feat: optimal-binary-search-tree challenge

Author: IbatoLionDev

3-HARD/Python/6-optimal-binary-search-tree/proposed-solution/__pycache__/optimal_bst.cpython-312.pyc

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+# Challenge: Given search frequencies for different elements, implement a dynamic programming algorithm
+# to construct an optimal binary search tree minimizing the average search cost.
+# Use object-oriented programming and follow the DRY principle.
+
+from optimal_bst import OptimalBST
+
+def main():
+    keys = [10, 12, 20]
+    freq = [34, 8, 50]
+    optimal_bst = OptimalBST(keys, freq)
+    cost = optimal_bst.optimal_cost()
+    print(f"Optimal BST cost: {cost}")
+
+if __name__ == "__main__":
+    main()

3-HARD/Python/6-optimal-binary-search-tree/proposed-solution/main.py

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+from typing import List
+
+class OptimalBST:
+    def __init__(self, keys: List[int], freq: List[int]):
+        self.keys = keys
+        self.freq = freq
+        self.n = len(keys)
+        self.dp = [[0] * self.n for _ in range(self.n)]
+
+    def optimal_cost(self) -> int:
+        # Precompute prefix sums for frequencies
+        prefix_sum = [0] * (self.n + 1)
+        for i in range(self.n):
+            prefix_sum[i + 1] = prefix_sum[i] + self.freq[i]
+
+        def get_sum(i, j):
+            return prefix_sum[j + 1] - prefix_sum[i]
+
+        def cost_for_root(i, j, r):
+            left = self.dp[i][r - 1] if r > i else 0
+            right = self.dp[r + 1][j] if r < j else 0
+            return left + right + get_sum(i, j)
+
+        for i in range(self.n):
+            self.dp[i][i] = self.freq[i]
+
+        for length in range(2, self.n + 1):
+            for i in range(self.n - length + 1):
+                j = i + length - 1
+                self.dp[i][j] = min(
+                    cost_for_root(i, j, r) for r in range(i, j + 1)
+                )
+
+        return self.dp[0][self.n - 1]