![Question: (a) Derive the operation count for the function F) below, assuming the worst case scenario. Show all the necessary details (5 marks) void F (float vl, int nargs: array il of size n int i; float ,-→ or int ! , , ,从 , while ( y[ i ] > x &&T넨。尤v[i+1]-v[i]: i = i-1; } }} イク](https://d2vlcm61l7u1fs.cloudfront.net/media%2Fce4%2Fce4c7d2e-b98f-4382-8cc6-faea49f54238%2FphpAIxEwR.png)
(a) Derive the operation count for the function F() below, assuming the worst case scenario. Show all the necessary details. void F (float [], int n) { // args: array v[] of size n int i: float x: for(int j = 1: j x && i > = 0) {v[i + 1] = v[i]: i = i – 1: } v[i + 1] = x: }}
In the function, the outer loop is running for n times and the inner loop is running for 1 to jth number of outer iteration (in the worst case).
So for j = 1 the inner loop runs for 1 time (worst case)
= 2 the inner loop runs for 2 time
=n the inner loop runs for n
Total complexity will be 1 +2 + 3+….+n = n(n+1)/2
So it is O(n^2)
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