Merge pull request #110 from jsca-kwok/jk-jewels-and-stones

DIRECTORY.md

@@ -12,6 +12,7 @@
     * [Find The Highest Altitude](https://github.com/TheAlgorithms/Ruby/blob/master/data_structures/arrays/find_the_highest_altitude.rb)
     * [Fizz Buzz](https://github.com/TheAlgorithms/Ruby/blob/master/data_structures/arrays/fizz_buzz.rb)
     * [Get Products Of All Other Elements](https://github.com/TheAlgorithms/Ruby/blob/master/data_structures/arrays/get_products_of_all_other_elements.rb)
+    * [Jewels & Stones](https://github.com/TheAlgorithms/Ruby/blob/master/data_structures/arrays/jewels_and_stones.rb)
     * [Remove Elements](https://github.com/TheAlgorithms/Ruby/blob/master/data_structures/arrays/remove_elements.rb)
     * [Single Number](https://github.com/TheAlgorithms/Ruby/blob/master/data_structures/arrays/single_number.rb)
     * [Sort Squares Of An Array](https://github.com/TheAlgorithms/Ruby/blob/master/data_structures/arrays/sort_squares_of_an_array.rb)

data_structures/arrays/jewels_and_stones.rb

@@ -0,0 +1,86 @@
+# 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
+
+#
+# Approach 2: Hash
+#
+# Time Complexity: O(n)
+#
+
+def find_jewels(jewels, stones)
+  jewels_array = jewels.split('')
+  stones_array = stones.split('')
+  result_hash = {}
+  result = 0
+
+  stones_array.each do |stone|
+    if result_hash[stone] 
+      result_hash[stone] += 1
+    else
+      result_hash[stone] = 1
+    end
+  end
+
+  jewels_array.each do |jewel|
+    if result_hash[jewel]
+      result += result_hash[jewel]
+    else
+      result
+    end
+  end
+
+  result
+end
+
+puts find_jewels("aA", "aAAbbbb")
+# => 3
+puts find_jewels("z", "ZZ")
+# => 0