Brute force approach

data_structures/arrays/find_all_duplicates_in_an_array.rb

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+# 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]
+
+def find_duplicates(array) 
+  current_num = array[0]
+  result_array = []
+  array.each_with_index do |num, i|
+    array.each_with_index do |num, j|
+      if i != j && current_num == array[j]
+        result_array.push(current_num)
+      end
+    end
+    current_num = array[i+1]
+  end
+  result_array.uniq
+end
+
+array = [4,3,2,7,8,2,3,1]
+long_array = [4,3,2,7,8,2,3,1]*100
+
+require 'benchmark'
+Benchmark.bmbm do |x|
+  x.report('execute algorithm') do
+    print find_duplicates(long_array)
+  end
+end
+
+def find_duplicates_2(array) 
+  sorted_array = array.sort
+  result_array = []
+
+  (1..sorted_array.count).each do |i|
+    if sorted_array[i] == sorted_array[i-1]
+      result_array.push(sorted_array[i])
+    end
+  end
+  result_array.uniq
+end
+
+require 'benchmark'
+Benchmark.bmbm do |x|
+  x.report('execute algorithm') do
+    print find_duplicates_2(long_array)
+  end
+end
+
+def find_duplicates_3(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|
+    if v > 1
+      result_array.push(k)
+    end
+  end
+  # return keys
+  result_array
+end
+
+require 'benchmark'
+Benchmark.bmbm do |x|
+  x.report('execute algorithm') do
+    print find_duplicates_3(array)
+  end
+end