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GlobalAndLocalInversions.java
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GlobalAndLocalInversions.java
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package math;
/**
* Created by gouthamvidyapradhan on 01/02/2018.
*
* We have some permutation A of [0, 1, ..., N - 1], where N is the length of A.
The number of (global) inversions is the number of i < j with 0 <= i < j < N and A[i] > A[j].
The number of local inversions is the number of i with 0 <= i < N and A[i] > A[i+1].
Return true if and only if the number of global inversions is equal to the number of local inversions.
Example 1:
Input: A = [1,0,2]
Output: true
Explanation: There is 1 global inversion, and 1 local inversion.
Example 2:
Input: A = [1,2,0]
Output: false
Explanation: There are 2 global inversions, and 1 local inversion.
Note:
A will be a permutation of [0, 1, ..., A.length - 1].
A will have length in range [1, 5000].
The time limit for this problem has been reduced.
Solution: O(N) For every i, Maintain a max value up until (i - 1). If the current element at i < max value return false
*/
public class GlobalAndLocalInversions {
/**
* Main method
* @param args
* @throws Exception
*/
public static void main(String[] args) throws Exception{
}
public boolean isIdealPermutation(int[] A) {
if(A.length == 0 || A.length == 1) return true;
int max = Integer.MIN_VALUE;
for(int i = 1; i < A.length; i ++) {
if(A[i] < max){
return false;
} else{
max = Math.max(max, A[i - 1]);
}
}
return true;
}
}