Merge branch 'master' into testing/MinStackUsingTwoStacks

Author: alxkm

src/main/java/com/thealgorithms/datastructures/disjointsetunion/DisjointSetUnion.java

@@ -1,53 +1,65 @@
 package com.thealgorithms.datastructures.disjointsetunion;
 
 /**
- * Disjoint Set Union or DSU is useful for solving problems related to connected components,
- * cycle detection in graphs, and maintaining relationships in disjoint sets of data.
- * It is commonly employed in graph algorithms and problems.
+ * Disjoint Set Union (DSU), also known as Union-Find, is a data structure that tracks a set of elements
+ * partitioned into disjoint (non-overlapping) subsets. It supports two primary operations efficiently:
  *
- * @see <a href="https://en.wikipedia.org/wiki/Disjoint-set_data_structure">Disjoint Set Union</a>
+ * <ul>
+ *     <li>Find: Determine which subset a particular element belongs to.</li>
+ *     <li>Union: Merge two subsets into a single subset.</li>
+ * </ul>
+ *
+ * @see <a href="https://en.wikipedia.org/wiki/Disjoint-set_data_structure">Disjoint Set Union (Wikipedia)</a>
  */
 public class DisjointSetUnion<T> {
 
     /**
-     * Creates a new node of DSU with parent initialised as same node
+     * Creates a new disjoint set containing the single specified element.
+     *
+     * @param value the element to be placed in a new singleton set
+     * @return a node representing the new set
      */
-    public Node<T> makeSet(final T x) {
-        return new Node<T>(x);
+    public Node<T> makeSet(final T value) {
+        return new Node<>(value);
     }
 
     /**
-     * Finds and returns the representative (root) element of the set to which a given element belongs.
-     * This operation uses path compression to optimize future findSet operations.
+     * Finds and returns the representative (root) of the set containing the given node.
+     * This method applies path compression to flatten the tree structure for future efficiency.
+     *
+     * @param node the node whose set representative is to be found
+     * @return the representative (root) node of the set
      */
     public Node<T> findSet(Node<T> node) {
-        while (node != node.parent) {
-            node = node.parent;
+        if (node != node.parent) {
+            node.parent = findSet(node.parent);
         }
-        return node;
+        return node.parent;
     }
 
     /**
-     * Unions two sets by merging their representative elements. The merge is performed based on the rank of each set
-     * to ensure efficient merging and path compression to optimize future findSet operations.
+     * Merges the sets containing the two given nodes. Union by rank is used to attach the smaller tree under the larger one.
+     * If both sets have the same rank, one becomes the parent and its rank is incremented.
+     *
+     * @param x a node in the first set
+     * @param y a node in the second set
      */
-    public void unionSets(final Node<T> x, final Node<T> y) {
-        Node<T> nx = findSet(x);
-        Node<T> ny = findSet(y);
+    public void unionSets(Node<T> x, Node<T> y) {
+        Node<T> rootX = findSet(x);
+        Node<T> rootY = findSet(y);
 
-        if (nx == ny) {
-            return; // Both elements already belong to the same set.
+        if (rootX == rootY) {
+            return; // They are already in the same set
         }
         // Merging happens based on rank of node, this is done to avoid long chaining of nodes and reduce time
         // to find root of the component. Idea is to attach small components in big, instead of other way around.
-        if (nx.rank > ny.rank) {
-            ny.parent = nx;
-        } else if (ny.rank > nx.rank) {
-            nx.parent = ny;
+        if (rootX.rank > rootY.rank) {
+            rootY.parent = rootX;
+        } else if (rootY.rank > rootX.rank) {
+            rootX.parent = rootY;
         } else {
-            // Both sets have the same rank; choose one as the parent and increment the rank.
-            ny.parent = nx;
-            nx.rank++;
+            rootY.parent = rootX;
+            rootX.rank++;
         }
     }
 }

src/main/java/com/thealgorithms/datastructures/hashmap/hashing/Intersection.java

@@ -8,60 +8,66 @@
 
 /**
  * The {@code Intersection} class provides a method to compute the intersection of two integer arrays.
- * The intersection is defined as the set of common elements present in both arrays.
  * <p>
- * This class utilizes a HashMap to efficiently count occurrences of elements in the first array,
- * allowing for an efficient lookup of common elements in the second array.
+ * This intersection includes duplicate values — meaning elements are included in the result
+ * as many times as they appear in both arrays (i.e., multiset intersection).
  * </p>
  *
  * <p>
- * Example:
- * <pre>
+ * The algorithm uses a {@link java.util.HashMap} to count occurrences of elements in the first array,
+ * then iterates through the second array to collect common elements based on these counts.
+ * </p>
+ *
+ * <p>
+ * Example usage:
+ * <pre>{@code
  * int[] array1 = {1, 2, 2, 1};
  * int[] array2 = {2, 2};
- * List<Integer> result = Intersection.intersection(array1, array2); // result will contain [2, 2]
- * </pre>
+ * List<Integer> result = Intersection.intersection(array1, array2); // result: [2, 2]
+ * }</pre>
  * </p>
  *
  * <p>
- * Note: The order of the returned list may vary since it depends on the order of elements
- * in the input arrays.
+ * Note: The order of elements in the returned list depends on the order in the second input array.
  * </p>
  */
 public final class Intersection {
 
