int find_peak (int n, int A []) {
int left, right, lmid, rmid;
left = 0;
right = n - 1;
while (left + 3 <= right) {
lmid = (2 * left + right) / 3;
rmid = (left + 2 * right) / 3;
if (A[lmid] <= A[rmid])
left = lmid;
else
right = rmid;
}
int x = left;
for (int i = left + 1; i <= right; i ++)
if (A[i] > A[x])
x = i;
return A[x];
}
I was trying to solve this function for its BigO notation, but I am very confused about it. Is it O(log n) or something else? I can somewhat solve it in my head but I can't do it on properly.
Yes, the loop is cut roughly in half at each iteration
lmid = (2 * left + right) / 3;
rmid = (left + 2 * right) / 3;
if (A[lmid] <= A[rmid])
left = lmid;
else
right = rmid;
To be precise it's log1.5(n), because the actual length right-left
decreases only by 1/3, not halving the iterations. The complexity is still O(log(n))
You can try it here https://onlinegdb.com/rkI5gn3Xd
Thanks to chqrlie for prompting me to give a more detailed answer