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.
bool isIdealPermutation(int* A, int ASize) {
if(ASize<3)
return true;
int i,j,k=0;
for(i=0;i<ASize-2;i++)
{
if(A[i]>k)
k=A[i];
if(k>A[i+2])
return false;
}
return true;
}