mirror of
https://github.com/TheAlgorithms/Ruby
synced 2024-12-25 21:58:57 +01:00
61 lines
1 KiB
Ruby
61 lines
1 KiB
Ruby
# Challenge name: Single Number
|
|
#
|
|
# Given a non-empty array of integers nums, every element appears twice
|
|
# except for one. Find that single one.
|
|
#
|
|
# Follow up: Could you implement a solution with a linear runtime
|
|
# complexity and without using extra memory?
|
|
#
|
|
# @param {Integer[]} nums
|
|
# @return {Integer}
|
|
|
|
#
|
|
# Approach 1: Hash map
|
|
#
|
|
# Time Complexity: O(n)
|
|
#
|
|
def single_number(nums)
|
|
result_hash = {}
|
|
nums.each do |num|
|
|
if result_hash[num]
|
|
result_hash[num] += 1
|
|
else
|
|
result_hash[num] = 1
|
|
end
|
|
end
|
|
|
|
result_hash.each do |k, v|
|
|
return k if v == 1
|
|
end
|
|
end
|
|
|
|
nums = [2, 2, 1]
|
|
puts(single_number(nums))
|
|
# Output: 1
|
|
nums = [4, 1, 2, 1, 2]
|
|
puts(single_number(nums))
|
|
# Output: 4
|
|
nums = [1]
|
|
puts(single_number(nums))
|
|
# Output: 1
|
|
|
|
#
|
|
# Approach 2: Use Ruby .count()
|
|
#
|
|
# Time Complexity: O(n^2)
|
|
#
|
|
def single_number(nums)
|
|
nums.find do |num|
|
|
nums.count(num) == 1
|
|
end
|
|
end
|
|
|
|
nums = [2, 2, 1]
|
|
puts(single_number(nums))
|
|
# Output: 1
|
|
nums = [4, 1, 2, 1, 2]
|
|
puts(single_number(nums))
|
|
# Output: 4
|
|
nums = [1]
|
|
puts(single_number(nums))
|
|
# Output: 1
|