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64. Minimum Path Sum

Medium

Given a m x n grid filled with non-negative numbers, find a path from top left to bottom right, which minimizes the sum of all numbers along its path.

Note: You can only move either down or right at any point in time.

Example 1:

Input: grid = [[1,3,1],[1,5,1],[4,2,1]]

Output: 7

Explanation: Because the path 1 → 3 → 1 → 1 → 1 minimizes the sum.

Example 2:

Input: grid = [[1,2,3],[4,5,6]]

Output: 12

Constraints:

• m == grid.length
• n == grid[i].length
• 1 <= m, n <= 200
• 0 <= grid[i][j] <= 100

Solution

# @param {Integer[][]} grid
# @return {Integer}
def min_path_sum(grid)
  return grid[0][0] if grid.length == 1 && grid[0].length == 1

  dm = Array.new(grid.length) {Array.new(grid[0].length, 0)}
  s = 0

  (grid.length - 1).downto(0) do |r|
    dm[r][grid[0].length - 1] = grid[r][grid[0].length - 1] + s
    s += grid[r][grid[0].length - 1]
  end

  s = 0
  (grid[0].length - 1).downto(0) do |c|
    dm[grid.length - 1][c] = grid[grid.length - 1][c] + s
    s += grid[grid.length - 1][c]
  end

  recur(grid, dm, 0, 0)
end

private

def recur(grid, dm, r, c)
  if dm[r][c].zero? && r != grid.length - 1 && c != grid[0].length - 1
    dm[r][c] = grid[r][c] + [recur(grid, dm, r + 1, c), recur(grid, dm, r, c + 1)].min
  end

  dm[r][c]
end

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