Jewels and Stones.md

Algorithm

1. Put jewels in a Set for O(1) access.
2. For each stone, check if it is a jewel.

Solution

class Solution {
    public int numJewelsInStones(String j, String s) {
        Set<Character> jewels = new HashSet();
        for (int i = 0; i < j.length(); i++) {
            jewels.add(j.charAt(i));
        }
        int count = 0;
        for (int i = 0; i < s.length(); i++) {
            if (jewels.contains(s.charAt(i))) {
                count++;
            }
        }
        return count;
    }
}

Time/Space Complexity

• Time Complexity: O(j + s)
• Space Complexity: O(j)

Links

• github.com/RodneyShag