🔥Beats 100%🔥✅Super Easy (C++/Java/Python) Solution With Detailed Explanation✅
Intuition\nThe intuition of code is to merge two singly linked lists by replacing a segment of the first list (from position a to b) with the entire second list
# Intuition\nThe intuition of code is to merge two singly linked lists by replacing a segment of the first list (from position a to b) with the entire second list. The approach involves three main steps: finding the node just before the start of the segment to be replaced in the first list, finding the node just after the end of this segment, and finally linking the second list between these two nodes. The last part of the process connects the end of the second list to the node just after the segment in the first list. This operation effectively splices the second list into the first, between the positions a and b, inclusive.\n\n![Screenshot 2024-03-20 070620.png](https://assets.leetcode.com/users/images/ad25ea0c-2157-49c3-8e25-89da96632d89_1710901239.6097941.png)\n\n## Step-by-Step Approach\n\n1. **Initialize Pointers and Counter**: The solution starts by initializing a pointer `point1` to the start of `list1` and a counter variable `count` to 1. This setup prepares for traversal through the list.\n\n2. **Traverse to the `a-1`th Node**: The loop continues until `point1` points to the node just before the `a`th node (1-indexed) or the `a-1`th node (0-indexed). This is the node after which `list2` will be inserted.\n\n3. **Traverse to the `b+1`th Node**: A second pointer `point2` starts at the node immediately after `point1` and moves forward until it points to the node just after the `b`th node. This node will be the point where `list1` continues after `list2` has been inserted.\n\n4. **Insert `list2` Between `a` and `b`**: The `next` pointer of the node at `a-1` is updated to point to the start of `list2`. This effectively inserts `list2` at position `a`.\n\n5. **Find the End of `list2`**: A temporary pointer traverses `list2` to find its last node. This step is necessary to link the end of `list2` back to `list1`.\n\n6. **Link the End of `list2` to `list1`**: The `next` pointer of the last node in `list2` is updated to point to `point2`, the `b+1`th node of `list1`. This completes the insertion process.\n\n7. **Return `list1`**: The function returns the head of the modified `list1`, which now includes `list2` between the `a`th and `b`th positions.\n\n## Complexity Analysis\n\n- **Time Complexity**: The algorithm traverses each list at most once. The traversal of `list1` to positions `a` and `b` takes \\(O(b)\\) time, as it stops after reaching the `b+1`th node. The traversal of `list2` to find its end takes \\(O(n)\\) time, where \\(n\\) is the length of `list2`. Thus, the total time complexity is \\(O(b + n)\\), where \\(b\\) is the position in `list1` where the merge ends, and \\(n\\) is the length of `list2`.\n\n- **Space Complexity**: The algorithm uses a constant amount of extra space for pointers and the counter variable, so the space complexity is \\(O(1)\\). Note that the space used by the lists themselves is not considered part of the space complexity for this analysis, as it is not additional space allocated by the algorithm.\n\n# Code\n```C++ []\n\nclass Solution {\npublic:\n    ListNode* mergeInBetween(ListNode* list1, int a, int b, ListNode* list2) {\n        ListNode *point1 = list1;\n        int count = 1;\n        while (count != a)\n        {\n            point1 = point1 -> next;\n            count++;\n        }\n\n        ListNode *point2 = point1 -> next;\n        while (count != b+1)\n        {\n            point2 = point2 -> next;\n            count++;\n        }\n\n        point1->next = list2;\n        ListNode *temp = list2;\n        while (temp -> next != NULL)\n            temp = temp -> next;\n\n        temp -> next = point2;\n        \n        return list1;\n\n    }\n};\n```\n```Java []\nclass Solution {\n    public ListNode mergeInBetween(ListNode list1, int a, int b, ListNode list2) {\n        ListNode point1 = list1;\n        int count = 1;\n        while (count != a) {\n            point1 = point1.next;\n            count++;\n        }\n\n        ListNode point2 = point1.next;\n        while (count != b + 1) {\n            point2 = point2.next;\n            count++;\n        }\n\n        point1.next = list2;\n        ListNode temp = list2;\n        while (temp.next != null)\n            temp = temp.next;\n\n        temp.next = point2;\n\n        return list1;\n    }\n};\n```\n```Python []\nclass Solution:\n    def mergeInBetween(self, list1, a, b, list2):\n        point1 = list1\n        count = 1\n        while count != a:\n            point1 = point1.next\n            count += 1\n        \n        point2 = point1.next\n        while count != b + 1:\n            point2 = point2.next\n            count += 1\n        \n        point1.next = list2\n        temp = list2\n        while temp.next is not None:\n            temp = temp.next\n        \n        temp.next = point2\n        \n        return list1\n\n```
['Linked List', 'C++', 'Java', 'Python3']