-
Notifications
You must be signed in to change notification settings - Fork 50
Leaves - Yitgop #28
New issue
Have a question about this project? Sign up for a free GitHub account to open an issue and contact its maintainers and the community.
By clicking “Sign up for GitHub”, you agree to our terms of service and privacy statement. We’ll occasionally send you account related emails.
Already on GitHub? Sign in to your account
base: master
Are you sure you want to change the base?
Leaves - Yitgop #28
Changes from all commits
f6d0e66
0f36d45
64fc37a
4970c16
File filter
Filter by extension
Conversations
Jump to
Diff view
Diff view
There are no files selected for viewing
Original file line number | Diff line number | Diff line change |
---|---|---|
@@ -1,8 +1,125 @@ | ||
// 1. Set up array data structure: | ||
const pool = Array(9).fill('A').concat( | ||
Array(2).fill('B'), | ||
Array(2).fill('C'), | ||
Array(4).fill('D'), | ||
Array(12).fill('E'), | ||
Array(2).fill('F'), | ||
Array(3).fill('G'), | ||
Array(2).fill('H'), | ||
Array(9).fill('I'), | ||
Array(1).fill('J'), | ||
Array(1).fill('K'), | ||
Array(4).fill('L'), | ||
Array(2).fill('M'), | ||
Array(6).fill('N'), | ||
Array(8).fill('O'), | ||
Array(2).fill('P'), | ||
Array(1).fill('Q'), | ||
Array(6).fill('R'), | ||
Array(4).fill('S'), | ||
Array(6).fill('T'), | ||
Array(4).fill('U'), | ||
Array(2).fill('V'), | ||
Array(2).fill('W'), | ||
Array(1).fill('X'), | ||
Array(2).fill('Y'), | ||
Array(1).fill('Z') | ||
); | ||
let POOL = pool; | ||
|
||
const Adagrams = { | ||
|
||
drawLetters() { | ||
// Implement this method for wave 1 | ||
let i = 0; | ||
const lettersInHand = new Array(10); | ||
while (i < 10) { | ||
There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. There's a small bug here. Consider how you could make sure that you don't choose the same letter 10 times (however unlikely) |
||
lettersInHand[i] = POOL[Math.floor(Math.random() * POOL.length)]; | ||
i += 1; | ||
} | ||
return lettersInHand; | ||
}, | ||
|
||
usesAvailableLetters(input, lettersInHand) { | ||
const lettersObj = new Object(); | ||
lettersInHand.forEach(function(element){ | ||
if (lettersObj[element]) { | ||
lettersObj[element] += 1; | ||
} else { | ||
lettersObj[element] = 1; | ||
} | ||
}); | ||
|
||
let j = 0; | ||
while (j < input.length) { | ||
if (lettersObj[input.charAt(j)]) { | ||
lettersObj[input.charAt(j)] -= 1; | ||
} else { | ||
return false; | ||
} | ||
j += 1; | ||
} | ||
|
||
let n = 0; | ||
while (n < lettersObj.length) { | ||
if (lettersObj[j] < 0) { | ||
return false; | ||
} | ||
n += 1; | ||
} | ||
return true | ||
}, | ||
|
||
scoreChart : { | ||
A : 1, | ||
E : 1, | ||
I : 1, | ||
O : 1, | ||
U : 1, | ||
L : 1, | ||
N : 1, | ||
R : 1, | ||
S : 1, | ||
T : 1, | ||
D : 2, | ||
G : 2, | ||
B : 3, | ||
C : 3, | ||
M : 3, | ||
P : 3, | ||
F : 4, | ||
H : 4, | ||
V : 4, | ||
W : 4, | ||
Y : 4, | ||
K : 5, | ||
J : 8, | ||
X : 8, | ||
Q : 10, | ||
Z : 10 | ||
}, | ||
|
||
scoreWord(word) { | ||
const final_word = word.toUpperCase(); | ||
let score = 0; | ||
let m = 0; | ||
while (m < final_word.length) { | ||
score += this.scoreChart[final_word.charAt(m)]; | ||
m += 1; | ||
} | ||
|
||
switch (final_word.length) { | ||
case 7: | ||
There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. Consider how you could refactor to DRY up this code. |
||
case 8: | ||
case 9: | ||
case 10: | ||
score += 8; | ||
break; | ||
} | ||
|
||
return score | ||
}, | ||
}; | ||
} | ||
|
||
// Do not remove this line or your tests will break! | ||
export default Adagrams; | ||
export default Adagrams; |
There was a problem hiding this comment.
Choose a reason for hiding this comment
The reason will be displayed to describe this comment to others. Learn more.
This data structure works, but it's not very DRY and would be tricky to change. For example, suppose I said you had the wrong number of "I"s - you would have to do a lot of counting to solve the problem.
Instead you might store a hash of letter frequencies like this
and use it to build a pool of letters dynamically.