complexity on recursive Big-O - recursion

Question

I have a Computer Science Midterm tomorrow and I need help determining the complexity of these recursive functions. I know how to solve simple cases, but I am still trying to learn how to solve these harder cases. Any help would be much appreciated and would greatly help in my studies, Thank you!
fonction F(n)
if n == 0
return 1
else
return F(n-1) * n
fonction UniqueElements(A[0..n-1])
for i=0 to i <= n-2 do
for j=i+1 to j <= n-1 do
if A[i] == A[j]
return false
return true
fonction BinRec(n)
if n == 1
return 1
else
return BinRec(floor(n/2)) + 1

Answer 1

For hands on learning, you can plug the functions into a program, and test their worst case scenario performance.
When trying to calculate O by hand, here are some things to remember
The +, -, *, and / offsets can be ignored. So 1 to n+5 and 1 to 5n is considered equivalent to 1 to n.
Also, Only the highest order of magnitude counts, so for O 2^n + n^2 + n, 2^n grows the fastest, so it is equivalent to O 2^n
With recursive functions, you are looking at how many times the function is called in the method (the split count) and how much it needs to be called (the depth, usually is equal to list length). So the final O will be depth_count^split_count
With loops, each nested loop multiplies to the one it's in, and sequential loops add, so (1-n){(1-n){}} (1-n){} is (n * n) + n) => n^2 + n =(only highest growth counts)> n^2
PRACTICE! You will need to practice to get the hang of the gatchas of growth rate and how control flows interact. (so do online practice quizs)
function F(n){
count++
if (n == 0)
return 1
else
return F(n-1) * n
}
function UniqueElements(A){
for (var i=0 ; i <= A.length-2; i++){
for (var j=i+1;j <= A.length-1; j++){
if (A[i] == A[j]){
return false
}
}
}
return true
}
function BinRec(n) {
count++
if (n == 1)
return 1
else
return BinRec(Math.floor(n/2)) + 1
}
count = 0;
console.log(F(10));
console.log(count);
count = 0;
console.log(UniqueElements([1,2,3,5]));
console.log(count);
count = 0;
console.log(BinRec(40));
console.log(count);