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Merge pull request #94 from jsca-kwok/jk-find-all-duplicates-in-array
Find All Duplicates in an Array
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data_structures/arrays/find_all_duplicates_in_an_array.rb
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data_structures/arrays/find_all_duplicates_in_an_array.rb
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# Find All Duplicates in an Array
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#
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# Given an array of integers, 1 ≤ a[i] ≤ n (n = size of array),
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# some elements appear twice and others appear once.
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#
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# Find all the elements that appear twice in this array.
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#
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# Could you do it without extra space and in O(n) runtime?
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#
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# Example:
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# Input:
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# [4,3,2,7,8,2,3,1]
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#
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# Output:
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# [2,3]
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require 'benchmark'
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array = [4, 3, 2, 7, 8, 2, 3, 1]
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long_array = [4, 3, 2, 7, 8, 2, 3, 1] * 100
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#
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# Approach 1: Brute force
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#
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#
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# Complexity Analysis
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#
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# Time complexity: O(n^2) average case.
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#
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def find_duplicates(array)
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current_num = array[0]
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result_array = []
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array.count.times do |i|
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array.count.times do |j|
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next if i == j || current_num != array[j]
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result_array.push(current_num)
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end
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current_num = array[i + 1]
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end
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result_array.uniq
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end
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Benchmark.bmbm do |x|
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x.report('execute algorithm 1') do
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print(find_duplicates(array))
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print(find_duplicates(long_array))
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end
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end
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#
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# Approach 2: Sort and Compare Adjacent Elements
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#
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# Intuition
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# After sorting a list of elements, all elements of equivalent value get placed together.
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# Thus, when you sort an array, equivalent elements form contiguous blocks.
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#
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# Complexity Analysis
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#
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# Time complexity: O(n log n)
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#
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def find_duplicates_2(array)
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sorted_array = array.sort
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result_array = []
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(1..sorted_array.count).each do |i|
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next if sorted_array[i] != sorted_array[i - 1]
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result_array.push(sorted_array[i])
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end
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result_array.uniq
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end
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Benchmark.bmbm do |x|
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x.report('execute algorithm 2') do
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print(find_duplicates(array))
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print(find_duplicates(long_array))
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end
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end
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#
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# Approach 3: Hash map
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#
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#
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# Complexity Analysis
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#
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# Time complexity: O(n) average case.
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#
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def find_duplicates_3(array)
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result_hash = {}
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result_array = []
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# loop through array and build a hash with counters
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# where the key is the array element and the counter is the value
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# increase counter when duplicate is found
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array.each do |num|
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if result_hash[num].nil?
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result_hash[num] = 1
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else
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result_hash[num] += 1
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end
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end
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# loop through hash and look for values > 1
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result_hash.each do |k, v|
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result_array.push(k) if v > 1
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end
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# return keys
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result_array
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end
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Benchmark.bmbm do |x|
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x.report('execute algorithm 3') do
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print(find_duplicates(array))
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print(find_duplicates(long_array))
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end
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end
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