feat: suffix-array-construction challenge

Author: IbatoLionDev

3-HARD/Python/3-suffix-array-construction/proposed-solution/__pycache__/suffix_array.cpython-312.pyc

@@ -0,0 +1,16 @@
+# Challenge: Construct a suffix array (and optionally an LCP array) for a given string,
+# enabling very efficient pattern searches. This is an advanced challenge in string processing.
+# Use object-oriented programming and follow the DRY principle.
+
+from suffix_array import SuffixArray
+
+def main():
+    text = "banana"
+    sa = SuffixArray(text)
+    suffix_arr = sa.build_suffix_array()
+    lcp_arr = sa.build_lcp_array()
+    print(f"Suffix Array: {suffix_arr}")
+    print(f"LCP Array: {lcp_arr}")
+
+if __name__ == "__main__":
+    main()

3-HARD/Python/3-suffix-array-construction/proposed-solution/main.py

@@ -0,0 +1,36 @@
+from typing import List
+
+class SuffixArray:
+    def __init__(self, text: str):
+        self.text = text
+        self.n = len(text)
+        self.suffix_array = []
+        self.lcp_array = []
+
+    def build_suffix_array(self) -> List[int]:
+        self.suffix_array = sorted(range(self.n), key=lambda i: self.text[i:])
+        return self.suffix_array
+
+    def build_lcp_array(self) -> List[int]:
+        if not self.suffix_array:
+            self.build_suffix_array()
+
+        self.lcp_array = [0] * self.n
+        inv_suffix = [0] * self.n
+
+        for i, suffix in enumerate(self.suffix_array):
+            inv_suffix[suffix] = i
+
+        k = 0
+        for i in range(self.n):
+            if inv_suffix[i] == self.n - 1:
+                k = 0
+                continue
+            j = self.suffix_array[inv_suffix[i] + 1]
+            while i + k < self.n and j + k < self.n and self.text[i + k] == self.text[j + k]:
+                k += 1
+            self.lcp_array[inv_suffix[i]] = k
+            if k:
+                k -= 1
+
+        return self.lcp_array