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