[LeetCode] Challenge log 771

Posted on | Edited on |
771. Jewels and Stones

https://leetcode.com/problems/jewels-and-stones/description/

  1. Use hashing to achieve O(n+m).
  2. Use set as a dict without value to save space

You’re given strings J representing the types of stones that are jewels, and S representing the stones you have. Each character in Sis a type of stone you have. You want to know how many of the stones you have are also jewels.

The letters in J are guaranteed distinct, and all characters in J and S are letters. Letters are case sensitive, so "a" is considered a different type of stone from "A".

Example 1:

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Input: J = "aA", S = "aAAbbbb"
Output: 3

Example 2:

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Input: J = "z", S = "ZZ"
Output: 0

Note:

  • S and J will consist of letters and have length at most 50.
  • The characters in J are distinct.
Soulution:

Both O(1)

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# use dict
class Solution:
    def numJewelsInStones(self, J, S):
        """
        :type J: str
        :type S: str
        :rtype: int
        """
        isJew = {}
        result = 0
        for i in J:
            isJew[i] = 1
        for i in S:
            if i in isJew:
                result += 1
        return result
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# use set
class Solution:
    def numJewelsInStones(self, J, S):
        """
        :type J: str
        :type S: str
        :rtype: int
        """
        return sum([1 for s in S if s in set(J)])
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