Count Inversions Problem in Kotlin

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In computer science, the count inversions problem is a classic algorithmic problem that involves counting the number of inversions in an array. An inversion is a pair of indices (i, j) in an array such that i < j and nums[i] > nums[j]. In other words, it represents two elements in the array that are in the wrong order relative to each other. The count inversions problem asks us to determine the total number of inversions in an array.

This problem is interesting because it has many real-world applications, such as in data analysis and sorting algorithms. For example, the number of inversions in an array is often used as a measure of its sortedness, and can be used to evaluate the performance of sorting algorithms. The count inversions problem can also be used to detect anomalies in data, such as detecting fraudulent transactions in financial data.

Solving the count inversions problem typically involves using a divide and conquer approach, such as merge sort or binary search tree, to efficiently count the number of inversions in the array.

Here’s a step-by-step algorithm for solving the count inversions problem using a modified merge sort approach:

  1. Implement a modified merge sort algorithm that counts the number of inversions in an array. This can be done by recursively dividing the input array in half until the subarrays have size 1, then merging the subarrays together while counting the number of inversions.
  2. In the merge step of the modified merge sort algorithm, while merging the two sorted subarrays, count the number of inversions that result from pairs of elements in the two subarrays that are out of order.
  3. Return the total number of inversions counted in the modified merge sort algorithm.

Here’s a Kotlin implementation of the modified merge sort algorithm for counting inversions:

Kotlin
fun countInversions(nums: IntArray): Long {
    val (sortedNums, inversions) = mergeSortAndCountInversions(nums)
    return inversions
}

fun mergeSortAndCountInversions(nums: IntArray): Pair<IntArray, Long> {
    if (nums.size <= 1) {
        return Pair(nums, 0)
    }

    // Divide the input array in half
    val mid = nums.size / 2
    val leftHalf = nums.sliceArray(0 until mid)
    val rightHalf = nums.sliceArray(mid until nums.size)

    // Recursively sort and count inversions in the two subarrays
    val (leftSorted, leftInversions) = mergeSortAndCountInversions(leftHalf)
    val (rightSorted, rightInversions) = mergeSortAndCountInversions(rightHalf)

    // Merge the sorted subarrays and count inversions
    var merged = IntArray(nums.size)
    var i = 0
    var j = 0
    var inversions = 0L
    while (i < leftSorted.size && j < rightSorted.size) {
        if (leftSorted[i] <= rightSorted[j]) {
            merged[i+j] = leftSorted[i]
            i++
        } else {
            merged[i+j] = rightSorted[j]
            j++
            inversions += leftSorted.size - i
        }
    }
    while (i < leftSorted.size) {
        merged[i+j] = leftSorted[i]
        i++
    }
    while (j < rightSorted.size) {
        merged[i+j] = rightSorted[j]
        j++
    }

    return Pair(merged, leftInversions + rightInversions + inversions)
}
fun main() {
    val nums = intArrayOf(1, 3, 5, 2, 4, 6)
    val inversions = countInversions(nums)
    println("Number of inversions: $inversions")
}

countInversions function

The countInversions function takes an array of integers and returns the number of inversions in the array. Here’s the step-by-step explanation:

  1. The function starts by calling the mergeSortAndCountInversions function, passing in the input array nums. This is a helper function that recursively sorts the input array and counts the number of inversions.
  2. The function then destructures the output of mergeSortAndCountInversions into two variables: sortedNums, which is the sorted version of the input array, and inversions, which is the number of inversions in the input array.
  3. Finally, the function returns the inversions variable, which contains the number of inversions in the input array.

mergeSortAndCountInversions function

The mergeSortAndCountInversions function is a helper function that recursively sorts the input array using a modified merge sort algorithm and counts the number of inversions in the array. Here’s the step-by-step explanation:

  1. The function starts by checking if the input array has only one element or is empty. If either of these conditions is true, the function simply returns the input array and the number of inversions, which is zero.
  2. If the input array has more than one element, the function divides the input array into two halves: leftHalf and rightHalf.
  3. The function then recursively calls mergeSortAndCountInversions on the leftHalf and rightHalf arrays, and destructures the output into leftSorted, leftInversions, rightSorted, and rightInversions. These variables contain the sorted versions of the left and right subarrays, as well as the number of inversions in each subarray.
  4. The function then creates an empty array merged to hold the sorted version of the input array, as well as variables i, j, and inversions. These variables are used to merge the two sorted subarrays and count the number of inversions.
  5. The function then enters a while loop that continues while both the leftSorted and rightSorted arrays have elements remaining to be merged. Within the loop, the function compares the first elements of the leftSorted and rightSorted arrays, and copies the smaller element into the merged array. If the element from the rightSorted array is smaller, the function also increments the inversions variable by the number of remaining elements in the leftSorted array. This is because the elements in the leftSorted array that come after the current element are all greater than the current element, and therefore form inversions with the current element.
  6. After the loop finishes, the function enters two more while loops to copy any remaining elements from the leftSorted and rightSorted arrays into the merged array.
  7. Finally, the function returns a pair of the merged array and the sum of leftInversions, rightInversions, and inversions. The merged array contains the sorted version of the input array, and the sum of inversions in the left and right subarrays plus the inversions formed by merging the subarrays is the total number of inversions in the input array.

The time complexity of the countInversions function is O(n log n), where n is the length of the input array. This is because the function uses a modified merge sort algorithm to sort the array, which has a time complexity of O(n log n) in the average and worst cases.

The space complexity of the countInversions function is O(n), where n is the length of the input array. This is because the function uses a temporary array to merge the sorted subarrays, which has a maximum size of n. Additionally, the function uses a constant amount of additional space for variables used in the merge sort algorithm.

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