Problem Statement
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"
.
Example
Example 1:
Input: J = "aA", S = "aAAbbbb"
Output: 3
Example 2:
Input: J = "z", S = "ZZ"
Output: 0
Note:

S
andJ
will consist of letters and have length at most 50.  The characters in
J
are distinct.
Solution Thought Process
This is a wellknown hash problem. First, we take the jewel string and record the frequencies in the hash set. Then we go through the S string to find out if this is a jewel by checking the set one by one, increasing the count of the result by one.
We can find the items in the unordered_set in O(1) time.
Solution
class Solution {
public:
int numJewelsInStones(string J, string S) {
unordered_set<char>jewels;
int result = 0;
for(int i=0;i<J.size();i++)
{
jewels.insert(J[i]);
}
for(int i=0;i<S.size();i++)
{
if(jewels.find(S[i])!=jewels.end())
{
result++;
}
}
return result;
}
};
Complexity
Time Complexity: O(m+n) where m = length of J, n = length of S
Space Complexity: O(m) where m = length of J
Discussion (1)