feat: add OptimalBinarySearchTree algorithm

Author: oleksii-tumanov

src/main/java/com/thealgorithms/dynamicprogramming/OptimalBinarySearchTree.java

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+package com.thealgorithms.dynamicprogramming;
+
+import java.util.Arrays;
+import java.util.Comparator;
+
+/**
+ * Computes the minimum search cost of an optimal binary search tree.
+ *
+ * <p>The algorithm sorts the keys, preserves the corresponding search frequencies, and uses
+ * dynamic programming with Knuth's optimization to compute the minimum weighted search cost.
+ *
+ * <p>Example: if keys = [10, 12] and frequencies = [34, 50], the best tree puts 12 at the root
+ * and 10 as its left child. The total cost is 50 * 1 + 34 * 2 = 118.
+ *
+ * <p>Reference:
+ * https://en.wikipedia.org/wiki/Optimal_binary_search_tree
+ */
+public final class OptimalBinarySearchTree {
+    private OptimalBinarySearchTree() {
+    }
+
+    /**
+     * Computes the minimum weighted search cost for the given keys and search frequencies.
+     *
+     * @param keys the BST keys
+     * @param frequencies the search frequencies associated with the keys
+     * @return the minimum search cost
+     * @throws IllegalArgumentException if the input is invalid
+     */
+    public static long findOptimalCost(int[] keys, int[] frequencies) {
+        validateInput(keys, frequencies);
+        if (keys.length == 0) {
+            return 0L;
+        }
+
+        int[][] sortedNodes = sortNodes(keys, frequencies);
+        int nodeCount = sortedNodes.length;
+        long[] prefixSums = buildPrefixSums(sortedNodes);
+        long[][] optimalCost = new long[nodeCount][nodeCount];
+        int[][] root = new int[nodeCount][nodeCount];
+
+        // Small example:
+        // keys        = [10, 12]
+        // frequencies = [34, 50]
+        // Choosing 12 as the root gives cost 50 * 1 + 34 * 2 = 118,
+        // which is better than choosing 10 as the root.
+
+        // Base case: a subtree containing one key has cost equal to its frequency,
+        // because that key becomes the root of the subtree and is searched at depth 1.
+        for (int index = 0; index < nodeCount; index++) {
+            optimalCost[index][index] = sortedNodes[index][1];
+            root[index][index] = index;
+        }
+
+        // Build solutions for longer and longer key ranges.
+        // optimalCost[start][end] stores the minimum search cost for keys in that range.
+        for (int length = 2; length <= nodeCount; length++) {
+            for (int start = 0; start <= nodeCount - length; start++) {
+                int end = start + length - 1;
+
+                // Every key in this range moves one level deeper when we choose a root,
+                // so the sum of frequencies is added once to the subtree cost.
+                long frequencySum = prefixSums[end + 1] - prefixSums[start];
+                optimalCost[start][end] = Long.MAX_VALUE;
+
+                // Knuth's optimization:
+                // the best root for [start, end] lies between the best roots of
+                // [start, end - 1] and [start + 1, end], so we search only this interval.
+                int leftBoundary = root[start][end - 1];
+                int rightBoundary = root[start + 1][end];
+                for (int currentRoot = leftBoundary; currentRoot <= rightBoundary; currentRoot++) {
+                    long leftCost = currentRoot > start ? optimalCost[start][currentRoot - 1] : 0L;
+                    long rightCost = currentRoot < end ? optimalCost[currentRoot + 1][end] : 0L;
+                    long currentCost = frequencySum + leftCost + rightCost;
+
+                    if (currentCost < optimalCost[start][end]) {
+                        optimalCost[start][end] = currentCost;
+                        root[start][end] = currentRoot;
+                    }
+                }
+            }
+        }
+
+        return optimalCost[0][nodeCount - 1];
+    }
+
+    private static void validateInput(int[] keys, int[] frequencies) {
+        if (keys == null || frequencies == null) {
+            throw new IllegalArgumentException("Keys and frequencies cannot be null");
+        }
+        if (keys.length != frequencies.length) {
+            throw new IllegalArgumentException("Keys and frequencies must have the same length");
+        }
+
+        for (int frequency : frequencies) {
+            if (frequency < 0) {
+                throw new IllegalArgumentException("Frequencies cannot be negative");
+            }
+        }
+    }
+
+    private static int[][] sortNodes(int[] keys, int[] frequencies) {
+        int[][] sortedNodes = new int[keys.length][2];
+        for (int index = 0; index < keys.length; index++) {
+            sortedNodes[index][0] = keys[index];
+            sortedNodes[index][1] = frequencies[index];
+        }
+
+        // Sort by key so the nodes can be treated as an in-order BST sequence.
+        Arrays.sort(sortedNodes, Comparator.comparingInt(node -> node[0]));
+
+        for (int index = 1; index < sortedNodes.length; index++) {
+            if (sortedNodes[index - 1][0] == sortedNodes[index][0]) {
+                throw new IllegalArgumentException("Keys must be distinct");
+            }
+        }
+
+        return sortedNodes;
+    }
+
+    private static long[] buildPrefixSums(int[][] sortedNodes) {
+        long[] prefixSums = new long[sortedNodes.length + 1];
+        for (int index = 0; index < sortedNodes.length; index++) {
+            // prefixSums[i] holds the total frequency of the first i sorted keys.
+            // This lets us get the frequency sum of any range in O(1) time.
+            prefixSums[index + 1] = prefixSums[index] + sortedNodes[index][1];
+        }
+        return prefixSums;
+    }
+}

src/test/java/com/thealgorithms/dynamicprogramming/OptimalBinarySearchTreeTest.java

