Leetcode 775. Find The Global and Local Inversions (Medium)

By /

Leetcode 775. Find The Global and Local Inversions :Hey there, coding enthusiasts! Welcome back to another exciting coding session. Today’s problem is a treat—literally! We’re going to solve the “Global and Local Inversions ” problem.

Leetcode 775
Leet code 775

Leet code 775 Code Explanation: Global and Local Inversions

  • 1: We’re given an array of numbers called nums. The goal is to determine if every “local inversion” is also a “global inversion.”
  • 2: A local inversion occurs when an element is larger than the next element.
    • For example, in [1, 3, 2], the pair 3, 2 is a local inversion because 3 is larger than 2.
  • 3: A global inversion is when the number of local inversions equals the number of global inversions.
    • For example, in [1, 3, 2], there are 2 local inversions (3, 2 and 1, 3) and 2 global inversions (3, 2 and 1, 3).
  • 4: The code iterates through each element in the nums array.
  • 5: For each element, it calculates the absolute difference between the element’s value and its index. If this difference is greater than 1, it means there’s a non-local inversion, so we return false.
  • 6: If we reach the end of the array without finding any non-local inversions, it means every local inversion is also a global inversion, and we return true.

Codes :

C++: Leetcode 775

#include <vector> #include <cstdlib> class Solution { public: bool isIdealPermutation(std::vector<int>& nums) { for (int i = 0; i < nums.size(); ++i) { if (std::abs(nums[i] - i) > 1) return false; } return true; } };

Leet Code : Validate Binary Search Tree Java | CPP | Python solution

Java:Leetcode 775

import java.util.*; class Solution { public boolean isIdealPermutation(int[] nums) { for (int i = 0; i < nums.length; i++) { if (Math.abs(nums[i] - i) > 1) return false; } return true; } }

Tackling Jump I , II , III , IV Game Problems on LeetCode | Cpp ,Java ,Python – Day 3

Python: Leetcode 775

class Solution: def isIdealPermutation(self, nums: List[int]) -> bool: for i in range(len(nums)): if abs(nums[i] - i) > 1: return False return True

JavaScript: Leetcode 775

class Solution { isIdealPermutation(nums) { for (let i = 0; i < nums.length; i++) { if (Math.abs(nums[i] - i) > 1) return false; } return true; } }

List of Some important Leet code Question :

Scroll to Top