From f6a84ea326f626869b7876fff5f564b2f49a6de1 Mon Sep 17 00:00:00 2001
From: Vitor Oliveira <vbrazo@gmail.com>
Date: Mon, 29 Mar 2021 15:37:02 -0700
Subject: [PATCH] Move more algos to hash table folders

---
 .../arrays/find_all_duplicates_in_an_array.rb | 41 -------------
 .../find_all_duplicates_in_an_array.rb        | 61 +++++++++++++++++++
 2 files changed, 61 insertions(+), 41 deletions(-)
 create mode 100644 data_structures/hash_table/find_all_duplicates_in_an_array.rb

diff --git a/data_structures/arrays/find_all_duplicates_in_an_array.rb b/data_structures/arrays/find_all_duplicates_in_an_array.rb
index 63e6517..7ee99ff 100644
--- a/data_structures/arrays/find_all_duplicates_in_an_array.rb
+++ b/data_structures/arrays/find_all_duplicates_in_an_array.rb
@@ -87,44 +87,3 @@ Benchmark.bmbm do |x|
     print(find_duplicates(long_array))
   end
 end
-
-#
-# Approach 3: Hash map
-#
-
-#
-# Complexity Analysis
-#
-# Time complexity: O(n) average case.
-#
-
-def find_duplicates(array)
-  result_hash = {}
-  result_array = []
-
-  # loop through array and build a hash with counters
-  # where the key is the array element and the counter is the value
-  # increase counter when duplicate is found
-  array.each do |num|
-    if result_hash[num].nil?
-      result_hash[num] = 1
-    else
-      result_hash[num] += 1
-    end
-  end
-
-  # loop through hash and look for values > 1
-  result_hash.each do |k, v|
-    result_array.push(k) if v > 1
-  end
-
-  # return keys
-  result_array
-end
-
-Benchmark.bmbm do |x|
-  x.report('execute algorithm 3') do
-    print(find_duplicates(array))
-    print(find_duplicates(long_array))
-  end
-end
diff --git a/data_structures/hash_table/find_all_duplicates_in_an_array.rb b/data_structures/hash_table/find_all_duplicates_in_an_array.rb
new file mode 100644
index 0000000..bc68423
--- /dev/null
+++ b/data_structures/hash_table/find_all_duplicates_in_an_array.rb
@@ -0,0 +1,61 @@
+# Find All Duplicates in an Array
+#
+# Given an array of integers, 1 ≤ a[i] ≤ n (n = size of array),
+# some elements appear twice and others appear once.
+#
+# Find all the elements that appear twice in this array.
+#
+# Could you do it without extra space and in O(n) runtime?
+#
+# Example:
+# Input:
+# [4,3,2,7,8,2,3,1]
+#
+# Output:
+# [2,3]
+
+require 'benchmark'
+
+array = [4, 3, 2, 7, 8, 2, 3, 1]
+long_array = [4, 3, 2, 7, 8, 2, 3, 1] * 100
+
+#
+# Approach: Hash table
+#
+
+#
+# Complexity Analysis
+#
+# Time complexity: O(n) average case.
+#
+
+def find_duplicates(array)
+  result_hash = {}
+  result_array = []
+
+  # loop through array and build a hash with counters
+  # where the key is the array element and the counter is the value
+  # increase counter when duplicate is found
+  array.each do |num|
+    if result_hash[num].nil?
+      result_hash[num] = 1
+    else
+      result_hash[num] += 1
+    end
+  end
+
+  # loop through hash and look for values > 1
+  result_hash.each do |k, v|
+    result_array.push(k) if v > 1
+  end
+
+  # return keys
+  result_array
+end
+
+Benchmark.bmbm do |x|
+  x.report('execute algorithm 3') do
+    print(find_duplicates(array))
+    print(find_duplicates(long_array))
+  end
+end