Solve  
 

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".

Input: J = "aA", S = "aAAbbbb"
Output: 3

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.

| Hint: For each stone, check if it is a jewel.

# @param {String} j
# @param {String} s
# @return {Integer}
def num_jewels_in_stones(j, s)
  #...
end

In a binary tree, the root node is at depth 0, and children of each depth k node are at depth k+1.

Two nodes of a binary tree are cousins if they have the same depth, but have different parents.

We are given the root of a binary tree with unique values, and the values x and y of two different nodes in the tree.

Return true if and only if the nodes corresponding to the values x and y are cousins.

Input: root = [1,2,3,4], x = 4, y = 3
Output: false

Input: root = [1,2,3,null,4,null,5], x = 5, y = 4
Output: true

Input: root = [1,2,3,null,4], x = 2, y = 3
Output: false

Constraints:

The number of nodes in the tree will be between 2 and 100. Each node has a unique integer value from 1 to 100.

Code reference:

# Definition for a binary tree node.
class TreeNode
    attr_accessor :val, :left, :right
    def initialize(val = 0, left = nil, right = nil)
        @val = val
        @left = left
        @right = right
    end
end

# @param {TreeNode} root
# @param {Integer} x
# @param {Integer} y
# @return {Boolean}
def is_cousins(root, x, y)
  # your code goes here...
end