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 1:
Input: J = "aA", S = "aAAbbbb"
Output: 3
Example 2:
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.
class Solution {
public int numJewelsInStones(String J, String S) {
Set<Character> Jset = new HashSet();
for (char j: J.toCharArray())
Jset.add(j);
int ans = 0;
for (char s: S.toCharArray())
if (Jset.contains(s))
ans++;
return ans;
}
}
Time Complexity:
O(J.length+S.length). The
O(J.length) part comes from creating J. The
O(S.length) part comes from searching S.
Space Complexity:
O(J.length).
