mirror of
https://github.com/TheAlgorithms/Ruby
synced 2024-12-26 21:58:56 +01:00
126 lines
2 KiB
Ruby
126 lines
2 KiB
Ruby
# Given an integer row_index, return the rowIndexth (0-indexed) row of the Pascal's triangle.
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# Example 1:
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#
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# Input: row_index = 3
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# Output: [1,3,3,1]
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#
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# Example 2:
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#
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# Input: row_index = 0
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# Output: [1]
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#
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# Example 3:
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#
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# Input: row_index = 1
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# Output: [1,1]
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#
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# Approach 1: Brute Force
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#
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# Complexity Analysis
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#
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# Time complexity: O(k^2).
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# Space complexity: O(k) + O(k) ~ O(k)
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def get_num(row, col)
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return 1 if row == 0 || col == 0 || row == col
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get_num(row - 1, col - 1) + get_num(row - 1, col)
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end
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def get_row(row_index)
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result = []
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(row_index + 1).times do |i|
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result.push(get_num(row_index, i))
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end
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result
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end
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row_index = 3
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print(get_row(row_index))
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# => [1,3,3,1]
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row_index = 0
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print(get_row(row_index))
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# => [1]
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row_index = 1
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print(get_row(row_index))
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# => [1,1]
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#
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# Approach 2: Dynamic Programming
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#
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# Complexity Analysis
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#
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# Time complexity: O(k^2).
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# Space complexity: O(k) + O(k) ~ O(k).
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# @param {Integer} row_index
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# @return {Integer[]}
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def get_row(row_index)
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result = generate(row_index)
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result[result.count - 1]
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end
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def generate(num_rows)
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return [[1]] if num_rows < 1
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result = [[1], [1, 1]]
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(2...num_rows + 1).each do |row|
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prev = result[row - 1]
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current = [1]
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med = prev.count / 2
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(1...prev.count).each do |i|
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current[i] = prev[i - 1] + prev[i]
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end
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current.push(1)
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result.push(current)
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end
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result
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end
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#
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# Approach 3: Memory-efficient Dynamic Programming
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#
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# Complexity Analysis
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#
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# Time complexity: O(k^2).
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# Space complexity: O(k).
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# @param {Integer} row_index
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# @return {Integer[]}
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def get_row(row_index)
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pascal = [[1]]
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(1..row_index).each do |i|
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pascal[i] = []
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pascal[i][0] = pascal[i][i] = 1
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(1...i).each do |j|
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pascal[i][j] = pascal[i - 1][j - 1] + pascal[i - 1][j]
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end
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end
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pascal[row_index]
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end
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row_index = 3
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print(get_row(row_index))
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# => [1,3,3,1]
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row_index = 0
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print(get_row(row_index))
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# => [1]
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row_index = 1
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print(get_row(row_index))
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# => [1,1]
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