Merge Sort

import java.util.Arrays;

public class MergeSort {

    // Main sort function — DIVIDE step
    static void mergeSort(int[] arr, int left, int right) {
        if (left >= right) return;  // base case: 0 or 1 element

        int mid = left + (right - left) / 2;  // avoid overflow
        mergeSort(arr, left, mid);             // sort left half
        mergeSort(arr, mid + 1, right);        // sort right half
        merge(arr, left, mid, right);          // COMBINE step
    }

    // Merge two sorted subarrays arr[left..mid] and arr[mid+1..right]
    static void merge(int[] arr, int left, int mid, int right) {
        int n1 = mid - left + 1;
        int n2 = right - mid;

        // Create temp arrays
        int[] L = new int[n1];
        int[] R = new int[n2];

        for (int i = 0; i < n1; i++) L[i] = arr[left + i];
        for (int j = 0; j < n2; j++) R[j] = arr[mid + 1 + j];

        // Merge back into arr
        int i = 0, j = 0, k = left;
        while (i < n1 && j < n2) {
            if (L[i] <= R[j]) arr[k++] = L[i++];
            else               arr[k++] = R[j++];
        }
        while (i < n1) arr[k++] = L[i++];
        while (j < n2) arr[k++] = R[j++];
    }

    // Count inversions using merge sort (bonus)
    static long countInversions(int[] arr, int left, int right) {
        long count = 0;
        if (left < right) {
            int mid = (left + right) / 2;
            count += countInversions(arr, left, mid);
            count += countInversions(arr, mid + 1, right);
            count += mergeCount(arr, left, mid, right);
        }
        return count;
    }

    static long mergeCount(int[] arr, int left, int mid, int right) {
        int[] L = Arrays.copyOfRange(arr, left, mid + 1);
        int[] R = Arrays.copyOfRange(arr, mid + 1, right + 1);
        int i = 0, j = 0, k = left;
        long count = 0;
        while (i < L.length && j < R.length) {
            if (L[i] <= R[j]) arr[k++] = L[i++];
            else { arr[k++] = R[j++]; count += L.length - i; }
        }
        while (i < L.length) arr[k++] = L[i++];
        while (j < R.length) arr[k++] = R[j++];
        return count;
    }

    public static void main(String[] args) {
        int[] arr = {38, 27, 43, 3, 9, 82, 10};
        System.out.println("Before: " + Arrays.toString(arr));
        mergeSort(arr, 0, arr.length - 1);
        System.out.println("After:  " + Arrays.toString(arr));
        // After: [3, 9, 10, 27, 38, 43, 82]

        int[] inv = {2, 4, 1, 3, 5};
        System.out.println("Inversions: " + countInversions(inv, 0, inv.length - 1));  // 3
    }
}

/*
 * Recurrence: T(n) = 2T(n/2) + O(n)
 * Master Theorem Case 2: T(n) = O(n log n)
 *
 * Advantages:
 * - Guaranteed O(n log n) — no worst case like Quick Sort
 * - Stable sort — preserves relative order of equal elements
 * - Good for linked lists (no random access needed)
 *
 * Disadvantage:
 * - O(n) extra space for temporary arrays
 */