mirror of
https://github.com/TheAlgorithms/Ruby
synced 2024-12-25 21:58:57 +01:00
117 lines
2.4 KiB
Ruby
117 lines
2.4 KiB
Ruby
# Arrays - Remove Elements
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#
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# Given an array nums and a value val, remove all instances of that value in-place and return the new length.
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# Do not allocate extra space for another array,
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# you must do this by modifying the input array in-place with O(1) extra memory.
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# The order of elements can be changed. It doesn't matter what you leave beyond the new length.
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#
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# Example
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#
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# Input: nums = [3,2,2,3], val = 3
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# Output: 2, nums = [2,2]
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#
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# Input: nums = [0,1,2,2,3,0,4,2], val = 2
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# Output: 5, nums = [0,1,4,0,3]
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#
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# Approach 1: Use `delete_if` Ruby method
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#
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# Time complexity: O(n)
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#
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def remove_elements(nums, val)
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nums.delete_if { |num| num == val }
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nums.length
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end
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puts remove_elements([3, 2, 2, 3], 3)
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# => 2
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puts remove_elements([0, 1, 2, 2, 3, 0, 4, 2], 2)
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# => 5
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#
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# Approach 2: Use `delete_at`, `unshift`, and `shift` Ruby method
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#
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# Time complexity: O(n)
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#
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def remove_elements(nums, val)
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result_length = nums.length
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shift_length = 0
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nums.each_with_index do |num, i|
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next unless num == val
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nums.delete_at(i)
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nums.unshift('removed')
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result_length -= 1
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shift_length += 1
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end
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nums.shift(shift_length)
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result_length
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end
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puts remove_elements([3, 2, 2, 3], 3)
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# => 2
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puts remove_elements([0, 1, 2, 2, 3, 0, 4, 2], 2)
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# => 5
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#
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# Approach 3: Two-pointers
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#
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# Complexity analysis
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#
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# Time complexity: O(n).
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# Assume the array has a total of n elements,
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# both pointer1 and pointer2 traverse at most 2n steps.
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#
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# Space complexity: O(1).
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def remove_element(nums, val)
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pointer1 = 0
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nums.each_with_index do |num, pointer2|
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if val != num
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nums[pointer1] = nums[pointer2]
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pointer1 += 1
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end
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end
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pointer1
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end
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puts remove_elements([3, 2, 2, 3], 3)
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# => 2
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puts remove_elements([0, 1, 2, 2, 3, 0, 4, 2], 2)
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# => 5
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#
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# Approach 4: Two-pointers (Optimized)
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#
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# Complexity analysis
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#
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# Time complexity: O(n). Both pointer1 and pointer2 traverse at most n steps.
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# In this approach, the number of assignment operations is equal to the
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# number of elements to remove.
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#
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# Space complexity: O(1)
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def remove_element(nums, val)
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pointer1 = 0
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pointer2 = nums.length
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while pointer1 < pointer2
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if nums[pointer1] == val
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pointer2 -= 1
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nums[pointer1] = nums[pointer2]
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else
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pointer1 += 1
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end
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end
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pointer1
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end
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puts remove_elements([3, 2, 2, 3], 3)
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# => 2
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puts remove_elements([0, 1, 2, 2, 3, 0, 4, 2], 2)
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# => 5
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