feat: Add SlidingWindowMaximum new algorithm with Junit tests

src/main/java/com/thealgorithms/datastructures/queues/SlidingWindowMaximum.java

@@ -0,0 +1,63 @@
+package com.thealgorithms.datastructures.queues;
+
+import java.util.Deque;
+import java.util.LinkedList;
+
+/**
+ * The {@code SlidingWindowMaximum} class provides a method to efficiently compute
+ * the maximum element within every sliding window of size {@code k} in a given array.
+ *
+ * <p>The algorithm uses a deque to maintain the indices of useful elements within
+ * the current sliding window. The time complexity of this approach is O(n) since
+ * each element is processed at most twice.
+ *
+ * @author Hardvan
+ */
+public final class SlidingWindowMaximum {
+    private SlidingWindowMaximum() {
+    }
+
+    /**
+     * Returns an array of the maximum values for each sliding window of size {@code k}.
+     * <p>If {@code nums} has fewer elements than {@code k}, the result will be an empty array.
+     * <p>Example:
+     * <pre>
+     * Input: nums = [1, 3, -1, -3, 5, 3, 6, 7], k = 3
+     * Output: [3, 3, 5, 5, 6, 7]
+     * </pre>
+     *
+     * @param nums the input array of integers
+     * @param k the size of the sliding window
+     * @return an array containing the maximum element for each sliding window
+     */
+    public static int[] maxSlidingWindow(int[] nums, int k) {
+        int n = nums.length;
+        if (n < k || k == 0) {
+            return new int[0];
+        }
+
+        int[] result = new int[n - k + 1];
+        Deque<Integer> deque = new LinkedList<>();
+        for (int i = 0; i < n; i++) {
+            // Remove elements from the front of the deque if they are out of the current window
+            if (!deque.isEmpty() && deque.peekFirst() < i - k + 1) {
+                deque.pollFirst();
+            }
+
+            // Remove elements from the back if they are smaller than the current element
+            while (!deque.isEmpty() && nums[deque.peekLast()] < nums[i]) {
+                deque.pollLast();
+            }
+
+            // Add the current element's index to the deque
+            deque.offerLast(i);
+
+            // Store the maximum element for the current window (starting from the k-1th element)
+            if (i >= k - 1) {
+                result[i - k + 1] = nums[deque.peekFirst()];
+            }
+        }
+
+        return result;
+    }
+}

src/test/java/com/thealgorithms/datastructures/queues/SlidingWindowMaximumTest.java

@@ -0,0 +1,50 @@
+package com.thealgorithms.datastructures.queues;
+
+import static org.junit.jupiter.api.Assertions.assertArrayEquals;
+
+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;
+
+public class SlidingWindowMaximumTest {
+
+    @ParameterizedTest
+    @MethodSource("provideTestCases")
+    public void testMaxSlidingWindow(int[] nums, int k, int[] expected) {
+        assertArrayEquals(expected, SlidingWindowMaximum.maxSlidingWindow(nums, k));
+    }
+
+    private static Stream<Arguments> provideTestCases() {
+        return Stream.of(
+            // Test case 1: Example from the problem statement
+            Arguments.of(new int[] {1, 3, -1, -3, 5, 3, 6, 7}, 3, new int[] {3, 3, 5, 5, 6, 7}),
+
+            // Test case 2: All elements are the same
+            Arguments.of(new int[] {4, 4, 4, 4, 4}, 2, new int[] {4, 4, 4, 4}),
+
+            // Test case 3: Window size equals the array length
+            Arguments.of(new int[] {2, 1, 5, 3, 6}, 5, new int[] {6}),
+
+            // Test case 4: Single element array with window size 1
+            Arguments.of(new int[] {7}, 1, new int[] {7}),
+
+            // Test case 5: Window size larger than the array length
+            Arguments.of(new int[] {1, 2, 3}, 4, new int[] {}),
+
+            // Test case 6: Decreasing sequence
+            Arguments.of(new int[] {9, 8, 7, 6, 5, 4}, 3, new int[] {9, 8, 7, 6}),
+
+            // Test case 7: Increasing sequence
+            Arguments.of(new int[] {1, 2, 3, 4, 5}, 2, new int[] {2, 3, 4, 5}),
+
+            // Test case 8: k is zero
+            Arguments.of(new int[] {1, 3, -1, -3, 5, 3, 6, 7}, 0, new int[] {}),
+
+            // Test case 9: Array with negative numbers
+            Arguments.of(new int[] {-4, -2, -5, -1, -3}, 3, new int[] {-2, -1, -1}),
+
+            // Test case 10: Empty array
+            Arguments.of(new int[] {}, 3, new int[] {}));
+    }
+}