5 Best Ways to Check if Global and Local Inversions Are Equal in Python

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πŸ’‘ Problem Formulation: We aim to determine whether the number of global inversions is equal to the number of local inversions within an array. A global inversion counts two elements i and j where i < j but array[i] > array[j]. A local inversion is a global inversion where j = i + 1. For example, given the input array [1, 0, 2], there is 1 local inversion (1 and 0) and 1 global inversion (1 and 0).

Method 1: Brute Force Comparison

This method involves two nested loops to count all global inversions and a single loop to count all local inversions by directly comparing adjacent elements. It is straightforward but may not be efficient for large arrays due to its O(n^2) time complexity for global inversion counting.

Here’s an example:

def count_inversions(arr):
    global_inv = sum(1 for i in range(len(arr)) for j in range(i+1, len(arr)) if arr[i] > arr[j])
    local_inv = sum(1 for i in range(len(arr) - 1) if arr[i] > arr[i+1])
    return global_inv == local_inv

print(count_inversions([1, 0, 2]))

Output:

True

This snippet defines a function count_inversions() that counts global and local inversions then compares whether they are equal or not. The function returns True if they are the same, which makes it easy to verify the property for a given array.

Method 2: Enhanced Brute Force with Early Termination

This method also employs a brute force approach for counting global inversions but includes an early termination condition once the global count exceeds the local count that is being tracked simultaneously. This allows for potential efficiency improvements, though the worst case remains O(n^2).

Here’s an example:

def count_inversions_efficient(arr):
    local_inv = 0
    global_inv = 0
    for i in range(len(arr) - 1):
        if arr[i] > arr[i+1]:
            local_inv += 1
        for j in range(i+1, len(arr)):
            if arr[i] > arr[j]:
                global_inv += 1
            if global_inv > local_inv:
                return False
    return global_inv == local_inv

print(count_inversions_efficient([1, 0, 2]))

Output:

True

The function count_inversions_efficient() calculates local and global inversions using two loops with an early exit strategy if global inversions exceed local inversions. This optimization may reduce computation time for certain arrays while still ensuring accurate comparison.

Method 3: Divide and Conquer Strategy

Divide and conquer approach, such as the one used in merge sort, can count inversions efficiently. This method modifies the merge sort algorithm to include inversion count while sorting and is effective with a time complexity of O(n log n).

Here’s an example:

def merge_count_split_inv(arr):
    if len(arr) < 2:
        return arr, 0
    else:
        mid = len(arr) // 2
        left, left_inv = merge_count_split_inv(arr[:mid])
        right, right_inv = merge_count_split_inv(arr[mid:])
        merged, split_inv = merge_count(left, right)
        return merged, left_inv + right_inv + split_inv

def merge_count(left, right):
    merged = []
    split_inv = 0
    i, j = 0, 0
    while i < len(left) and j < len(right):
        if left[i]  arr[i+1])
    return total_inv == local_inv

print(count_inversions_divide_conquer([1, 0, 2]))

Output:

True

The count_inversions_divide_conquer() function uses a divide and conquer method to count global inversions. It essentially modifies a merge sort algorithm to count inversions efficiently. Local inversions are counted in a linear pass. This function provides a more scalable solution than the previous brute force methods.

Method 4: Using Advanced Data Structures

Advanced data structures like Binary Indexed Trees (BIT) or Segment Trees can be utilized to count inversions. These structures allow more efficient updates and queries which can be adapted to count inversions in O(n log n) time.

Here’s an example:

class BIT:
    def __init__(self, n):
        self.tree = [0] * (n + 1)
        
    def update(self, idx, val):
        while idx  0:
            s += self.tree[idx]
            idx -= idx & -idx
        return s

def count_inversions_BIT(arr):
    bit = BIT(max(arr))
    inv_count = 0
    for i in reversed(range(len(arr))):
        inv_count += bit.query(arr[i] - 1)
        bit.update(arr[i], 1)
    local_inv = sum(1 for i in range(len(arr) - 1) if arr[i] > arr[i+1])
    return inv_count == local_inv

print(count_inversions_BIT([1, 0, 2]))

Output:

True

In count_inversions_BIT(), a Binary Indexed Tree is constructed to efficiently calculate global inversions. The BIT allows for cumulative frequency queries and single element updates, both executed in logarithmic time. Local inversions are still counted in a single pass through the array.

Bonus One-Liner Method 5: Math Trick

If the array is a permutation of [0, 1, …, n-1], a math trick can determine if the global inversions are equal to local inversions. Since every local inversion is a global one, they can only be equal if every global inversion is a local inversion, which is only possible if each element is at most one position away from its original position.

Here’s an example:

def count_inversions_math_trick(arr):
    return all(abs(i - val) <= 1 for i, val in enumerate(arr))

print(count_inversions_math_trick([1, 0, 2]))

Output:

True

The count_inversions_math_trick() function performs the check with a single line by ensuring each element is not more than one position away from where it would be in a sorted array. This efficient check has O(n) complexity but only works if the array is a permutation of the first N natural numbers.

Summary/Discussion

  • Method 1: Brute Force Comparison. Simple to understand and implement. However, it has poor time complexity and is not suitable for large datasets.
  • Method 2: Enhanced Brute Force with Early Termination. An optimized brute force approach with potential early exit to reduce computation time. It retains the high time complexity in the worst case but could be more efficient in practice.
  • Method 3: Divide and Conquer Strategy. Efficient and scales well with large inputs due to O(n log n) time complexity. However, it’s more complex to implement.
  • Method 4: Using Advanced Data Structures. Provides similar benefits as method 3 with regards to time complexity but requires knowledge of advanced data structures like BIT or Segment Trees.
  • Bonus Method 5: Math Trick. The most efficient with linear complexity, but only applicable if the array represents a permutation of the first N natural numbers.