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https://github.com/TheAlgorithms/Ruby
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86 lines
1.8 KiB
Ruby
86 lines
1.8 KiB
Ruby
# Challenge name: Two Sum
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#
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# Given an array of integers nums and an integer target, return indices of the two numbers such that they add up to target.
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#
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# You may assume that each input would have exactly one solution, and you may not use the same element twice.
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#
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# You can return the answer in any order.
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#
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#
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# Examples
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#
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# Input: nums = [2, 7, 11, 15], target = 9
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# Output: [0,1]
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# Explanation: Because nums[0] + nums[1] == 9, we return [0, 1].
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#
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# Input: nums = [3, 2, 4], target = 6
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# Output: [1,2]
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#
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# Input: nums = [3, 3], target = 6
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# Output: [0,1]
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# Explanation: Because nums[0] + nums[1] == 9, we return [0, 1].
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#
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# @param {Integer[]} nums
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# @param {Integer} target
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# @return {Integer[]}
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#
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# Approach 1: Brute Force with Addition
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#
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# Complexity analysis
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# Time Complexity: O(n^2). For each element, we try to find its complement
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# by looping through the rest of the array which takes O(n) time.
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# Therefore, the time complexity is O(n^2)
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# Space complexity: O(1)
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def two_sum(nums, target)
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result_array = []
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nums.count.times do |i|
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nums.count.times do |j|
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next unless i != j && i < j
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current_sum = nums[i] + nums[j]
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return [i, j] if current_sum == target
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end
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end
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end
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print(two_sum([2, 7, 11, 15], 9))
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# => [0,1]
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print(two_sum([3, 2, 4], 6))
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# => [1,2]
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print(two_sum([3, 3], 6))
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# => [0,1]
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#
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# Approach 2: Brute Force with Difference
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#
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# Complexity analysis
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#
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# Time Complexity: O(N^2), where N is the length of the array
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#
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def two_sum(nums, target)
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nums.each_with_index do |num, i|
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target_difference = target - num
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nums.each_with_index do |num, j|
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return [i, j] if i != j && num == target_difference
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end
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end
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end
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print(two_sum([2, 7, 11, 15], 9))
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# => [0,1]
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print(two_sum([3, 2, 4], 6))
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# => [1,2]
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print(two_sum([3, 3], 6))
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# => [0,1]
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