Show how to get the big-Oh for the following code: void CountSort (int A[N], int range)...
Show how to get the big-Oh for the following code:
void CountSort (int A[N], int range) {
// assume 0 <= A[i] < range for any element A[i] int *pi =
new int[range];
for ( int i = 0; i < N; i++ ) pi[A[i]]++;
for ( int j = 0; j < range; j++ )
for ( int k = 1; k <= pi[j]; k++ ) cout << j << endl;
}
Homework Answers
`Hey,
Note: Brother if you have any queries related the answer please do comment. I would be very happy to resolve all your queries.
Since there are 2 loops but, they are not nested. So, first for loop has complexity of
O(n)
Second for loop will have complexity of O(N+range)
So,
total complexity is O(N+range)
Kindly revert for any queries
Thanks.
Add Answer to:
Show how to get the big-Oh for the following code:
void CountSort (int A[N], int range)...
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