Traverse to the node after \'b\'\n    for (let i=a; i<=b+1; i++) {\n      current=current.next;\n    }\n    \n    // Connect list2 at the position of \'a\' to \'b\'\n    startNode.next=list2;\n    \n   // Traverse to the end of lis t21\n   while(list2.next!=null){\n       list2=list2.next;\n   }\n\n   // Connect the end of list21 to the node after\'b"\n   list2.next=current;\n\n   return list1//Return the head of modified linkedlist (list1) \n};\n\n```\n```python3 []\n# Definition for singly-linked list.\n# class ListNode:\n#     def __init__(self, val=0, next=None):\n#         self.val = val\n#         self.next = next\n\nclass Solution:\n    def mergeInBetween(self, list1: ListNode, a: int, b: int, list2: ListNode) -> ListNode:\n        # Traverse to the node before \'a\'\n        current = list1\n        for _ in range(a - 1):\n            current = current.next\n        \n        # Mark the node before \'a\'\n        startNode = current\n        \n        # Traverse to the node after \'b\'\n        for _ in range(b - a + 2):\n            current = current.next\n        \n        # Connect list2 at the position of \'a\' to \'b\'\n        startNode.next = list2\n        \n        # Traverse to the end of list2\n        while(list2.next != None):\n            list2 =list2.next\n        \n        # Connect the end of list2 to the node after \'b\'\n        list2.next = current\n        \n        return list1  # Return the head of modified linked list\n\n```\n
['Linked List', 'C++', 'Java', 'Python3', 'JavaScript']