Algorithm - Python - 帽子を全員で揃える(条件分岐、if/else、入れ子)

問題解決のPythonプログラミング
原著
楽天ブックス(Kobo) Yahoo!

開発環境

• macOS Mojave - Apple
• Emacs (Text Editor)
• Python 3.7 (プログラミング言語)

問題解決のPythonプログラミング ―数学パズルで鍛えるアルゴリズム的思考 (Srini Devadas (著)、黒川 利明 (翻訳)、オライリージャパン)の1章(帽子を全員で揃える)、練習問題、問題1.を取り組んでみる。

コード(Emacs)

Python 3

#!/usr/bin/env python3

caps1 = ['F', 'F', 'B', 'B', 'B', 'F', 'B', 'B', 'B', 'F', 'F', 'B', 'F']
caps2 = ['F', 'F', 'B', 'B', 'B', 'F', 'B', 'B', 'B', 'F', 'F', 'F', 'F']


def please_conform(caps):
    start = 0
    forward = 0
    backward = 0
    intervals = []
    for i in range(1, len(caps)):
        if caps[start] != caps[i]:
            intervals.append((start, i - 1, caps[start]))
            if caps[start] == 'F':
                forward += 1
            else:
                backward += 1
            start = i
    intervals.append((start, len(caps) - 1, caps[start]))
    if caps[start] == 'F':
        forward += 1
    else:
        backward += 1
    if forward < backward:
        flip = 'F'
    else:
        flip = 'B'

    # 修正箇所
    for t in intervals:
        if t[2] == flip:
            if t[0] == t[1]:
                print(f'Person at position {t[0]} flip your cap!')
            else:
                print(
                    f'People in positions {t[0]} through {t[1]} flip your caps!'
                )


for cap in [caps1, caps2]:
    please_conform(cap)
    print()

入出力結果(Terminal, Jupyter(IPython))

$ ./sample1.py
People in positions 2 through 4 flip your caps!
People in positions 6 through 8 flip your caps!
Person at position 11 flip your cap!

People in positions 2 through 4 flip your caps!
People in positions 6 through 8 flip your caps!

$