计算复杂性和递归——这个分析正确吗？

Question

Hi, I have a doubt about this resolution of this algorithm analysis, specifically referred to the return min(L[i:j+1]) : why is it considered O(n)?: it always refers to a defined slice, with a limited possible dimension (j<=i+2)

Answer 1

For simplicity, consider size of the list to be a power of 3.

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Algorithm

if j-i+1 <= 3:
    # Compute their minimum
    return min(i:j+1)

The if statement forms the base case. Time complexity of above statements is O(1) . However, the if statement will be executed n times.

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Recursion Tree

                                     T(n)
                                       |
                 ______________________|______________________
                |                      |                      |
                |                      |                      |
             T(n/3)                 T(n/3)                 T(n/3)
                |                      |                      |
         _______|_______        _______|_______        _______|_______
        |       |       |      |       |       |      |       |       |
        |       |       |      |       |       |      |       |       |
     T(n/9)  T(n/9)  T(n/9) T(n/9)  T(n/9)  T(n/9) T(n/9)  T(n/9)  T(n/9)

        .       .       .      .       .       .      .       .       .
        .       .       .      .       .       .      .       .       .
        .       .       .      .       .       .      .       .       . 
        .       .       .      .       .       .      .       .       .    
      T(1)    T(1)    T(1)   T(1)    T(1)    T(1)   T(1)    T(1)    T(1)    --- (n/3) * T(1)      

It should be obvious from the recursion tree, that for an array of size n , if statement will be executed n/3 times. Hence, overall complexity of if statement is O(n) .

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The reason if is executed n/3 times and not n is that recursion ends when we encountered an sub-array of size 3 . If the the recursion came to an end when we encountered an sub-array of size 1, then it would have been executed for n times.