Jewels and Stones

Introduction

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.

Solution

We need to cache Jewels for O(1) check. Array works better than map, because it is faster and we can use it (limited max 256 variations of Jewels).

func numJewelsInStones(J string, S string) int {
    cache := make([]bool, 256)
    for _, ch := range J {
        cache[int(ch)] = true
    }
    cnt := 0
    for _, ch := range S {
        if cache[int(ch)] {
            cnt++
        }
    }
    return cnt
}