unreasonably effective

link From palindrome’s mirror symmetry point of view, there are two types of elements: We find the longest length in two passes: (1) Build {element -> count} map (2) Depending on if it is a “repeated” element or not, we include them appropriately. If there are elements, time: , space: .

link An anagram of p is a window in s, so we can use sliding window. Variable-length window We expand the window until we get at least the required count for each character in p. Then we shrink to get the window size right. If the shrunk window still has all characters by the required…

link The first character in the string that appears once is our answer. So, we use two passes: (1) Create the map {char -> count} (2) Find first character having count = 1. Time: , space: .

link Since a tie for the maximum frequency must be broken in favor of recent-most value, we should maintain LIFO order per frequency. So, we need to maintain a stack per color. We shall use the below three pieces of information: Say there are numbers in the FreqStack having distinct values. So, . Operation Time…

link To group anagrams, we need a group id against which we can collect the anagrams. Sort Since all anagrams are identical after sorting, we can use sorted anagram as the group id. If there are strings and the longest string has length , time: , space: . Character frequency Sequence of character counts, in…

link For each player we can keep four counts: (1) per row (2) per col (3) the major diagonal (4) the minor diagonal. After the move, if any of these four counts is , the player won. Operation Time Space __init__ move For better readability we could encapsulate counts in a Player class.

link Sort Sorted, anagrams must be identical. Time: , space: . Character map Every character of t must appear in s possibly in different order. Time: , space: .

link As we build {char -> count} map, we can find out the count of characters that appear odd number of times. For a palindrome there can be at most 1 such character. Time: , space: . We can also maintain a set of characters that appear odd number of times.