Degree of symmetry (Editorial)

Written By : MathModel

The problem is asking us to count the indices $i$ such that $i_{th}$ digit from left is equal to $i_{th}$ from the right , this can be done by brute force , You should be careful that when n=\text{size}(s) is odd then the middle digit is mirror to itself which means it'll be always counted otherwise we count the digits where

(0-based indexing)

This can be done by brute forcing the indices from 0 \rightarrow \lfloor \frac{n}{2} \rfloor and updating the answer.

If $\text{size}(s)$ is odd then we add $1$ to the answer since middle digit is mirrored to itself.

s=input()
ans=0
for i in range(len(s)//2):
    if(s[i]==s[len(s)-1-i]):
        ans+=1
print(ans)

Complexity $\mathcal{O}(\text{size}(s))$.