feat(strings): add suffix array algorithm

strings/suffix_array.py

@@ -0,0 +1,79 @@
+"""
+https://en.wikipedia.org/wiki/Suffix_array
+https://cp-algorithms.com/string/suffix-array.html
+
+A suffix array lists the starting positions of every suffix of a string in
+lexicographic order. It needs one integer per suffix, where a suffix tree
+needs a node per character, so it is the compact structure used for pattern
+search and for the LCP-based algorithms built on top of it.
+
+Construction here uses prefix doubling (Manber-Myers).
+
+Time Complexity: O(n log² n) - where n is the length of the text
+Space Complexity: O(n)
+"""
+
+
+def suffix_array(text: str) -> list[int]:
+    """
+    Return the starting indices of all suffixes of `text`, in lexicographic
+    order.
+
+    >>> suffix_array("banana")
+    [5, 3, 1, 0, 4, 2]
+    >>> suffix_array("mississippi")
+    [10, 7, 4, 1, 0, 9, 8, 6, 3, 5, 2]
+    >>> suffix_array("abababab")
+    [6, 4, 2, 0, 7, 5, 3, 1]
+    >>> suffix_array("aaaa")
+    [3, 2, 1, 0]
+    >>> suffix_array("aça")
+    [2, 0, 1]
+    >>> suffix_array("a")
+    [0]
+    >>> suffix_array("")
+    []
+    """
+    n = len(text)
+    if n == 0:
+        return []
+
+    suffixes = list(range(n))
+    # Ranking by code point is enough to start: Python orders strings by code
+    # point too, so one comparison of the first character is already right.
+    rank = list(map(ord, text))
+    k = 1
+
+    while k < n:
+        # Sort by the first 2k characters of each suffix, which are exactly the
+        # pair (rank of the first k, rank of the next k). A second half that
+        # runs past the end gets -1: ranks are never negative, so a shorter
+        # suffix sorts before a longer one that shares its prefix.
+        keys = [(rank[i], rank[i + k] if i + k < n else -1, i) for i in range(n)]
+        keys.sort()
+        suffixes = [i for _, _, i in keys]
+
+        # Renumber so that equal 2k-prefixes share a rank and every distinct
+        # prefix gets the next one. This keeps ranks dense in range(n).
+        new_rank = [0] * n
+        for position in range(1, n):
+            previous, current = suffixes[position - 1], suffixes[position]
+            previous_key = (
+                rank[previous],
+                rank[previous + k] if previous + k < n else -1,
+            )
+            current_key = (
+                rank[current],
+                rank[current + k] if current + k < n else -1,
+            )
+            new_rank[current] = new_rank[previous] + int(previous_key != current_key)
+
+        rank = new_rank
+        # Dense ranks ending at n - 1 means every suffix has a unique rank, so
+        # the 2k-prefixes already order them completely; doubling cannot change
+        # the order, and the next round always breaks too, since 2k >= n.
+        if rank[suffixes[-1]] == n - 1:
+            break
+        k *= 2
+
+    return suffixes