There are at most 26 different lengths of the complete substrings: k *1, k * 2, … k * 26.****
提示 2
For each length, we can use sliding window to count the frequency of each letter in the window.
提示 3
We still need to check for all characters in the window that abs(word[i] - word[i - 1]) <= 2. We do this by maintaining the values of abs(word[i] - word[i - 1]) in the sliding window dynamically in an ordered multiset or priority queue, so that we know the maximum value at each iteration.
