refactor: Enhance docs, code, add tests in `MaximumSumOfDistinctSubar… (#6649)

* refactor: Enhance docs, code, add tests in `MaximumSumOfDistinctSubarraysWithLengthK`

* Fix

* Fix spotbug

* Fix

* Fix

* Fix

* Fix

Author: Hardvan

src/main/java/com/thealgorithms/others/MaximumSumOfDistinctSubarraysWithLengthK.java

@@ -1,69 +1,89 @@
 package com.thealgorithms.others;
 
-import java.util.HashSet;
-import java.util.Set;
+import java.util.HashMap;
+import java.util.Map;
 
 /**
- * References: https://en.wikipedia.org/wiki/Streaming_algorithm
+ * Algorithm to find the maximum sum of a subarray of size K with all distinct
+ * elements.
  *
- * This model involves computing the maximum sum of subarrays of a fixed size \( K \) from a stream of integers.
- * As the stream progresses, elements from the end of the window are removed, and new elements from the stream are added.
+ * This implementation uses a sliding window approach with a hash map to
+ * efficiently
+ * track element frequencies within the current window. The algorithm maintains
+ * a window
+ * of size K and slides it across the array, ensuring all elements in the window
+ * are distinct.
  *
+ * Time Complexity: O(n) where n is the length of the input array
+ * Space Complexity: O(k) for storing elements in the hash map
+ *
+ * @see <a href="https://en.wikipedia.org/wiki/Streaming_algorithm">Streaming
+ *      Algorithm</a>
+ * @see <a href="https://en.wikipedia.org/wiki/Sliding_window_protocol">Sliding
+ *      Window</a>
  * @author Swarga-codes (https://github.com/Swarga-codes)
  */
 public final class MaximumSumOfDistinctSubarraysWithLengthK {
     private MaximumSumOfDistinctSubarraysWithLengthK() {
     }
 
     /**
-     * Finds the maximum sum of a subarray of size K consisting of distinct elements.
+     * Finds the maximum sum of a subarray of size K consisting of distinct
+     * elements.
      *
-     * @param k The size of the subarray.
-     * @param nums The array from which subarrays will be considered.
+     * The algorithm uses a sliding window technique with a frequency map to track
+     * the count of each element in the current window. A window is valid only if
+     * all K elements are distinct (frequency map size equals K).
      *
-     * @return The maximum sum of any distinct-element subarray of size K. If no such subarray exists, returns 0.
+     * @param k    The size of the subarray. Must be non-negative.
+     * @param nums The array from which subarrays will be considered.
+     * @return The maximum sum of any distinct-element subarray of size K.
+     *         Returns 0 if no such subarray exists or if k is 0 or negative.
+     * @throws IllegalArgumentException if k is negative
      */
     public static long maximumSubarraySum(int k, int... nums) {
-        if (nums.length < k) {
+        if (k <= 0 || nums == null || nums.length < k) {
             return 0;
         }
-        long masSum = 0; // Variable to store the maximum sum of distinct subarrays
-        long currentSum = 0; // Variable to store the sum of the current subarray
-        Set<Integer> currentSet = new HashSet<>(); // Set to track distinct elements in the current subarray
 
-        // Initialize the first window
+        long maxSum = 0;
+        long currentSum = 0;
+        Map<Integer, Integer> frequencyMap = new HashMap<>();
+
+        // Initialize the first window of size k
         for (int i = 0; i < k; i++) {
             currentSum += nums[i];
-            currentSet.add(nums[i]);
+            frequencyMap.put(nums[i], frequencyMap.getOrDefault(nums[i], 0) + 1);
         }
-        // If the first window contains distinct elements, update maxSum
-        if (currentSet.size() == k) {
-            masSum = currentSum;
+
+        // Check if the first window has all distinct elements
+        if (frequencyMap.size() == k) {
+            maxSum = currentSum;
         }
+
         // Slide the window across the array
-        for (int i = 1; i < nums.length - k + 1; i++) {
-            // Update the sum by removing the element that is sliding out and adding the new element
-            currentSum = currentSum - nums[i - 1];
-            currentSum = currentSum + nums[i + k - 1];
-            int j = i;
-            boolean flag = false; // flag value which says that the subarray contains distinct elements
-            while (j < i + k && currentSet.size() < k) {
-                if (nums[i - 1] == nums[j]) {
-                    flag = true;
-                    break;
-                } else {
-                    j++;
-                }
-            }
-            if (!flag) {
-                currentSet.remove(nums[i - 1]);
+        for (int i = k; i < nums.length; i++) {
+            // Remove the leftmost element from the window
+            int leftElement = nums[i - k];
+            currentSum -= leftElement;
+            int leftFrequency = frequencyMap.get(leftElement);
+            if (leftFrequency == 1) {
+                frequencyMap.remove(leftElement);
+            } else {
+                frequencyMap.put(leftElement, leftFrequency - 1);
             }
-            currentSet.add(nums[i + k - 1]);
-            // If the current window has distinct elements, compare and possibly update maxSum
-            if (currentSet.size() == k && masSum < currentSum) {
-                masSum = currentSum;
+
+            // Add the new rightmost element to the window
+            int rightElement = nums[i];
+            currentSum += rightElement;
+            frequencyMap.put(rightElement, frequencyMap.getOrDefault(rightElement, 0) + 1);
+
+            // If all elements in the window are distinct, update maxSum if needed
+            if (frequencyMap.size() == k && currentSum > maxSum) {
+                maxSum = currentSum;
             }
         }
-        return masSum; // the final maximum sum
+
+        return maxSum;
     }
 }

