Program to check number of global and local inversions are same or not in Python

Suppose we have a list of distinct numbers called nums. Here a global inversion is when there are indices i < j such that nums[i] > nums[j]. And a local inversion is when there is an index i such that nums[i] > nums[i + 1]. We have to check whether the number of global inversions is equal to the number of local inversions or not.

So, if the input is like nums = [3, 2, 4], then the output will be True, as the indices 0 and 1 are both a global and local inversion.

Key Insight

Every local inversion is also a global inversion. So if the counts are equal, there should be no "extra" global inversions beyond the local ones. This means we only need to check if there's any element at position i that is greater than any element at position j where j > i + 1.

Algorithm

To solve this, we will follow these steps −

• Get the length of nums
• For each index i from 0 to length - 3, do
  • For each index j from i + 2 to length - 1, do
    • If nums[i] > nums[j], then return False
• Return True

Example

class Solution:
    def solve(self, nums):
        l = len(nums)
        for i in range(l - 2):
            for j in range(i + 2, l):
                if nums[i] > nums[j]:
                    return False
        return True

# Test the solution
ob = Solution()
nums = [3, 2, 4]
print(ob.solve(nums))

True

How It Works

Let's trace through the example [3, 2, 4] ?

nums = [3, 2, 4]

# Check all pairs where j > i + 1
# i=0, j=2: nums[0]=3, nums[2]=4 ? 3 < 4 ?
# No violations found, so return True

print("Global inversions:", [(i, j) for i in range(len(nums)) for j in range(i+1, len(nums)) if nums[i] > nums[j]])
print("Local inversions:", [(i, i+1) for i in range(len(nums)-1) if nums[i] > nums[i+1]])

Global inversions: [(0, 1)]
Local inversions: [(0, 1)]

Additional Examples

ob = Solution()

# Example 1: Equal counts
test1 = [1, 0, 2]
print(f"nums = {test1}: {ob.solve(test1)}")

# Example 2: Different counts  
test2 = [1, 2, 0]
print(f"nums = {test2}: {ob.solve(test2)}")

nums = [1, 0, 2]: True
nums = [1, 2, 0]: False

Conclusion

The solution efficiently checks if global and local inversion counts are equal by identifying any "extra" global inversions. If no element at position i is greater than any element at position j > i + 1, then the counts are equal.