Goldman Sachs新鲜电面

完美世界

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of 2 votes

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Almost.
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Let's define ascii values: . y* H" C$ w8 |5 x. z
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A= 65
B = 66
C = 67
D = 68 ' U1 |; L( u. b' e4 h( n) l3 D
E = 69 / A1 x) E/ ]) n; J4 t2 u
... etc. 7 a2 m" y, f1 U7 V

If we have two strings "BE" and "DC", they both sum to the same value, but they have different characters. - w/ N& Q& k8 ]/ D5 ~3 @: {: [8 z
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I think instead of transforming based on the sum of characters, when you iterate over the list, the hash code should be the hashcode produced when the particular string is lexicographically ordered. ) H) f9 x6 x" C- n: ^( s4 I

so if you have the following strings: % g0 s- D+ y" c5 F3 q8 k$ } x$ D7 b8 |

"BAC", "CAB"
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we would first find the lexicographically sorted version of the string , (for both this is "ABC") and in the hash location for "ABC", we would add both of them

lol

1

of 1 vote

You are right, definitely missed that detail, your solution would cover that issue.

Powerful

def anagramQuestion(list1, input):
	dict1 = {}
9 x' l5 c2 b8 B, _* W	for x in range(0,len(list1)):
/ Y( |2 V; E+ M( k! ]		key = ''.join(sorted(list1[x]))
7 X& U. P" k. `+ p		if key in dict1:
			dict1[key].append(list1[x])
		else:
  b6 ]8 [% x/ M# B/ b			dict1[key] = [list1[x]]
	result = []
	for x in dict1:
		inputsorted = ''.join(sorted(input))
		if x == inputsorted:
) ?( F) X0 u( E9 u			dict1[x].remove(input)
			result.append(dict1[x])

, F8 N' p, P7 d) s	return result