6 ms·
I am not clear why the proof is so long. By induction, the first i elements are already smaller than the A[i], so it's essentially an unoptimised version of fo
by ac42 5y ago
I am not clear why the proof is so long. By induction, the first i elements are already smaller than the A[i], so it's essentially an unoptimised version of
for i: 0..n-1
A[i] = min(A[i:n])
- areyousure 5y agoAfter the first iteration of the loop (and every other one too), A[i] is the maximum of the array, not the minimum.
- ac42 5y agoOh crap, now I see it. Thanks!