What is the worst case time complexity for this algorithm? - algorithm

Question

procedure matrixvector(n:integer);
var i,j:integer;
begin
for i<-1 to n do begin
B[i] = 0;
C[i] = 0;
for j<-1 to i do
B[i]<- B[i]+ A[i,j];
for j<-n down to i+1 do
C[i]<-C[i] + A[i,j]
end
end;

Answer 1

O(n^2), if I read it right.
Why you need two inner loops is beyond me. Why not sum B and C in the same loop?

Answer 2

Let us trace the number of times each loop executes in each iteration.
procedure matrixvector(n : integer);
var i, j : integer;
begin
for i<-1 to n do begin // OuterLoop
B[i] = 0;
C[i] = 0;
for j <- 1 to i do // InnerLoop_1
B[i] <- B[i] + A[i, j];
for j <- n down to (i + 1) do // InnerLoop_2
C[i] <- C[i] + A[i, j]
end
end;
InnerLoop_1
In the first iteration of OuterLoop (i = 1), InnerLoop_1 executes once.
In the second iteration of OuterLoop (i = 2), InnerLoop_1 executes twice.
In the third iteration of OuterLoop (i = 3), InnerLoop_1 executes thrice.
.
.
.
In the last iteration of OuterLoop (i = n), InnerLoop_1 executes n times.
Therefore, the total number of times this code executes is
1 + 2 + 3 + … + n
= (n(n + 1) / 2) (Sum of Natural Numbers Formula)
= (((n^2) + n) / 2)
= O(n^2)
InnerLoop_2
In the first iteration of OuterLoop (i = 1), InnerLoop_2 executes n - 1 times.
In the second iteration of OuterLoop (i = 2), InnerLoop_2 executes n - 2 times.
In the third iteration of OuterLoop (i = 3), InnerLoop_2 executes n - 3 times.
.
.
.
In the n - 2th iteration of OuterLoop (i = n - 2), InnerLoop_2 executes 2 times.
In the n - 1th iteration of OuterLoop (i = n - 1), InnerLoop_2 executes 1 time.
In the last (nth) iteration of OuterLoop (i = n), InnerLoop_2 executes 0 times.
Therefore, the total number of times this code executes is
n - 1 + n - 2 + n - 3 + … + 2 + 1 + 0
= 0 + 1 + 2 + … + n - 3 + n - 2 + n - 1
= (n - 1)((n - 1) + 1) / 2 (Sum of Natural Numbers Formula)
= (n - 1)(n) / 2
= (((n^2) - n) / 2)
= O(n^2)
Time Complexity
Number of times InnerLoop_1 executes : (((n^2) + n) / 2)
Number of times InnerLoop_2 executes : (((n^2) - n) / 2)
Adding, we get
(((n^2) + n) / 2) + (((n^2) - n) / 2)
= ((((n^2) + n) + ((n^2) - n)) / 2)
= (((n^2) + n + (n^2) - n) / 2)
= (((n^2) + (n^2)) / 2)
= ((2(n^2)) / 2)
= (n^2)
= O(n^2)
——————
Also, do take a look at these
https://stackoverflow.com/a/71537431/17112163
https://stackoverflow.com/a/71146522/17112163
https://stackoverflow.com/a/69821878/17112163
https://stackoverflow.com/a/72046825/17112163
https://stackoverflow.com/a/72046933/17112163

Answer 3

Just explaining in detail for beginners:
Outermost for loop will run n times (0 to n)
Then there are two for loops within the out ermost for loop.
First for loop will go from 1 to n (1+2+3+4+.....+n)
And the second for loop will go from n to 1 (n+n-1+n-2+....+1)
The summation formula for (1+2+3+4+5+....+n ) is n(n+1)/2
so the total running time can be computed as n + n(n+1)/2 + n(n+1)/2
Observe the highest polynomial in this equation, it is n^2.
We can further simplify this equation and drop the constants and ignore the linear part, which will give us a run time of n^2.

Answer 4

worst case is O(n²).
there are indeed three loops, but not all inside each other, thus giving O(n²).
also, you can clearly see that the inner loops won't go from 1 to n (like the outer loop does). But because this would only change the time complexity by some constant, we can ignore this and say that it is just O(n^2).
This shows that time complexity is a measure saying: your algorithm will scale with this order, and it won't ever take any longer. (faster is however always possible)
for more information about "calculating" the worst case complexity of any algorithm, I can point you to a related question I asked earlier