Count Inversions

Intuition

An inversion is a pair (i, j) where i < j but arr[i] > arr[j]. Merge sort counts inversions during the merge step: when an element from the right half is placed before elements from the left half, each of those left-half elements forms an inversion with it.

Algorithm

1Merge sort recursively. Count inversions during merge.
2During merge: whenever right[j] < left[i], all elements left[i..mid] are inversions with right[j]. inv_count += (mid âˆ’ i + 1).

Example Walkthrough

Input: arr = [2,4,1,3,5]

1.Merge [2,4] and [1,3]: 1<2 â†’ 2 inversions. 3>2, 3>4: 0.
2.Merge [2,4,1,3] and [5]: 0 inversions.
3.Total = 3 (pairs: (2,1),(4,1),(4,3)).

Output: 3

Common Pitfalls

•Count inversions during merge (not during split) â€” only then do you know the relative positions.

Count Inversions.java

Java

// Approach: Modified merge sort. Count inversions during the merge step when a right element is placed before left elements.
// Time: O(n log n) Space: O(n)
import java.util.*;

class Solution {
    // Function to count inversions in the array.
    static int inversionCount(int arr[]) {
        return mergeSort(arr, 0, arr.length - 1);
    }

    private static int mergeSort(int[] arr, int left, int right) {
        int count = 0;
        if (left < right) {
            int mid = left + (right - left) / 2;
            count += mergeSort(arr, left, mid);
            count += mergeSort(arr, mid + 1, right);
            count += merge(arr, mid, left, right);
        }
        return count;
    }

    private static int merge(int[] arr, int mid, int left, int right) {
        int l = left, r = mid + 1;
        int count = 0;
        ArrayList<Integer> temp = new ArrayList<>();
        while (l <= mid && r <= right) {
            if (arr[l] <= arr[r]) {
                temp.add(arr[l]);
                l++;
            } else {
                count += (mid - l + 1);
                temp.add(arr[r]);
                r++;
            }
        }
        while (l <= mid) {
            temp.add(arr[l]);
            l++;
        }
        while (r <= right) {
            temp.add(arr[r]);
            r++;
        }
        for (int i = left; i <= right; i++)
            arr[i] = temp.get(i - left);

        return count;
    }
}

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