Question

Mergesort combining phase:

24     S = []  
25     while L != [] and R != []:  
26       if head(L) <= head(R):  
27         S = S + [head(L)]  
28         L = tail(L)  
29       else:  
30         S = S + [head(R)]  
31         R = tail(R)  
32     S = S + L + R  
33     return S

Find the run time of mergesort’s combining phase (lines 24–33). Do not find the run time for the rest of mergesort. Do not use O, Ω, or Θ. Your answer must define T(n), where n is the number of elements to be sorted.
      Assume that head, tail, list concatenation ‘+’, and list creation ‘[x]’, work in constant time. (This is probably not true in Python, but assume it anyway!) Like the example in Cormen’s text, also assume that the run times of lines 24, 25 ..., 33 are constants c₂₄, c₂₅ ..., c₃₃. All these constants are greater than or equal to 0.

Answer 1

let us have a look at how the code is working

suppose we need to merge two lists L = [1.4.6] and R = [2,5,7]

so we do it as first take s = [ ], now the steps are:

1. s = [1] , L = [4,6] , R = [2,5,7]
2. s = [1,2] , L = [4,6] , R = [5,7]
3. s = [1,2,4] , L = [6] , R = [5,7]
4. s = [1,2,4,5] , L = [6] , R = [7]
5. s = [1,2,4,5,6] , L = [ ] , R = [7]
6. s = [1,2,4,5,6,7] , L = [ ] , R = [ ]

So if you intutively see it, you know it takes n times to merge the n elements.

Writing it in the form of T(n)

T(n) = T(n - k) + T(k) // assuming the two parts are of length k and n-k length

We decrement one from either of T(n-k) or T(k), randomly until either of it is finished.

1. T(n) = T(n - k - 1) + T(k) + c1
2. T(n) = T(n - k - 1) + T(k-1) + c1 + c2
   .
   .
   .
   .
3. at the nth step T(n) = c1 + c2 + c3 + ...cn

so t(n) = n x constant, so the time complexity is O(n)

EDIT

line 24 - 1 time

line 25 - 31 will run n times in total, you cannot accurately say which one of if or else will run exactly for how many times, if the if part runs for k times, the else part will run for n-k times

line 32 and 33 will run 1 time each

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