🚀Day 3. Count Inversions 🧠

The problem can be found at the following link: Problem Link

💡 Problem Description:

Given an array of integers arr[]. Find the Inversion Count in the array.

Two elements arr[i] and arr[j] form an inversion if arr[i] > arr[j] and i < j.

Inversion Count indicates how far (or close) the array is from being sorted:

• If the array is already sorted, the inversion count is 0.

• If the array is sorted in reverse order, the inversion count is maximum.

🔍 Example Walkthrough:

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

Explanation: The sequence 2, 4, 1, 3, 5 has three inversions: (2, 1), (4, 1), (4, 3).

Input: arr[] = [2, 3, 4, 5, 6] Output: 0

Explanation: The array is already sorted, so there are no inversions.

Input: arr[] = [10, 10, 10] Output: 0

Explanation: All elements of the array are the same, so there are no inversions.

Constraints:

• $1 ≤ arr.size() ≤ 10^5$

• $1 ≤ arr[i] ≤ 10^4$

🎯 My Approach:

1. Merge Sort-Based Counting:

   • The problem can be efficiently solved using a modified Merge Sort algorithm.

   • During the merge step, count the number of inversions based on the positions of elements in the two halves.

2. Steps:

   • Divide: Recursively divide the array into two halves.

   • Merge: Merge the two halves while counting inversions.

   • Count Inversions: For every pair (i, j) where arr[i] > arr[j] and i < j, increment the count.

3. Advantages:

   • This approach efficiently counts inversions with a time complexity of O(n log n), compared to the naive O(n^2) method.

🕒 Time and Auxiliary Space Complexity

• Expected Time Complexity: O(n log n), as the algorithm performs a logarithmic number of merge steps, with each step taking linear time.

• Expected Auxiliary Space Complexity: O(n), as we use an additional array for temporary storage during the merge process.

📝 Solution Code

Code (C)

int mergeAndCount(int arr[], int temp[], int left, int mid, int right) {
    int i = left, j = mid + 1, k = left, inversions = 0;

    while (i <= mid && j <= right) {
        if (arr[i] <= arr[j]) {
            temp[k++] = arr[i++];
        } else {
            temp[k++] = arr[j++];
            inversions += (mid - i + 1);
        }
    }

    while (i <= mid) temp[k++] = arr[i++];
    while (j <= right) temp[k++] = arr[j++];

    for (i = left; i <= right; i++) {
        arr[i] = temp[i];
    }

    return inversions;
}

int mergeSortAndCount(int arr[], int temp[], int left, int right) {
    int inversions = 0;
    if (left < right) {
        int mid = left + (right - left) / 2;

        inversions += mergeSortAndCount(arr, temp, left, mid);
        inversions += mergeSortAndCount(arr, temp, mid + 1, right);
        inversions += mergeAndCount(arr, temp, left, mid, right);
    }
    return inversions;
}
int countInversions(int arr[], int n) {
    int temp[n];
    return mergeSortAndCount(arr, temp, 0, n - 1);
}

Code (Cpp)

class Solution {
public:
    int mergeAndCount(vector<int>& arr, int left, int mid, int right) {
        int i = left, j = mid + 1, k = 0, inversions = 0;
        vector<int> temp(right - left + 1);

        while (i <= mid && j <= right) {
            if (arr[i] <= arr[j]) {
                temp[k++] = arr[i++];
            } else {
                temp[k++] = arr[j++];
                inversions += (mid - i + 1);
            }
        }

        while (i <= mid) temp[k++] = arr[i++];
        while (j <= right) temp[k++] = arr[j++];
        for (i = left, k = 0; i <= right; ++i, ++k) {
            arr[i] = temp[k];
        }

        return inversions;
    }

    int mergeSortAndCount(vector<int>& arr, int left, int right) {
        int inversions = 0;
        if (left < right) {
            int mid = left + (right - left) / 2;

            inversions += mergeSortAndCount(arr, left, mid);
            inversions += mergeSortAndCount(arr, mid + 1, right);
            inversions += mergeAndCount(arr, left, mid, right);
        }
        return inversions;
    }

    int inversionCount(vector<int>& arr) {
        return mergeSortAndCount(arr, 0, arr.size() - 1);
    }
};

Code (Java)

class Solution {
    private int mergeAndCount(int[] arr, int[] temp, int left, int mid, int right) {
        int i = left, j = mid + 1, k = left, inversions = 0;

        while (i <= mid && j <= right) {
            if (arr[i] <= arr[j]) {
                temp[k++] = arr[i++];
            } else {
                temp[k++] = arr[j++];
                inversions += (mid - i + 1);
            }
        }

        while (i <= mid) temp[k++] = arr[i++];
        while (j <= right) temp[k++] = arr[j++];

        for (i = left; i <= right; i++) {
            arr[i] = temp[i];
        }

        return inversions;
    }

    private int mergeSortAndCount(int[] arr, int[] temp, int left, int right) {
        int inversions = 0;
        if (left < right) {
            int mid = left + (right - left) / 2;

            inversions += mergeSortAndCount(arr, temp, left, mid);
            inversions += mergeSortAndCount(arr, temp, mid + 1, right);
            inversions += mergeAndCount(arr, temp, left, mid, right);
        }
        return inversions;
    }

    public int inversionCount(int[] arr) {
        int[] temp = new int[arr.length];
        return mergeSortAndCount(arr, temp, 0, arr.length - 1);
    }
}

Code (Python)

class Solution:
    def mergeAndCount(self, arr, temp, left, mid, right):
        i, j, k = left, mid + 1, left
        inversions = 0

        while i <= mid and j <= right:
            if arr[i] <= arr[j]:
                temp[k] = arr[i]
                i += 1
            else:
                temp[k] = arr[j]
                j += 1
                inversions += (mid - i + 1)
            k += 1

        while i <= mid:
            temp[k] = arr[i]
            i += 1
            k += 1

        while j <= right:
            temp[k] = arr[j]
            j += 1
            k += 1

        for i in range(left, right + 1):
            arr[i] = temp[i]

        return inversions

    def mergeSortAndCount(self, arr, temp, left, right):
        inversions = 0
        if left < right:
            mid = left + (right - left) // 2

            inversions += self.mergeSortAndCount(arr, temp, left, mid)
            inversions += self.mergeSortAndCount(arr, temp, mid + 1, right)
            inversions += self.mergeAndCount(arr, temp, left, mid, right)
        return inversions

    def inversionCount(self, arr):
        n = len(arr)
        temp = [0] * n
        return self.mergeSortAndCount(arr, temp, 0, n - 1)

🎯 Contribution and Support:

For discussions, questions, or doubts related to this solution, feel free to connect on LinkedIn: Any Questions. Let’s make this learning journey more collaborative!

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