Counting Inversions Coding Problem

May 19, 2019

** Inversion There is an array(a) and two indexes i and j. Inversion is the element for whom i < j and a[i] > a[j]

Example:

2 6 5 4 8

# Inversion pairs
(6, 5) (6, 4) (5, 4)

# Total inversion = 3

Counting Inversion Algorithm - Brute Force - O(n^2)

Lets look at simple brute force algorithm:

public void calculate_bad(int[] arr) {
    int count = 0;
    for (int i=0; i<arr.length-1; i++) {
        for (int j=i+1; j<arr.length; j++) {
            if (arr[i] > arr[j]) {
                count ++;
                System.out.println(arr[i] + " " + arr[j]);
            }
        }
    }
    System.out.println("Total inversions: " + count);
}

Above is a simple algorithm. Lets think about optimizing this.

Thoughts about Optimizing Algorithm

Next better number than O(n^2) is O(n log n) Can we do it in O(n log n)? Can we apply divide and conquer algorithm in this? If somehow, we can think what we will achieve by dividing the input array?

Hint Can we use Merge Sort? Yes

How In function where we merge two sorted sub-problems. Before sorting, their positions are actually relatively left and right which we want. The advantage of using merge sort is that, we know each of left and right array is sorted.

So, while comparing, one thing is for sure. For every element in left array, the index-i is always lesser than the index from right sorted sub-array. If we know, that one element in left sub-array which is greater than an element from right array.

Inversion Count algorithm

In above image, we know in left array that index=4 element is greater than element in right array at index=2 WHat it means, we found one inversion pair. In addition to that, we know every element in left array from index=4 onwards will be greater than element in right array where we found at index=2 Which means, in above example, we found total 3 inversion pairs for this condition.

And, we keep on continuing like this.

The code of algorithm taking O(n log n)

private void calculate_good_recursive(int[] arr, int l, int m, int r) {
    int l1 = m-l+1;
    int l2 = r-m;
    
    int[] left = new int[l1];
    for (int i=0; i<l1; i++) {
        left[i] = arr[l+i];
    }
    
    int[] right = new int[l2];
    for (int i=0; i<l2; i++) {
        right[i] = arr[m+i+1];
    }
    
    int lindex = 0;
    int rindex = 0;
    while (lindex < l1 && rindex < l2) {
        if (right[rindex] < left[lindex]) {
            for (int jj=lindex; jj<l1; jj++) {
                System.out.println(left[jj] + " " + right[rindex]);
            }
        }
        
        if (left[lindex] <= right[rindex]) {
            arr[l] = left[lindex];
            lindex ++;
        }
        else {
            arr[l] = right[rindex];
            rindex ++;
        }
        
        l++;
    }
    while (lindex < l1) {
        arr[l] = left[lindex];
        l++; lindex ++;
    }
    while (rindex < l2) {
        arr[l] = right[rindex];
        l++; rindex ++;
    }
}

public void inversions_merge_sort_method(int[] arr, int l, int r) {
    if (l < r) {
        int m = (l+r)/2;
        inversions_merge_sort_method(arr, l, m);
        inversions_merge_sort_method(arr, m+1, r);
        
        calculate_good_recursive(arr, l, m, r);
    }
}