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Leetcode 775. Global and Local Inversions

Problem Statement

You are given an integer array nums of length n which represents a permutation of all the integers in the range [0, n - 1].

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

• 0 <= i < j < n
• nums[i] > nums[j]

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

• 0 <= i < n - 1
• nums[i] > nums[i + 1]

Return true if the number of global inversions is equal to the number of local inversions.

Clarifying Questions

1. Can the length of the array n be 0 or 1?
   • For length n = 0, there are no elements, so global and local inversions are trivially equal.
   • For length n = 1, no inversions are possible because there’s only one element.
2. Is the array guaranteed to include all integers from 0 to n-1 without any duplicates?
   • Yes, the array is a permutation of the integers from 0 to n-1.

Strategy

To solve the problem, consider the definitions of global and local inversions:

• Every local inversion is a global inversion, but not all global inversions are local inversions.

Using this, we can deduce that for the number of global inversions to be equal to the number of local inversions, there should be no “non-local” global inversions.

A “non-local” global inversion (i < j, j > i + 1) would be detected if for any index i, there exists a j such that nums[i] > nums[j] and (j - i) > 1.

To determine this efficiently, we can keep track of the maximum value encountered up to the element just before the current one. If at any point this maximum value is greater than the current element, it indicates a “non-local” global inversion.

Code

#include <vector>
#include <algorithm>

class Solution {
public:
    bool isIdealPermutation(std::vector<int>& nums) {
        int max_val = -1;
        
        // Traverse the array
        for (int i = 0; i < nums.size() - 2; ++i) {
            max_val = std::max(max_val, nums[i]);
            if (max_val > nums[i + 2]) {
                return false;
            }
        }
        return true;
    }
};

Time Complexity

• Time Complexity: The loop runs in O(n) time, where n is the length of the input array nums.
  • Inside the loop, the max_val update is a constant-time operation.
• Space Complexity: O(1), as we are using only a constant amount of extra space for variables.

This solution efficiently checks the condition required to determine whether all global inversions are local inversions, ensuring that the algorithm runs in linear time suitable for large inputs.

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