Jewels and Stones
Leetcode Easy
Problem
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
andJ
will consist of letters and have length at most 50.- The characters in
J
are distinct.
Solution
class Solution {
public int numJewelsInStones(String J, String S) {
HashSet<Character> jewels = new HashSet<>();
// Add all the elements to the hashSet
for (int i = 0; i < J.length(); i++) {
jewels.add(J.charAt(i));
}
// Count the
int numJewels = 0;
for (int i = 0; i < S.length(); i++) {
if (jewels.contains(S.charAt(i))) {
numJewels++;
}
}
return numJewels;
}
}
How it works
First we loop through the string containing jewels and add them to a hashSet. This allows for us to loop through the second string and check if those stones are jewels in constant time, as the contains()
method is O(1) for a hashSet.