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Count Inversions

Medium Arrays & Hashing Merge Sort Like Approach Original on GeeksforGeeks

Count Inversions is a medium arrays & hashing problem solved with the merge sort like approach pattern. The best approach, optimal (merge sort and count), runs in O(n log n) time and O(n) space. Below are 2 approaches in Java, from brute force up.

Problem

Two indices i < j form an inversion when arr[i] > arr[j]. Count the inversions in the array. The count measures how far the array is from being sorted.

Examples

Example 1

Input
arr = [5, 3, 2, 4, 1]
Output
8
Why
(5,3) (5,2) (5,4) (5,1) (3,2) (3,1) (2,1) (4,1).

Example 2

Input
arr = [1, 2, 3]
Output
0

Constraints

  • 1 <= arr.length <= 10^5
  • 1 <= arr[i] <= 10^4
  • The count can exceed 32 bits; use a long.

Animated walkthrough

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Solutions

Try it yourself first. Then compare: each approach lists its idea, the steps, its time and space, and the Java code.

ApproachTimeSpace
Brute forceO(n²)O(1)
Optimal (merge sort and count)O(n log n)O(n)

1Brute force

TimeO(n²)
SpaceO(1)

Check every pair i < j and count those with arr[i] > arr[j].

  1. Two nested loops, count arr[i] > arr[j].
Java
class Solution {
    static long inversionCount(int[] arr) {
        long count = 0;
        for (int i = 0; i < arr.length; i++)
            for (int j = i + 1; j < arr.length; j++)
                if (arr[i] > arr[j]) count++;
        return count;
    }
}

2Optimal (merge sort and count)

TimeO(n log n)Standard merge sort: log n levels, O(n) work each.
SpaceO(n)Temporary array used during merging.

Merge sort the array. While merging two sorted halves, if the right element is smaller than the left element, it is also smaller than every remaining element in the left half, so it adds mid - i + 1 inversions in one step.

  1. Recursively sort and count each half.
  2. Merge; when right[j] < left[i], add (mid - i + 1).
  3. Sum the counts from both halves and the merge.
Java
class Solution {
    static long inversionCount(int[] arr) {
        return sort(arr, new int[arr.length], 0, arr.length - 1);
    }

    private static long sort(int[] a, int[] tmp, int lo, int hi) {
        if (lo >= hi) return 0;
        int mid = (lo + hi) >>> 1;
        long count = sort(a, tmp, lo, mid) + sort(a, tmp, mid + 1, hi);
        int i = lo, j = mid + 1, k = lo;
        while (i <= mid && j <= hi) {
            if (a[i] <= a[j]) tmp[k++] = a[i++];
            else {
                count += mid - i + 1;
                tmp[k++] = a[j++];
            }
        }
        while (i <= mid) tmp[k++] = a[i++];
        while (j <= hi) tmp[k++] = a[j++];
        for (k = lo; k <= hi; k++) a[k] = tmp[k];
        return count;
    }
}

Edge cases to test

  • Sorted array (0) and reverse-sorted array (n(n - 1) / 2)
  • Equal values are not inversions

Hints

Hint 1

During a merge, when you take an element from the right half, how many left-half elements does it jump over?

FAQ

What is the best time complexity for Count Inversions?

Optimal (merge sort and count) runs in O(n log n) time and O(n) extra space. Standard merge sort: log n levels, O(n) work each.

Which pattern does Count Inversions use?

It is a arrays & hashing problem that uses the merge sort like approach pattern. Other problems with the same pattern: Arranging the array.

Is there a brute force solution for Count Inversions?

Yes. Brute force takes O(n²) time and O(1) space. Check every pair i < j and count those with arr[i] arr[j].

Which edge cases should I test for Count Inversions?

Sorted array (0) and reverse-sorted array (n(n - 1) / 2); Equal values are not inversions.