grid47 Exploring patterns and algorithms
Aug 21, 2024
5 min read
Solution to LeetCode 775: Global and Local Inversions Problem
You are given an integer array nums of length n, which is a permutation of all integers in the range [0, n - 1]. A global inversion is a pair of indices (i, j) such that 0 <= i < j < n and nums[i] > nums[j]. A local inversion is a pair where 0 <= i < n - 1 and nums[i] > nums[i + 1]. Return true if the number of global inversions is equal to the number of local inversions.
Problem
Approach
Steps
Complexity
Input: The input consists of a single array nums, which represents a permutation of integers in the range [0, n - 1].
Example: Input: nums = [2, 0, 1]
Constraints:
• 1 <= nums.length <= 10^5
• 0 <= nums[i] < n
• All elements in nums are unique.
Output: Return true if the number of global inversions is equal to the number of local inversions.
Example: Output: true
Constraints:
• The number of global inversions must equal the number of local inversions.
Goal: The goal is to check if the number of global inversions is equal to the number of local inversions.
Steps:
• Iterate through the array, maintaining the current maximum value encountered.
• Check if any element is out of place beyond its local inversion (i.e., check for global inversions).
• If a global inversion is found that does not qualify as a local inversion, return false.
Goal: The constraints ensure that the array is a permutation of all integers from 0 to n-1 and the size of the array is manageable for the solution approach.
Steps:
• The array is a permutation of integers in the range [0, n - 1].
• The array size n is at most 10^5.
Assumptions:
• The input array will always be a valid permutation of integers from 0 to n-1.
• Input: Example 1: Input: nums = [2, 0, 1]
• Explanation: There is 1 global inversion (pair (0, 1)) and 1 local inversion (pair (1, 2)). Since the counts are equal, the result is true.
• Input: Example 2: Input: nums = [3, 2, 1, 0]
• Explanation: There are 3 global inversions (pairs (0, 1), (0, 2), (0, 3)) but only 2 local inversions (pairs (0, 1), (1, 2)). The counts are not equal, so the result is false.
Approach: The approach involves iterating over the array and comparing the elements to detect global inversions that are not local inversions. The maximum value encountered up to each index can be used to determine whether an inversion is a global or local one.
Observations:
• Global inversions can be detected by checking pairs (i, j) where i < j and nums[i] > nums[j].
• The maximum value encountered so far helps in identifying whether an inversion is local or global.
Steps:
• Iterate through the array while keeping track of the maximum value encountered.
• Check for any global inversion that does not qualify as a local inversion and return false if found.
• Return true if all global inversions are local inversions.
Empty Inputs:
• The array will always have at least one element, as n >= 1.
Large Inputs:
• Ensure the solution handles arrays with up to 10^5 elements.
Special Values:
• If the array is sorted, there are no inversions, hence the result will be true.
Constraints:
• The array is a permutation of integers from 0 to n-1.
boolisIdealPermutation(vector<int>& nums) {
int cmax =0, n = nums.size();
for(int i =0; i < n -2; i++) {
cmax = max(cmax, nums[i]);
if(cmax > nums[i +2]) returnfalse;
}
returntrue;
}
1 : Function Definition
boolisIdealPermutation(vector<int>& nums) {
Define the function that checks whether a given permutation of integers is 'ideal'.
2 : Variable Initialization
int cmax =0, n = nums.size();
Initialize the 'cmax' variable to store the maximum value encountered so far and 'n' to store the size of the input array.
3 : Loop Start
for(int i =0; i < n -2; i++) {
Start a loop that iterates through the array from index 0 to n-3.
4 : Max Update
cmax = max(cmax, nums[i]);
Update the 'cmax' variable to hold the maximum value found so far between 'cmax' and the current element 'nums[i]'.
5 : Condition Check
if(cmax > nums[i +2]) returnfalse;
Check if the maximum value encountered so far ('cmax') is greater than the element two positions ahead ('nums[i + 2]'). If true, return false, indicating the permutation is not ideal.
6 : Return Statement
returntrue;
Return true, indicating the permutation is ideal if the loop completes without returning false.
Best Case: O(n), where n is the length of the array.
Average Case: O(n), since we iterate over the array once.
Worst Case: O(n), as we only perform one pass through the array.
Description: The time complexity is linear because we only iterate through the array once.
Best Case: O(1), since we only need constant space to track the maximum value.
Worst Case: O(1), as no additional space is required except for the variables used in the logic.
Description: The space complexity is constant because we only use a few variables for tracking.
