refactor: Intersection (#6379)

alxkm · web-flow · commit 287a708c7fec · 2025-07-15T17:04:58.000+02:00
refactor: Intersection improvement
diff --git a/src/main/java/com/thealgorithms/datastructures/hashmap/hashing/Intersection.java b/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;
     }
 }
