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Move more algos to hash table folders
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2 changed files with 61 additions and 41 deletions
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@ -87,44 +87,3 @@ Benchmark.bmbm do |x|
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print(find_duplicates(long_array))
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print(find_duplicates(long_array))
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end
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end
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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(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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@ -0,0 +1,61 @@
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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: Hash table
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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(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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