Analyzing Algorithms Part III
In this lesson, we will work out a generalization for the number of instructions executed for an array of length n
We'll cover the following...
Generalizing Instruction Count
Let's try to generalize the running time for an array of size n.
| Code Statement | Cost |
|---|---|
| 1. for (int i = 0; i < input.length; i++) { | executed (2 x n) + 2 times. |
| 2. int key = input[i]; | executed n times |
| 3. int j = i - 1; | executed n times |
| 4. while (j >= 0 && input[j] > key) { | We already calculated the cost of each iteration of the inner loop. We sum up the costs across all the iterations. Note that the inner loop will be executed n times and each execution will result in (iterations x 7)+2 instructions being executed. And the number of iterations will successively increase from 0, 1, 2 all the way up to(n-1). See the dry run in the previous lesson to convince yourself of the validity of this result. |
| 5. Inner loop statments | [(0 x 7) + 2] + [(1 x 7) + 2] + [(2 x 7) + 2] + ... [( (n-1) x 7) + 2] = 2n + 7 * [ 0 + 1 + 2 + ... (n-1) ] = 2n + 7 [ n(n-1)⁄2 ] |
| 11. } //inner loop ends | |
| 12. } |
If we use the above formulas and apply an array of size 5 to them, then the cost should match with what we calculated earlier.
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