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0771_Jewels_And_Stones.py
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0771_Jewels_And_Stones.py
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'''
You're given strings J representing the types of stones that are jewels, and S representing the
stones you have. Each character in S is 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".
Input: J = "aA", S = "aAAbbbb"
Output: 3
Input: J = "z", S = "ZZ"
Output: 0
'''
class Solution:
''' Approach #1 (Hash Table) '''
'''
Store every Jewel as a key in the Hash Table
Iterate over stones and in the jewel count if stone is present in the Hash
'''
def numJewelsInStones(self, J: str, S: str) -> int:
Hash = dict()
jewelCount = 0
for jewel in J:
Hash[jewel] = True
for stone in S:
jewelCount += 1 if(stone in Hash) else 0
return jewelCount
if __name__ == '__main__':
sol = Solution()
J = "aA"; S = "aAAbbbb"
print(sol.numJewelsInStones(J,S))