+    private Intersection() {
+        // Utility class; prevent instantiation
+    }
+
     /**
-     * Computes the intersection of two integer arrays.
+     * Computes the intersection of two integer arrays, preserving element frequency.
+     * For example, given [1,2,2,3] and [2,2,4], the result will be [2,2].
+     *
      * Steps:
-     * 1. Count the occurrences of each element in the first array using a HashMap.
-     * 2. Iterate over the second array and check if the element is present in the HashMap.
-     * If it is, add it to the result list and decrement the count in the HashMap.
-     * 3. Return the result list containing the intersection of the two arrays.
+     * 1. Count the occurrences of each element in the first array using a map.
+     * 2. Iterate over the second array and collect common elements.
      *
      * @param arr1 the first array of integers
      * @param arr2 the second array of integers
-     * @return a list containing the intersection of the two arrays, or an empty list if either array is null or empty
+     * @return a list containing the intersection of the two arrays (with duplicates),
+     *         or an empty list if either array is null or empty
      */
     public static List<Integer> intersection(int[] arr1, int[] arr2) {
         if (arr1 == null || arr2 == null || arr1.length == 0 || arr2.length == 0) {
             return Collections.emptyList();
         }
 
-        Map<Integer, Integer> cnt = new HashMap<>(16);
-        for (int v : arr1) {
-            cnt.put(v, cnt.getOrDefault(v, 0) + 1);
+        Map<Integer, Integer> countMap = new HashMap<>();
+        for (int num : arr1) {
+            countMap.put(num, countMap.getOrDefault(num, 0) + 1);
         }
 
-        List<Integer> res = new ArrayList<>();
-        for (int v : arr2) {
-            if (cnt.containsKey(v) && cnt.get(v) > 0) {
-                res.add(v);
-                cnt.put(v, cnt.get(v) - 1);
+        List<Integer> result = new ArrayList<>();
+        for (int num : arr2) {
+            if (countMap.getOrDefault(num, 0) > 0) {
+                result.add(num);
+                countMap.computeIfPresent(num, (k, v) -> v - 1);
             }
         }
-        return res;
-    }
 
-    private Intersection() {
+        return result;
     }
 }

src/main/java/com/thealgorithms/datastructures/hashmap/hashing/MajorityElement.java

@@ -1,8 +1,10 @@
 package com.thealgorithms.datastructures.hashmap.hashing;
 
 import java.util.ArrayList;
+import java.util.Collections;
 import java.util.HashMap;
 import java.util.List;
+import java.util.Map;
 
 /**
  * This class provides a method to find the majority element(s) in an array of integers.
@@ -18,13 +20,18 @@ private MajorityElement() {
      * Returns a list of majority element(s) from the given array of integers.
      *
      * @param nums an array of integers
-     * @return a list containing the majority element(s); returns an empty list if none exist
+     * @return a list containing the majority element(s); returns an empty list if none exist or input is null/empty
      */
     public static List<Integer> majority(int[] nums) {
-        HashMap<Integer, Integer> numToCount = new HashMap<>();
+        if (nums == null || nums.length == 0) {
+            return Collections.emptyList();
+        }
+
+        Map<Integer, Integer> numToCount = new HashMap<>();
         for (final var num : nums) {
             numToCount.merge(num, 1, Integer::sum);
         }
+
         List<Integer> majorityElements = new ArrayList<>();
         for (final var entry : numToCount.entrySet()) {
             if (entry.getValue() >= nums.length / 2) {

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

@@ -1,35 +1,39 @@
 package com.thealgorithms.dynamicprogramming;
-/*
-The Sum of Subset problem determines whether a subset of elements from a
-given array sums up to a specific target value.
-*/
+
+/**
+ * Utility class for solving the Subset Sum problem using a space-optimized dynamic programming approach.
+ *
+ * <p>This algorithm determines whether any subset of a given array sums up to a specific target value.</p>
+ *
+ * <p><b>Time Complexity:</b> O(n * sum)</p>
+ * <p><b>Space Complexity:</b> O(sum)</p>
+ */
 public final class SubsetSumSpaceOptimized {
     private SubsetSumSpaceOptimized() {
     }
+
     /**
-     * This method checks whether the subset of an array
-     * contains a given sum or not. This is an space
-     * optimized solution using 1D boolean array
-     * Time Complexity: O(n * sum), Space complexity: O(sum)
+     * Determines whether there exists a subset of the given array that adds up to the specified sum.
+     * This method uses a space-optimized dynamic programming approach with a 1D boolean array.
      *
-     * @param arr An array containing integers
-     * @param sum The target sum of the subset
-     * @return True or False
+     * @param nums The array of non-negative integers
+     * @param targetSum The desired subset sum
+     * @return {@code true} if such a subset exists, {@code false} otherwise
      */
-    public static boolean isSubsetSum(int[] arr, int sum) {
-        int n = arr.length;
-        // Declare the boolean array with size sum + 1
-        boolean[] dp = new boolean[sum + 1];
+    public static boolean isSubsetSum(int[] nums, int targetSum) {
+        if (targetSum < 0) {
+            return false; // Subset sum can't be negative
+        }
 
