Solve ABC098-C in Python

Introduction

The person from PyBegi solved it, so I'll piggyback on it.

** Thoughts ** Due to the constraints of the problem, it must be stored at the maximum of about $ O (N) $, so it is not enough to calculate for each leader. Therefore, we use the cumulative sum. Now you can calculate for each reader without having to $ sum $ each time. By the way, the computational complexity of $ sum $ in Python is $ O (N) $. If you understand this far, just do it.

n = int(input())
s = input()

count_e = [0] * n
count_w = [0] * n
for i in range(n):
    if i == 0:
        if s[i] == 'E':
            count_e[i] = 1
        else:
            count_w[i] = 1
        continue
    if s[i] == 'E':
        count_e[i] = count_e[i-1] + 1
        count_w[i] = count_w[i-1]
    else:
        count_e[i] = count_e[i-1]
        count_w[i] = count_w[i-1] + 1

ans = 10 ** 9 #e+w is at most 3*10**It's about 5, but I have a margin
for i in range(n):
    if s[i] == 'E':
        e = count_e[-1] - count_e[i] #Use cumulative sum
        w = count_w[i]
    else:
        e = count_e[-1] - count_e[i] #Use cumulative sum
        w = count_w[i] - 1 #Subtract your share
    ans = min(ans,e+w)
print(ans)

Summary

I'm glad that I've come up with algorithms that can be used by looking at the constraints. See you again, good night.