# Merge Recurrence Analysis [duplicate]

Say I have a dynamic array of the proper length $$n$$. A sort pivot is given.

I run the sort algorithm, wait, and get $$j$$ unbalanced pivots. Is the time complexity $$O(n\log n)$$ or has it destabilized to quadratic in the best case? I believe that it will be useful to use recurrences to model each scenario, but not sure how to set up from there.

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• I am unclear on the question. Do you mean $j$ pivots total or per recursive part after $k$ levels of recursion? – HackerBoss Oct 12 '18 at 3:12
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Say I have an array sized $$n$$. A quicksort pivot is balanced when each side of the partition ends up with $$n/c$$ ($$c$$ is a fixed constant). Else, the partition in unbalanced.
In both these cases, partitioning takes $$pn$$ time ($$p=O(1)$$). Let's say I run quicksort and after one balanced pivot, I get $$j$$ unbalanced pivots. Is the runtime still $$O(n\log n)$$? I believe that it will be useful to use recurrences, but not sure how to set up from there.
The running time should still be $$O(n\log n)$$. Intuitively, the size of the array decreases by a factor of $$c$$ every $$j+1$$ recursion levels, so the total number of levels would be $$(j+1)\log_c n = O(\log n)$$. The work at every level is $$O(n)$$, for a total of $$O(n\log n)$$.
How would a recurrence look? We will have $$j+1$$ different functions: $$T_j$$ for the balanced levels, and $$T_0,\ldots,T_{j-1}$$ for the unbalanced levels. The recurrences are $$T_j(n) = \max_{n/c \leq m \leq n-n/c} T_0(m) + T_0(n-1-m) + pn, \\ T_i(n) = \max_{1 \leq m < n/c} T_{i+1}(m) + T_{i+1}(n-1-m) + pn.$$ Actually solving the recurrences could be a chore, so it's better to formalize the intuitive argument outlined above.
It is because we can use two recurrences, one for balanced and one for unbalanced, and see that summing them yields $$O(nlogn)$$.