How To Calculate The Time Complexity Of The Algorithm? About Best Case, Worst Case And Average Case.

You have to know

AS PER 1ST algorithm,
inside teh first while loop----->

we need to find out , what all basic operation which takes constant amount of time.
Here
k ← 0   ------ is one basic operation
k < n*n  ---- assume , n*n = N ( any way it will be one number) one basic opertion
  y ← y + j - k   ------ anotherbasic operation
k ← k + 1 ------ is one more basic operation
j ← j - 1 ------ last basic operation inside 1st while loop

Now consider the time,

1+ N + N -1 + N-1 + 1 = 3N

inside while loop , takes 3N ( N is just a big number if we assume a big value of n)
Y ← 0
j ← n
while j >= 1 do {
  k ← 0
  while k < n*n do {
  y ← y + j - k
  k ← k + 1
  }
j ← j - 1
}

Y ← 0
j ← n

now take 1st algorithm as a whole,

now for n value 1 , the total alogorithm will take 3N times,
now for n value 2, the total algorithm will take 2*3N times
so on...
For value   n
n * 3N =  3 N2 (three N square) so finally we can say, the oerder of this algorithm is n square.

For(I=1;I

Happy new year and peace to you all! Can you please explain to me what is the execution time T(n) of the 3 algorithms?

Algorithm 1:
Y ← 0
j ← n
while j >= 1 do {
  k ← 0
  while k < n*n do {
  y ← y + j - k
  k ← k + 1
  }
j ← j - 1
}
return y

Algorithm 2:
For I ← 1 to n do
  for j ← 1 to I do
  for k ← j to I+j do
  a ← a + 1
  end for
  end for
end for

Algorithm 3:
For I ← 1 to m do
  for j ← 1 to i2 do
  for k ← 1 to j do
  b ← b + 1
  end for
  end for
end for

@Oddman, you mean if only n is not divisible by 2 .. The loop will go for ever .. =)

X=1;
I=1;
while(I