# Leetcode: Jewels and Stones

This week is just a rundown on a Leetcode problem I went through.

Here’s the solution:

`/**`

* @param {string} J

* @param {string} S

* @return {number}

*/

var numJewelsInStones = function(J, S) {

let jewels = J.split("");

let stones = S.split("");

let ans = 0;

for(let i = 0; i < jewels.length; i++){

for(let j = 0; j < stones.length; j++){

// if (jewels[i] === stones[j]){

// ans++;

// }

(jewels[i] === stones[j]) ? ans++ : null;

}

}

return ans;

// let ans = 0;

// for(let char of S){

// // J contains char stone

// // Chars is J are distinct from one another

// if (J.lastIndexOf(char) > -1){

// ans++;

// }

// }

// return ans;

};

It’s a fairly simple bit. The question at hand was:

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

I first split jewels and stones into an array so I could work with them. I then have 2 for loops to see if jewels were in stones and vice versa. If they were, I add 1 to the answer.

This is a fairly simple question, one that’s easy to talk about in a pair programming interview. I may use this same question in an upcoming practice interview with some fellow job searchers.

Stay safe and stay cool everyone! Practice good social distancing.