Leetcode: Jewels and Stones

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

Jewels and Stones - LeetCode

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.