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+package com.thealgorithms.dynamicprogramming;
+
+import static org.junit.jupiter.api.Assertions.assertEquals;
+import static org.junit.jupiter.api.Assertions.assertThrows;
+
+import java.util.Arrays;
+import java.util.stream.Stream;
+import org.junit.jupiter.params.ParameterizedTest;
+import org.junit.jupiter.params.provider.Arguments;
+import org.junit.jupiter.params.provider.MethodSource;
+
+class OptimalBinarySearchTreeTest {
+
+    @ParameterizedTest
+    @MethodSource("validTestCases")
+    void testFindOptimalCost(int[] keys, int[] frequencies, long expectedCost) {
+        assertEquals(expectedCost, OptimalBinarySearchTree.findOptimalCost(keys, frequencies));
+    }
+
+    private static Stream<Arguments> validTestCases() {
+        return Stream.of(Arguments.of(new int[] {}, new int[] {}, 0L), Arguments.of(new int[] {15}, new int[] {9}, 9L), Arguments.of(new int[] {10, 12}, new int[] {34, 50}, 118L), Arguments.of(new int[] {20, 10, 30}, new int[] {50, 34, 8}, 134L),
+            Arguments.of(new int[] {12, 10, 20, 42, 25, 37}, new int[] {8, 34, 50, 3, 40, 30}, 324L), Arguments.of(new int[] {1, 2, 3}, new int[] {0, 0, 0}, 0L));
+    }
+
+    @ParameterizedTest
+    @MethodSource("crossCheckTestCases")
+    void testFindOptimalCostAgainstBruteForce(int[] keys, int[] frequencies) {
+        assertEquals(bruteForceOptimalCost(keys, frequencies), OptimalBinarySearchTree.findOptimalCost(keys, frequencies));
+    }
+
+    private static Stream<Arguments> crossCheckTestCases() {
+        return Stream.of(Arguments.of(new int[] {3, 1, 2}, new int[] {4, 2, 6}), Arguments.of(new int[] {5, 2, 8, 6}, new int[] {3, 7, 1, 4}), Arguments.of(new int[] {9, 4, 11, 2}, new int[] {1, 8, 2, 5}));
+    }
+
+    @ParameterizedTest
+    @MethodSource("invalidTestCases")
+    void testFindOptimalCostInvalidInput(int[] keys, int[] frequencies) {
+        assertThrows(IllegalArgumentException.class, () -> OptimalBinarySearchTree.findOptimalCost(keys, frequencies));
+    }
+
+    private static Stream<Arguments> invalidTestCases() {
+        return Stream.of(Arguments.of(null, new int[] {}), Arguments.of(new int[] {}, null), Arguments.of(new int[] {1, 2}, new int[] {3}), Arguments.of(new int[] {1, 1}, new int[] {2, 3}), Arguments.of(new int[] {1, 2}, new int[] {3, -1}));
+    }
+
+    private static long bruteForceOptimalCost(int[] keys, int[] frequencies) {
+        int[][] sortedNodes = new int[keys.length][2];
+        for (int index = 0; index < keys.length; index++) {
+            sortedNodes[index][0] = keys[index];
+            sortedNodes[index][1] = frequencies[index];
+        }
+        Arrays.sort(sortedNodes, java.util.Comparator.comparingInt(node -> node[0]));
+
+        int[] sortedFrequencies = new int[sortedNodes.length];
+        for (int index = 0; index < sortedNodes.length; index++) {
+            sortedFrequencies[index] = sortedNodes[index][1];
+        }
+
+        return bruteForceOptimalCost(sortedFrequencies, 0, sortedFrequencies.length - 1, 1);
+    }
+
+    private static long bruteForceOptimalCost(int[] frequencies, int start, int end, int depth) {
+        if (start > end) {
+            return 0L;
+        }
+
+        long minimumCost = Long.MAX_VALUE;
+        for (int root = start; root <= end; root++) {
+            long currentCost = (long) depth * frequencies[root] + bruteForceOptimalCost(frequencies, start, root - 1, depth + 1) + bruteForceOptimalCost(frequencies, root + 1, end, depth + 1);
+            minimumCost = Math.min(minimumCost, currentCost);
+        }
+        return minimumCost;
+    }
+}