JAVA Solution Explained in HINDI
https://youtu.be/clHuUMe0UeM\n\nFor explanation, please watch the above video and do like, share and subscribe the channel. \u2764\uFE0F Also, please do upvote 
https://youtu.be/clHuUMe0UeM\n\nFor explanation, please watch the above video and do like, share and subscribe the channel. \u2764\uFE0F Also, please do upvote the solution if you liked it.\n\nSubscribe link:- https://www.youtube.com/@reelcoding?sub_confirmation=1\n\nSubscribe Goal:- 300\nCurrent Subscriber:- 242\n\n# Complexity\n- Time complexity: O(m+n)\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: O(1)\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\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        \n        ListNode start = null;\n        ListNode end = list1;\n\n        for(int i = 0; i < b; i++) {\n            if(i == a - 1) {\n                start = end;\n            }\n            end = end.next;\n        }\n\n        start.next = list2;\n\n        while(list2.next != null) {\n            list2 = list2.next;\n        }\n\n        list2.next = end.next;\n        end.next = null;\n\n        return list1;\n    }\n}\n```
['Java']