見習村20 - Scramblies

20 - Scramblies

Don’t say so much, just coding…

Instruction

Complete the function scramble(str1, str2) that returns true if a portion of str1 characters can be rearranged to match str2, otherwise returns false.

Notes:

  • Only lower case letters will be used (a-z). No punctuation or digits will be included.
  • Performance needs to be considered
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Input strings s1 and s2 are null terminated.

Examples

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scramble('rkqodlw', 'world') ==> True
scramble('cedewaraaossoqqyt', 'codewars') ==> True
scramble('katas', 'steak') ==> False

Ruby

Init

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def scramble(s1,s2)
  #your code here
end

Sample Testing

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Test.assert_equals(scramble('rkqodlw','world'),true)
Test.assert_equals(scramble('cedewaraaossoqqyt','codewars'),true)
Test.assert_equals(scramble('katas','steak'),false)
Test.assert_equals(scramble('scriptjava','javascript'),true)
Test.assert_equals(scramble('scriptingjava','javascript'),true)

Javascript

Init

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function scramble(str1, str2) {
 //code me
}

Sample Testing

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describe('Example Tests', function(){
  Test.assertEquals(scramble('rkqodlw','world'),true);
  Test.assertEquals(scramble('cedewaraaossoqqyt','codewars'),true);
  Test.assertEquals(scramble('katas','steak'),false);
  Test.assertEquals(scramble('scriptjava','javascript'),true);
  Test.assertEquals(scramble('scriptingjava','javascript'),true);
  Test.assertEquals(scramble('scriptsjava','javascripts'),true);
  Test.assertEquals(scramble('jscripts','javascript'),false);
  Test.assertEquals(scramble('aabbcamaomsccdd','commas'),true);
});

Thinking

想法(1): 後者的值，跟前者比較時，要所有的字都有在前者且數量也要ㄧ樣
想法(2): 將後者的值透過 uniq 後，跑迴圈來比較 count 有無 小於等於 前者

https://ithelp.ithome.com.tw/upload/images/20201005/20120826IN9SHwnVlw.jpg
圖片來源：Unsplash Buse Doga Ay

Hint & Reference

Solution

Ruby

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# Solution 1
def scramble(s1,s2)
  s3 = s1.split("")
  s2.split("").each do |x|
    s3.each_with_index{ |y,i| (s3.delete_at i ; break) if y == x }
  end
  s1.length - s2.length == s3.join.length ? true : false
end

# Solution 2
def scramble(s1,s2)
  s2.chars.uniq.all?{ |x| s2.count(x) <= s1.count(x) }
end

Javascript

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// Solution 1
function scramble(str1, str2) {
  return [...str2].every(val => str2.split(val).length <= str1.split(val).length);
}