Add brute force solution

data_structures/arrays/jewels_and_stones.rb

@@ -0,0 +1,50 @@
+# Challenge name: Jewels and Stones
+#
+# You're given strings jewels representing the types of stones that are jewels,
+# and stones representing the stones you have. Each character in stones is a type
+# of stone you have. You want to know how many of the stones you have are also
+# jewels.
+#
+# Letters are case sensitive, so "a" is considered a different type of stone from "A".
+#
+# Example 1:
+#
+# Input: jewels = "aA", stones = "aAAbbbb"
+# Output: 3
+#
+# Example 2:
+#
+# Input: jewels = "z", stones = "ZZ"
+# Output: 0
+#
+#
+# Constraints:
+#
+# 1 <= jewels.length, stones.length <= 50
+# jewels and stones consist of only English letters.
+# All the characters of jewels are unique.
+
+#
+# Approach 1: Brute Force
+#
+# Time Complexity: O(n^2)
+#
+
+def find_jewels(jewels, stones) 
+  jewels_array = jewels.split('')
+  stones_array = stones.split('')
+  result = 0
+  jewels_array.each do |jewel|
+    stones_array.each do |stone|
+      if jewel == stone
+        result += 1
+      end
+    end
+  end
+  result
+end
+
+puts find_jewels("aA", "aAAbbbb")
+# => 3
+puts find_jewels("z", "ZZ")
+# => 0