The palindrome condition boils down to: at most one character has an odd frequency across both strings combined.
Reduce each string to a bitmask representing the parity of character counts, since a palindrome can have at most one character with odd frequency. Then count pairs of strings whose bitmasks differ in at most one bit (including identical masks). Use a hash map to count frequencies of each mask and compute the total efficiently.
Pro tip: Mention that the bitmask approach works because there are only 26 lowercase letters, so the mask fits in an integer. Also, clarify that pairs with identical masks are valid because their combined character counts are all even.
Explain that a multiset of characters can form a palindrome if and only if at most one character has an odd count. This is the key insight.
For each string, compute a 26-bit integer where the i-th bit is 1 if the count of the i-th letter is odd, else 0. This mask captures the parity of character frequencies.
Two strings can form a palindrome together if their masks differ in at most one bit (i.e., XOR has at most one set bit). Use a hash map to count frequencies of each mask.
For each mask, add pairs with the same mask (choose 2) and pairs with masks that differ by exactly one bit. Iterate over all masks and their possible single-bit flips.
Time: O(N * 26) to compute masks and O(N * 26) to count pairs, so O(26N) = O(N). Space: O(N) for the hash map. Mention that 26 is constant.
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