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Java 8, (削除) 231 (削除ここまで) (削除) 193 (削除ここまで) (削除) 185 (削除ここまで) (削除) 122 (削除ここまで) (削除) 103 (削除ここまで) 78 bytes

s->s.length==5&&(s[1]-s[0])*(s[3]-s[2])<0&(s[2]-s[1])*(s[4]-s[3])<0&s[4]==s[0]

Try it here.

-38 bytes thanks to @dpa97 for reminding me to use char[] instead of String.
-63 bytes thanks to @KarlNapf's derived formula.
-25 bytes by converting it from Java 7 to Java 8 (and now returning a boolean instead of integer).

193 bytes answer:

int c(char[]s){if(s.length!=5)return 0;int a=s[0],b=s[1],c=s[2],d=s[3],e=s[4],z=b-a,y=c-b,x=d-c,w=e-d;return e!=a?0:(z>0&y>0&x<0&w<0)|(z<0&y>0&x>0&w<0)|(z>0&y<0&x<0&w>0)|(z<0&y<0&x>0&w>0)?1:0;}

Explanation:

• If the length of the string isn't 5, we return false
• If the first character doesn't equal the last character, we return false
• Then we check the four valid cases one by one (let's indicate the five characters as 1 through 5), and return true if it complies to any of them (and false otherwise):
  1. If the five characters are distributed like: 1<2<3>4>5 (i.e. ALPHA)
  2. If the five characters are distributed like: 1>2<3<4>5 (i.e. EAGLE, HARSH, NINON, PINUP)
  3. If the five characters are distributed like: 1<2>3>4<5 (i.e. RULER)
  4. If the five characters are distributed like: 1>2>3<4<5 (i.e. THEFT, WIDOW)

These four rules can be simplified to 1*3<0 and 2*4<0 (thanks to @KarlNapf's Python 2 answer).