-        // Initialize the first element as true
-        dp[0] = true;
+        boolean[] dp = new boolean[targetSum + 1];
+        dp[0] = true; // Empty subset always sums to 0
 
-        // Find the subset sum using 1D array
-        for (int i = 0; i < n; i++) {
-            for (int j = sum; j >= arr[i]; j--) {
-                dp[j] = dp[j] || dp[j - arr[i]];
+        for (int number : nums) {
+            for (int j = targetSum; j >= number; j--) {
+                dp[j] = dp[j] || dp[j - number];
             }
         }
-        return dp[sum];
+
+        return dp[targetSum];
     }
 }

src/main/java/com/thealgorithms/maths/Convolution.java

@@ -23,24 +23,21 @@ public static double[] convolution(double[] a, double[] b) {
         double[] convolved = new double[a.length + b.length - 1];
 
         /*
-    The discrete convolution of two signals A and B is defined as:
-
-          A.length
-    C[i] = Σ (A[k]*B[i-k])
-          k=0
-
-    It's obvious that:  0 <= k <= A.length , 0 <= i <= A.length + B.length - 2  and  0 <= i-k <=
-    B.length - 1 From the last inequality we get that:  i - B.length + 1 <= k <= i and thus we get
-    the conditions below.
+         * Discrete convolution formula:
+         * C[i] = Σ A[k] * B[i - k]
+         * where k ranges over valid indices so that both A[k] and B[i-k] are in bounds.
          */
+
         for (int i = 0; i < convolved.length; i++) {
-            convolved[i] = 0;
-            int k = Math.max(i - b.length + 1, 0);
+            double sum = 0;
+            int kStart = Math.max(0, i - b.length + 1);
+            int kEnd = Math.min(i, a.length - 1);
 
-            while (k < i + 1 && k < a.length) {
-                convolved[i] += a[k] * b[i - k];
-                k++;
+            for (int k = kStart; k <= kEnd; k++) {
+                sum += a[k] * b[i - k];
             }
+
+            convolved[i] = sum;
         }
 
         return convolved;

src/main/java/com/thealgorithms/maths/Mode.java

@@ -3,46 +3,49 @@
 import java.util.ArrayList;
 import java.util.Collections;
 import java.util.HashMap;
+import java.util.List;
+import java.util.Map;
 
-/*
- * Find the mode of an array of numbers
- *
- * The mode of an array of numbers is the most frequently occurring number in the array,
- * or the most frequently occurring numbers if there are multiple numbers with the same frequency
+/**
+ * Utility class to calculate the mode(s) of an array of integers.
+ * <p>
+ * The mode of an array is the integer value(s) that occur most frequently.
+ * If multiple values have the same highest frequency, all such values are returned.
+ * </p>
  */
 public final class Mode {
     private Mode() {
     }
 
-    /*
-     * Find the mode of an array of integers
+    /**
+     * Computes the mode(s) of the specified array of integers.
+     * <p>
+     * If the input array is empty, this method returns {@code null}.
+     * If multiple numbers share the highest frequency, all are returned in the result array.
+     * </p>
      *
-     * @param numbers array of integers
-     * @return mode of the array
+     * @param numbers an array of integers to analyze
+     * @return an array containing the mode(s) of the input array, or {@code null} if the input is empty
      */
     public static int[] mode(final int[] numbers) {
         if (numbers.length == 0) {
             return null;
         }
 
-        HashMap<Integer, Integer> count = new HashMap<>();
+        Map<Integer, Integer> count = new HashMap<>();
 
         for (int num : numbers) {
-            if (count.containsKey(num)) {
-                count.put(num, count.get(num) + 1);
-            } else {
-                count.put(num, 1);
-            }
+            count.put(num, count.getOrDefault(num, 0) + 1);
         }
 
         int max = Collections.max(count.values());
-        ArrayList<Integer> modes = new ArrayList<>();
+        List<Integer> modes = new ArrayList<>();
 
         for (final var entry : count.entrySet()) {
             if (entry.getValue() == max) {
                 modes.add(entry.getKey());
             }
         }
-        return modes.stream().mapToInt(n -> n).toArray();
+        return modes.stream().mapToInt(Integer::intValue).toArray();
     }
 }