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Global and Local Inversions in C++
Suppose we have some permutation A of [0, 1, ..., N - 1], where N is the length of A. Now the number of (global) inversions is the number of i < j with 0 <= i < j < N and A[i] > A[j]. And the number of local inversions is the number of i with 0 <= i < N and A[i] > A[i+1]. We have to return true if and only if the number of global inversions is equal to the number of local inversions. So if the input is like [1,0,2], then return true, as there is only one local inversion and one global inversion.
To solve this, we will follow these steps −
- maxVal := -1, n := size of A
- for i in range 0 to n – 3
- maxVal := max of A[i] and maxVal
- if maxVal > A[i + 2], then return false
- return true
Let us see the following implementation to get better understanding −
Example
#include <bits/stdc++.h>
using namespace std;
class Solution {
public:
bool isIdealPermutation(vector<int>& A) {
int maxVal = -1;
int n = A.size();
for(int i = 0; i < n - 2; i++){
maxVal = max(A[i], maxVal);
if(maxVal > A[i + 2])
return false;
}
return true;
}
};
main(){
vector<int> v = {1,0,2};
Solution ob;
cout << (ob.isIdealPermutation(v));
}Input
[1,0,2]
Output
1
Updated on: 04-May-2020
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