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You are given an integer array nums of length n, where nums is a permutation of the numbers in the range [0, n - 1].

The number of global inversions is the number of the different pairs (i, j) where:

The number of local inversions is the number of indices i where:

Given an array nums, return true if the number of global inversions is equal to the number of local inversions.

Clarifying Questions

  1. Input Size: What is the typical size of the input array nums?
    • We need to know for assessing the efficiency.
  2. Permutations: Can we assume that nums will always be a permutation of numbers [0, n-1]?
    • This assumption simplifies the logic as each number is unique and we can leverage its properties.
  3. Edge Cases: Are there any edge cases we should be considerate of?
    • E.g., smallest possible arrays [0] or [1, 0].

Code

def isIdealPermutation(nums):
    n = len(nums)
    # Checking the main condition for ideal permutation
    for i in range(n):
        # If a number is found whose position difference is more than 1
        # it means there is a global inversion which is not local 
        # since local inversions must be between adjacent numbers
        if abs(nums[i] - i) > 1:
            return False
    return True

# Example Usage
print(isIdealPermutation([1, 0, 2]))  # Output should be True
print(isIdealPermutation([1, 2, 0]))  # Output should be False

Strategy

The key insight here is to recognize that for the number of global inversions to be equal to the number of local inversions, there cannot be any global inversions that are not local. In other words, every inversion must be between adjacent elements.

For each element nums[i] in the array nums, the ideal position it should be in is i because it’s a permutation of [0, n-1]. If nums[i] is off by more than 1 position from i, then there is a global inversion that is not a local inversion.

Time Complexity

The time complexity of this solution is O(n) where n is the number of elements in nums. This is because we are iterating through the array once performing a constant time check for each element. The space complexity is O(1) as we are using a constant amount of additional space.

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