Time Complexity of Recursive Algorithm Array

Question

I have a recursive algorithm like that which computes the smallest integer in the array.

 ALGORITHM F_min1(A[0..n-1])
//Input: An array A[0..n-1] of real numbers
If n = 1
return A[0]
else
temp ← F_minl(A[0..n-2])
If temp ≤ A[n-1]
return temp
else
return A[n-1]

What I think that the reccurence relation should be

T(n)=T(n-1) + n

However I am not sure about the + n part. I want to be sure in which cases the recurrence is T(n)=T(n-1) + 1 and in which cases the recurrence is T(n)=T(n-1) + n.

Its T(n)=T(n-1) + 1 so it's Theta(n) globally. At least, at first glance. Where "+1" is an umbrella term to indicate a finite set of constant time operations. — Margaret Bloom
– Margaret Bloom, Commented Feb 24, 2017 at 10:32

Codor · Accepted Answer · 2017-02-24 10:33:34Z

3

The recurrence should be

T(1) = 1,
T(n) = T(n-1) + 1

because besides the recursive call to the smaller array, all computational effort (reading the last entry of A and doing the comparison) take constant time in unit cost measure. The algorithm can be understood as divide-and-conquer where the divide part is splitting the array into a prefix and the last entry; the conquer part, which is a comparison, cannot take more than constant time here. In total, a case where there is linear work after the recursive call does not exist.

answered Feb 24, 2017 at 10:33

Codor

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4 Comments

ayt_cem Over a year ago

Thank you very much, what I tought is that after the recursion there will be n comparions so o(n).

Codor Over a year ago

@ayt_cem Indeed, there are n comparisons in total; however, there is only one comparison in each recursive call.

ayt_cem Over a year ago

So in order to be + o(n) in each comparison after the recursive call, there should be n comparisons ?

Codor Over a year ago

Yes, but there is only one.

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