Implement anagram check and odd-even linked list

You are given two independent coding tasks. ### Task 1: Check if two strings are anagrams Implement a function that determines whether two given strings are anagrams of each other. - Input: Two strings `s` and `t` consisting only of lowercase English letters (`'a'`-`'z'`). - Output: A boolean value indicating whether `t` is an anagram of `s`. Two strings are anagrams if and only if they contain exactly the same characters with the same multiplicities, in any order. Requirements: - Design a function, for example `bool areAnagrams(string s, string t)`. - Time complexity should be O(n), where n is the total length of the strings. - Use only O(1) additional space (you may assume a fixed-size alphabet of 26 lowercase letters). You may assume: - `0 <= len(s), len(t) <= 10^5`. --- ### Task 2: Reorder a linked list by node index parity You are given the head of a singly linked list of integers. The nodes are indexed starting from 1 at the head, then 2 for the next node, and so on. Reorder the list so that: - All nodes that were originally at odd indices (1, 3, 5, ...) appear first, in their original relative order. - All nodes that were originally at even indices (2, 4, 6, ...) appear after all odd-index nodes, also in their original relative order. The reordering must be done in-place by changing the `next` pointers; do not allocate or copy list nodes. - Input: `ListNode* head` pointing to the head of a singly linked list. - Output: The head of the reordered list. Example: - Input list: `1 -> 2 -> 3 -> 4 -> 5 -> null` - Output list: `1 -> 3 -> 5 -> 2 -> 4 -> null` Constraints: - The number of nodes in the list is in the range `[0, 10^5]`. - Node values are arbitrary integers. - The algorithm should run in O(n) time and O(1) extra space.