src/test/java/com/thealgorithms/others/MaximumSumOfDistinctSubarraysWithLengthKTest.java

@@ -3,20 +3,157 @@
 import static org.junit.jupiter.api.Assertions.assertEquals;
 
 import java.util.stream.Stream;
+import org.junit.jupiter.api.Test;
 import org.junit.jupiter.params.ParameterizedTest;
 import org.junit.jupiter.params.provider.Arguments;
 import org.junit.jupiter.params.provider.MethodSource;
 
-public class MaximumSumOfDistinctSubarraysWithLengthKTest {
+/**
+ * Test class for {@link MaximumSumOfDistinctSubarraysWithLengthK}.
+ *
+ * This class contains comprehensive test cases to verify the correctness of the
+ * maximum subarray sum algorithm with distinct elements constraint.
+ */
+class MaximumSumOfDistinctSubarraysWithLengthKTest {
 
+    /**
+     * Parameterized test for various input scenarios.
+     *
+     * @param expected the expected maximum sum
+     * @param k        the subarray size
+     * @param arr      the input array
+     */
     @ParameterizedTest
     @MethodSource("inputStream")
-    void testMaximumSubarraySum(int expected, int k, int[] arr) {
+    void testMaximumSubarraySum(long expected, int k, int[] arr) {
         assertEquals(expected, MaximumSumOfDistinctSubarraysWithLengthK.maximumSubarraySum(k, arr));
     }
 
