Jewels and Stones

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.

METHOD 1:

class Solution {
    public int numJewelsInStones(String J, String S) {
       int res=0;
       for(char c : S.toCharArray()){
           if(J.indexOf(c) != -1){
               res++;
           }
       }
       return res;
    }
}


Try it on Leetcode

Here, in the problem they mention that J string has unique characters. So, we are traversing each character from S string and if we found index of that character in J string [i.e., whatever the index we can end up that character is present in J string, if that character is not found, it will be -1] , then increase the result by 1. Finally return the result.


METHOD 2:


class Solution {
    public int numJewelsInStones(String J, String S) {
        int len = S.length(), len1=J.length();
        for(int i=0;i<len1;i++)
        {
            S = S.replaceAll(J.charAt(i)+"","");
        }
        return len-S.length();
    }
}



Try it on Leetcode

Here, we are going to traverse J string. For each character found in S string, we are removing multiple occurrence of those characters from S string. Finally, subtraction of total length and current S string length will be the result.


Below image runtime and memory comparison of both methods.
First row accepted solution is METHOD 1
Second row accepted solution is METHOD 2

Runtime and Memory comparison

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