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An array with bubblesort time $\Theta(n)$ is nothing but a sorted array like: A = 1 2 3 4 5
No swaps are done so only $n - 1$ comparisons.
An array with insertionsort run time $\Theta(n^2)$ is a reverse sorted array: A = 5 4 3 2 1
Each element is swapped in every iteration.
There cannot be any array that can have IS time as $\Theta(n^2)$ and BS time as $\Theta(n)$.
But how can i prove this in terms of their inversions?

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Let's consider the swaps performed by insertion sort. In the $i$-th iteration, insertion sort repeatedly swaps the element $x$ originally in $A[i]$ with some element $y$ originally in $A[1:i-1]$ until the subarray $A[1:i]$ is sorted. Then, since we must have $x<y$ for the swap to occur, we have that each swap induces an inversion in the original array. Notice that the $i$-th iteration takes time proportional to $1+s_i$ where $s_i$ is the number of swaps in the iteration.

If follows that insertion sort takes time $O(n+\sum_{i=1}^n s_i) = O(n + s)$, where $s$ is the number of inversions of $A$. Thus, when insertion sort takes time $\Omega(n^2)$ we must have $s=\Omega(n^2)$.

Finally, notice that at most one inversion can be removed by each swap performed by bubble sort, implying that bubble sort takes time $\Omega(s)=\Omega(n^2)$.

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