+    /**
+     * Provides test cases for the parameterized test.
+     *
+     * Test cases cover:
+     * - Normal cases with distinct and duplicate elements
+     * - Edge cases (empty array, k = 0, k > array length)
+     * - Single element arrays
+     * - Arrays with all duplicates
+     * - Negative numbers
+     * - Large sums
+     *
+     * @return stream of test arguments
+     */
     private static Stream<Arguments> inputStream() {
-        return Stream.of(Arguments.of(15, 3, new int[] {1, 5, 4, 2, 9, 9, 9}), Arguments.of(0, 3, new int[] {4, 4, 4}), Arguments.of(12, 3, new int[] {9, 9, 9, 1, 2, 3}), Arguments.of(0, 0, new int[] {9, 9, 9}), Arguments.of(0, 5, new int[] {9, 9, 9}), Arguments.of(9, 1, new int[] {9, 2, 3, 7}),
-            Arguments.of(15, 5, new int[] {1, 2, 3, 4, 5}), Arguments.of(6, 3, new int[] {-1, 2, 3, 1, -2, 4}), Arguments.of(10, 1, new int[] {10}), Arguments.of(0, 2, new int[] {7, 7, 7, 7}), Arguments.of(0, 3, new int[] {}), Arguments.of(0, 10, new int[] {1, 2, 3}));
+        return Stream.of(
+            // Normal case: [5, 4, 2] has distinct elements with sum 11, but [4, 2, 9] also
+            // distinct with sum 15
+            Arguments.of(15L, 3, new int[] {1, 5, 4, 2, 9, 9, 9}),
+            // All elements are same, no distinct subarray of size 3
+            Arguments.of(0L, 3, new int[] {4, 4, 4}),
+            // First three have duplicates, but [1, 2, 3] are distinct with sum 6, wait
+            // [9,1,2] has sum 12
+            Arguments.of(12L, 3, new int[] {9, 9, 9, 1, 2, 3}),
+            // k = 0, should return 0
+            Arguments.of(0L, 0, new int[] {9, 9, 9}),
+            // k > array length, should return 0
+            Arguments.of(0L, 5, new int[] {9, 9, 9}),
+            // k = 1, single element (always distinct)
+            Arguments.of(9L, 1, new int[] {9, 2, 3, 7}),
+            // All distinct elements, size matches array
+            Arguments.of(15L, 5, new int[] {1, 2, 3, 4, 5}),
+            // Array with negative numbers
+            Arguments.of(6L, 3, new int[] {-1, 2, 3, 1, -2, 4}),
+            // Single element array
+            Arguments.of(10L, 1, new int[] {10}),
+            // All duplicates with k = 2
+            Arguments.of(0L, 2, new int[] {7, 7, 7, 7}),
+            // Empty array
+            Arguments.of(0L, 3, new int[] {}),
+            // k much larger than array length
+            Arguments.of(0L, 10, new int[] {1, 2, 3}));
+    }
+
+    /**
+     * Test with a larger array and larger k value.
+     */
+    @Test
+    void testLargerArray() {
+        int[] arr = new int[] {1, 2, 3, 4, 5, 6, 7, 8, 9, 10};
+        long result = MaximumSumOfDistinctSubarraysWithLengthK.maximumSubarraySum(5, arr);
+        // Maximum sum with 5 distinct elements: [6,7,8,9,10] = 40
+        assertEquals(40L, result);
+    }
+
+    /**
+     * Test with negative k value.
+     */
+    @Test
+    void testNegativeK() {
+        int[] arr = new int[] {1, 2, 3, 4, 5};
+        long result = MaximumSumOfDistinctSubarraysWithLengthK.maximumSubarraySum(-1, arr);
+        assertEquals(0L, result);
+    }
+
+    /**
+     * Test with null array.
+     */
+    @Test
+    void testNullArray() {
+        int[] nullArray = null;
+        long result = MaximumSumOfDistinctSubarraysWithLengthK.maximumSubarraySum(3, new int[][] {nullArray}[0]);
+        assertEquals(0L, result);
+    }
+
+    /**
+     * Test with array containing duplicates at boundaries.
+     */
+    @Test
+    void testDuplicatesAtBoundaries() {
+        int[] arr = new int[] {1, 1, 2, 3, 4, 4};
+        // [2, 3, 4] is the only valid window with sum 9
+        long result = MaximumSumOfDistinctSubarraysWithLengthK.maximumSubarraySum(3, arr);
+        assertEquals(9L, result);
+    }
+
+    /**
+     * Test with large numbers to verify long return type.
+     */
+    @Test
+    void testLargeNumbers() {
+        int[] arr = new int[] {1000000, 2000000, 3000000, 4000000};
+        // All elements are distinct, max sum with k=3 is [2000000, 3000000, 4000000] =
+        // 9000000
+        long result = MaximumSumOfDistinctSubarraysWithLengthK.maximumSubarraySum(3, arr);
+        assertEquals(9000000L, result);
+    }
+
+    /**
+     * Test where multiple windows have the same maximum sum.
+     */
+    @Test
+    void testMultipleMaxWindows() {
+        int[] arr = new int[] {1, 2, 3, 4, 3, 2, 1};
+        // Windows [1,2,3], [2,3,4], [4,3,2], [3,2,1] - max is [2,3,4] = 9
+        long result = MaximumSumOfDistinctSubarraysWithLengthK.maximumSubarraySum(3, arr);
+        assertEquals(9L, result);
+    }
+
+    /**
+     * Test with only two elements and k=2.
+     */
+    @Test
+    void testTwoElementsDistinct() {
+        int[] arr = new int[] {5, 10};
+        long result = MaximumSumOfDistinctSubarraysWithLengthK.maximumSubarraySum(2, arr);
+        assertEquals(15L, result);
+    }
+
+    /**
+     * Test with only two elements (duplicates) and k=2.
+     */
+    @Test
+    void testTwoElementsDuplicate() {
+        int[] arr = new int[] {5, 5};
+        long result = MaximumSumOfDistinctSubarraysWithLengthK.maximumSubarraySum(2, arr);
+        assertEquals(0L, result);
     }
 }