Optimize inversion for in-place

Question

Find an efficient solution for inversion given the O(n) time constraint.

Codemia · Accepted Answer

To solve the problem of optimizing inversions in an array, we need to identify the number of inversions, which are pairs (i, j) such that i < j and array[i] > array[j]. Given the O(n) time constraint, a naive O(n^2) approach using nested loops is not feasible. Instead, we can utilize a modified merge sort algorithm, which allows us to count inversions in O(n log n) time. This approach involves recursively dividing the array and counting inversions during the merging process, making it efficient for larger datasets.

1. **Divide the Array**: Recursively split the array into two halves until each sub-array contains a single element.
2. **Merge and Count Inversions**: While merging two sorted halves, count how many elements from the right half are less than the current element from the left half. This count represents the number of inversions for those elements.
3. **Combine Counts**: The total number of inversions is the sum of inversions found in the left half, right half, and during the merge.

**Example**: For the array [3, 1, 2, 4], we recursively split into [3, 1] and [2, 4]. For [3, 1], during merging, we find that 1 is less than 3, thus counting one inversion. The process continues until we have counted all inversions.

```python
def merge_and_count(arr, temp_arr, left, mid, right):
    i = left    # Starting index for left subarray
    j = mid + 1 # Starting index for right subarray
    k = left    # Starting index to be sorted
    inv_count = 0

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

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

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

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

return inv_count

def merge_sort_and_count(arr, temp_arr, left, right):
    inv_count = 0
    if left < right:
        mid = (left + right) // 2
        inv_count += merge_sort_and_count(arr, temp_arr, left, mid)
        inv_count += merge_sort_and_count(arr, temp_arr, mid + 1, right)
        inv_count += merge_and_count(arr, temp_arr, left, mid, right)
    return inv_count

def count_inversions(arr):
    temp_arr = [0] * len(arr)
    return merge_sort_and_count(arr, temp_arr, 0, len(arr) - 1)

# Example usage:
arr = [3, 1, 2, 4]
print(count_inversions(arr))  # Output: 3
```

The time complexity of this solution is O(n log n) because the merge sort divides the array into halves (log n levels) and the merge step takes O(n) time for each level. The space complexity is O(n) due to the temporary array used for merging. This is efficient given the constraints, as it handles larger arrays within the required time.

1. **Empty Array**: An empty input should return 0 inversions. 
2. **Single Element**: An array with a single element also has 0 inversions. 
3. **Sorted Array**: An already sorted array should yield 0 inversions. 
4. **Reverse Sorted Array**: An array sorted in reverse should yield the maximum number of inversions, which is n(n-1)/2 for n elements. 
5. **Duplicate Elements**: An array with duplicates should still count inversions correctly, as the definition of inversions relies on the relative ordering of values.

Optimize inversion for in-place

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