# 771. Jewels and Stones

## Description

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

Letters are case sensitive, so "a" is considered a different type of stone from "A".

Example 1:

Input: jewels = "aA", stones = "aAAbbbb"
Output: 3


Example 2:

Input: jewels = "z", stones = "ZZ"
Output: 0


Constraints:

• 1 <= jewels.length, stones.length <= 50
• jewels and stones consist of only English letters.
• All the characters of jewels are unique.

## Solutions

• class Solution {
public int numJewelsInStones(String jewels, String stones) {
int[] s = new int[128];
for (char c : jewels.toCharArray()) {
s[c] = 1;
}
int ans = 0;
for (char c : stones.toCharArray()) {
ans += s[c];
}
return ans;
}
}

• class Solution {
public:
int numJewelsInStones(string jewels, string stones) {
int s[128] = {0};
for (char c : jewels) s[c] = 1;
int ans = 0;
for (char c : stones) ans += s[c];
return ans;
}
};

• class Solution:
def numJewelsInStones(self, jewels: str, stones: str) -> int:
s = set(jewels)
return sum(c in s for c in stones)


• func numJewelsInStones(jewels string, stones string) (ans int) {
s := [128]int{}
for _, c := range jewels {
s[c] = 1
}
for _, c := range stones {
ans += s[c]
}
return
}

• function numJewelsInStones(jewels: string, stones: string): number {
const s = new Set([...jewels]);
let ans = 0;
for (const c of stones) {
s.has(c) && ans++;
}
return ans;
}


• /**
* @param {string} jewels
* @param {string} stones
* @return {number}
*/
var numJewelsInStones = function (jewels, stones) {
const s = new Set(jewels.split(''));
return stones.split('').reduce((prev, val) => prev + s.has(val), 0);
};


• use std::collections::HashSet;
impl Solution {
pub fn num_jewels_in_stones(jewels: String, stones: String) -> i32 {
let mut set = jewels.as_bytes().iter().collect::<HashSet<&u8>>();
let mut ans = 0;
for c in stones.as_bytes() {
if set.contains(c) {
ans += 1;
}
}
ans
}
}