Add CountingInversions algorithm (#5745)

src/main/java/com/thealgorithms/divideandconquer/CountingInversions.java

@@ -0,0 +1,101 @@
+package com.thealgorithms.divideandconquer;
+
+/**
+ * A utility class for counting the number of inversions in an array.
+ * <p>
+ * An inversion is a pair (i, j) such that i < j and arr[i] > arr[j].
+ * This class implements a divide-and-conquer approach, similar to merge sort,
+ * to count the number of inversions efficiently.
+ * <p>
+ * Time Complexity: O(n log n)
+ * Space Complexity: O(n) (due to temporary arrays during merge step)
+ *
+ * <p>Applications:
+ * - Used in algorithms related to sorting and permutation analysis.
+ * - Helps in determining how far an array is from being sorted.
+ * - Applicable in bioinformatics and signal processing.
+ *
+ * <p>This class cannot be instantiated, as it is intended to provide
+ * only static utility methods.
+ *
+ * @author Hardvan
+ */
+public final class CountingInversions {
+    private CountingInversions() {
+    }
+
+    /**
+     * Counts the number of inversions in the given array.
+     *
+     * @param arr The input array of integers.
+     * @return The total number of inversions in the array.
+     */
+    public static int countInversions(int[] arr) {
+        return mergeSortAndCount(arr, 0, arr.length - 1);
+    }
+
+    /**
+     * Recursively divides the array into two halves, sorts them, and counts
+     * the number of inversions. Uses a modified merge sort approach.
+     *
+     * @param arr  The input array.
+     * @param left The starting index of the current segment.
+     * @param right The ending index of the current segment.
+     * @return The number of inversions within the segment [left, right].
+     */
+    private static int mergeSortAndCount(int[] arr, int left, int right) {
+        if (left >= right) {
+            return 0;
+        }
+
+        int mid = left + (right - left) / 2;
+        int inversions = 0;
+
+        inversions += mergeSortAndCount(arr, left, mid);
+        inversions += mergeSortAndCount(arr, mid + 1, right);
+        inversions += mergeAndCount(arr, left, mid, right);
+        return inversions;
+    }
+
+    /**
+     * Merges two sorted subarrays and counts the cross-inversions between them.
+     * A cross-inversion occurs when an element from the right subarray is
+     * smaller than an element from the left subarray.
+     *
+     * @param arr The input array.
+     * @param left The starting index of the first subarray.
+     * @param mid The ending index of the first subarray and midpoint of the segment.
+     * @param right The ending index of the second subarray.
+     * @return The number of cross-inversions between the two subarrays.
+     */
+    private static int mergeAndCount(int[] arr, int left, int mid, int right) {
+        int[] leftArr = new int[mid - left + 1];
+        int[] rightArr = new int[right - mid];
+
+        System.arraycopy(arr, left, leftArr, 0, mid - left + 1);
+        System.arraycopy(arr, mid + 1, rightArr, 0, right - mid);
+
+        int i = 0;
+        int j = 0;
+        int k = left;
+        int inversions = 0;
+
+        while (i < leftArr.length && j < rightArr.length) {
+            if (leftArr[i] <= rightArr[j]) {
+                arr[k++] = leftArr[i++];
+            } else {
+                arr[k++] = rightArr[j++];
+                inversions += mid + 1 - left - i;
+            }
+        }
+
+        while (i < leftArr.length) {
+            arr[k++] = leftArr[i++];
+        }
+        while (j < rightArr.length) {
+            arr[k++] = rightArr[j++];
+        }
+
+        return inversions;
+    }
+}