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Leetcode 775. Global and Local Inversions

Problem Statement

You are given an integer array A of length n, containing all the integers from 0 to n-1 in some order. A global inversion is defined as a pair (i, j) where 0 <= i < j < n and A[i] > A[j].

A local inversion is defined as a pair (i, j) where 0 <= i < j < n, j = i + 1, and A[i] > A[j].

Return true if the number of global inversions is equal to the number of local inversions, and false otherwise.

Clarifying Questions

1. Is the input array guaranteed to contain all integers from 0 to n-1 without duplicates?
   • Yes, the problem statement guarantees it.
2. What is the maximum value of n?
   • The problem does not state this explicitly, but typical constraints would allow values up to around 10^5 for performance reasons.
3. Should we assume the input is valid, or do we need to validate it?
   • Assume the input is valid as per the problem constraints.

Strategy

1. Observation:
   • Every local inversion is a global inversion.
   • If the number of global inversions equals the number of local inversions, there must be no inversions that are “non-local”.
2. Key Insight:
   • For the conditions to hold true, any inversion (i, j) where j > i + 1 should not exist. This means that the array should almost be sorted except for the local inversions.
3. Approach:
   • Traverse the array while maintaining the maximum value found up to index i - 2.
   • If at any point A[i] < max(A[0]...A[i-2]), this means there is a global inversion that is not a local inversion.
4. Algorithm:
   • Initialize max_value to track the maximum value up to A[i-2].
   • Traverse from index 2 to n-1 and check if there exists any A[i] that is less than max_value.

Code

public class Solution {
    public boolean isIdealPermutation(int[] A) {
        int n = A.length;
        int max_value = A[0];
        
        // We start from the index 2 onwards
        for (int i = 2; i < n; i++) {
            if (A[i] < max_value) {
                return false;
            }
            max_value = Math.max(max_value, A[i-1]);
        }
        return true;
    }
}

Time Complexity

• Time Complexity: O(n)
  • We traverse the array exactly once, performing constant time operations for each element.
• Space Complexity: O(1)
  • We use only a few extra variables for tracking purposes, making the space complexity constant.

This approach ensures that we efficiently check the required conditions, leveraging the insight that any global inversion that isn’t local invalidates our condition.

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