<p>We define the string <code>base</code> to be the infinite wraparound string of <code>&quot;abcdefghijklmnopqrstuvwxyz&quot;</code>, so <code>base</code> will look like this:</p>

<ul>
	<li><code>&quot;...zabcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyzabcd....&quot;</code>.</li>
</ul>

<p>Given a string <code>s</code>, return <em>the number of <strong>unique non-empty substrings</strong> of </em><code>s</code><em> are present in </em><code>base</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;a&quot;
<strong>Output:</strong> 1
<strong>Explanation:</strong> Only the substring &quot;a&quot; of s is in base.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;cac&quot;
<strong>Output:</strong> 2
<strong>Explanation:</strong> There are two substrings (&quot;a&quot;, &quot;c&quot;) of s in base.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;zab&quot;
<strong>Output:</strong> 6
<strong>Explanation:</strong> There are six substrings (&quot;z&quot;, &quot;a&quot;, &quot;b&quot;, &quot;za&quot;, &quot;ab&quot;, and &quot;zab&quot;) of s in base.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= s.length &lt;= 10<sup>5</sup></code></li>
	<li><code>s</code> consists of lowercase English letters.</li>
</ul>

class Solution:
    def findSubstringInWraproundString(self, s: str) -> int:
        return self._sol2(s)

    def _sol2(self, s: str) -> int:
        ord_a = ord("a")
        s = [ord(c) - ord_a for c in s]
        counts_ending_at_c = [0] * 26
        running_count = 1
        for i, c in enumerate(s):
            counts_ending_at_c[c] = max(counts_ending_at_c[c], running_count)
            if i < len(s) - 1 and s[i + 1] != (c + 1) % 26:
                running_count = 1
            else:
                running_count += 1
        return sum(counts_ending_at_c)





















    # Timing out, but otherwise succeeding on test cases
    def _sol1(self, s: str) -> int:
        if not s:
            return 0
        # ord("a") = 97, ord("z") = 122
        s_ord = [ord(char) - 97 for char in s]
        # the longest continuous substring that begins at this alphabet character, 0 = a, 25 = z
        longest_by_start = [0] * 26
        i = 0
        while i < len(s):
            j = i + 1
            while j < len(s) and (s_ord[j] == (s_ord[j - 1] + 1) or (s_ord[j] == 0 and s_ord[j - 1] == 25)):
                j += 1
            # j is now 1 higher than our large substring (i.e. j is an exclusive upper bound)
            longest_by_start[s[i]] = max(longest_by_start[s[i]], j - i)
            i = j
        # Process these chains to see which we can ignore because they're subsumed by others
        max_length = 0
        for i in range(26):
            if longest_by_start[i] > max_length:
                max_length = longest_by_start[i]
                longest_chain_i = i
        # There are potentially several chains we can ignore.
        # One is if the length of this chain is <= (max_length - difference in starts).
        #   For example a chain of 3 starting at c is subsumed by a chain of 5 starting at a.
        for i in range(26):
            start_diff = (i - longest_chain_i) % 26
            if longest_by_start[i] + start_diff <= max_length:
                longest_by_start[i] = 0
        # This is obviously not a full solution, but for now I'm going to see if it hits the test cases: just treat the largest
        # substring.
        # Ugh, actually this math is brutal. Huge number and would have to work with like "wheel" calculations to calculate number of
        # smaller duplicates.
        print(longest_by_start)
        return self.substring_count

    def _old_code(self) -> None:
        self.substrings = {(s_ord[0],)} # set of tuples. Any single character is always a substring, preload index 0 for simplicity.
        self.substring_count = 1
        def count_substrings(lower: int, upper: int) -> None:
            if lower == upper:
                return
            if tuple(s_ord[lower:upper]) in self.substrings:
                return
            self.substrings.add(tuple(s_ord[lower:upper]))
            self.substring_count += 1
            for i in range(lower + 1, upper):
                count_substrings(lower, i)
                count_substrings(i, upper)
            return

        # add every internal substring of s[i:j], which will also be a substring of the "base" string
        count_substrings(i, j)

    # I was very stumped and (correctly) guessed there is some kind of clever algorithm for this, found this in the solutions tab.
    # The below is not the original code, I've been decompressing and explaining it (while ensuring it still works after changes)
    # Perf: 28.5% time, 96 ms, 26.1% mem, 17.1 mb.
    #   Okay now 80.8% time 75 ms, 32.6% mem 17.1, after switching to arry and ord() rather than strings and substrings
    def found_code(self, s: str) -> int:
        endings_count_at_char = [0] * 26
        # We begin, with these 2 lines outside the loop, by manually processing the first character of s. Otherwise we'd have an out
        # -of-range index in the loop when trying to index on s[i - 1: i + 1]
        prev = ord(s[0]) - 97
        running_substring_length = 1
        endings_count_at_char[prev] = 1
        for i in range(1, len(s)):
            current = ord(s[i]) - 97
            if current - prev == 1 or (current == 0 and prev == 25):
                running_substring_length += 1
            else:
                # This character (i) does not form a valid connection with the previous (i - 1) and so the chain is broken
                running_substring_length = 1
            # Okay, LE is a dictionary holding the longest chain (possibly none or just 1) ending with the letter at s[i]
            endings_count_at_char[current] = max(endings_count_at_char[current], running_substring_length)
            prev = current
        return sum(endings_count_at_char)
    # Okay I think I'm coming to understand the above code. It was helpful to think through cases.
    # Consider 'abc'. Now as we step through we find 'a' as a substring, and record the longest known chain ending at letter a as 1.
    # Then we consider/add 'b', which adds 2 new substrings: 'ab' and 'b', corresponding to the 2 we record as longest chain at 'b'.
    # Then we add 'c', adding 'c', 'bc', and 'abc'.
    # What we are doing is that the length of the ongoing chain counts how many ending points there are to start strings that end at our
    #   current point. So with 'c', there are 3, corresponding to the 3 start points of 'a', 'b', and 'c'.
    # To go a bit more general still, that is to speak about more than just comparing to the repeating alphabet "base" string, we are
    #   determing the number of substrings that end at a given letter, and there are several reasons we can do this in these ways.
    #   One is that our processing for counting a sequence as 'valid' is due to a comparison that ... well what specifically matters is
    #   that for one long substring, every interior substring is also valid. So, just merely counting all substrings of the larger works.
    #   Another element deals with the repeating base string. We are unable to reference anything on the indices of the comparison string
    #   because it has no start or end, but the relation between all characters is identical for a given letter. In other words the
    #   letters you find going left or right are exactly the same starting from 'a', or 'f', etc, regardless of which of infinite 'a's or
    #   'f's you start from. This means that keying on the ending character gives us all we need, it is both sufficient and efficient for
    #   counting only unique substrings.
        
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