Question

Given a matrix with m rows and n columns, m adjacent numbers are chosen from m rows, where two numbers are adjacent to each other if they are directly connected vertically or diagonally and only one number is taken from one row. Design a dynamic programming algorithm to find the smallest sum of these m numbers.

For example, given a 3 by 3 matrix

1 2 3
4 5 6
7 0 2

The sum of three numbers 1, 4, and 0 from three rows is the smallest sum 5.

Example 1:

Input:

• board = [[1,2,3],[4,5,6],[7,0,2]]

Output:

• 5

Example 2:

Input:

• board = [[1]]

Output:

• 1

Example 3:

Input:

• board = [[1,2,3]]

Output:

• 1

Example 4:

Input:

• board = [[1,2,3,0],[4,5,1,6],[7,0,2,0]]

Output:

• 1

here is some starting code

def matrix(board):

sum = ...

return sum

Answer 1

All the explanations is in the code comments. Please take care of indentation by referring to code screenshots.

Hope this helps! Comment in case of a doubt.

Code:

# recursive function using dynamic programming
# as each location is board is traversed only once, the time complexity is O(n^2)
def matrix(board, dp, i, j):
  
# first row
if(i == 0):
dp[i][j] = board[i][j]
  
# other rows
else:
a = 100000 # largest value possible, can take it as required
b = dp[i-1][j] # vertically up
c = 100000 # largest value possible, can take it as required
  
# left diagonal, if present
if(j > 0):
a = dp[i-1][j-1]
# right diagonal, if present
if((j+1) < len(board[0])):
c = dp[i-1][j+1]
  
# calculate dp which is
# value of present cell + minimum value from reachable cells of the above row
dp[i][j] = board[i][j] + min(a, b, c)
  
# end of column
if(j == (len(board[0]) -1)):
# reached the end of board
if(i == (len(board) -1)):
# return the minimum value from the last row
return min(dp[i])
else:
# move to the next row
return matrix(board, dp, i+1, 0)
else:
# move to the next column
return matrix(board, dp, i, j+1)

# sample run
# example 1
board = [[1,2,3],[4,5,6],[7,0,2]]
# dp denotes the minimum sum for that cell in row upto that column
# initailly all are -1
dp = [[-1 for i in range(len(board[0]))] for j in range(len(board))]
print('Example 1:', matrix(board, dp, 0, 0))

# example 2
board = [[1]]
dp = [[-1 for i in range(len(board[0]))] for j in range(len(board))]
print('Example 2:', matrix(board, dp, 0, 0))

# example 3
board = [[1,2,3]]
dp = [[-1 for i in range(len(board[0]))] for j in range(len(board))]
print('Example 3:', matrix(board, dp, 0, 0))

# example 4
board = [[1,2,3,0],[4,5,1,6],[7,0,2,0]]
dp = [[-1 for i in range(len(board[0]))] for j in range(len(board))]
print('Example 4:', matrix(board, dp, 0, 0))

Sample run:

Example 1: 5 Example 2: 1 Example 3: 1 Example 4: 1

Code screenshots:

i # recursive function using dynamic programming # as each location is board is traversed only once, the time complexity is O(n^2) def matrix(board, dp, i, i): # first row if(i == 0): dp[i][j] = board[i][j] # other rows else: a = 100000 # largest value possible, can take it as required b = dp[i-1][j] # vertically up C = 100000 # largest value possible, can take it as required # left diagonal, if present if(j > 0): a = dp[i-1][j-1] # right diagonal, if present if((+1) < len(board[0])): C = dp[i-1][j+1] # calculate dp which is # value of present cell + minimum value from reachable cells of the above row dp[i][j] = board[i][j] + min(a, b, c) # end of column if(j == (len(board[@]) -1)): # reached the end of board if(i == (len(board) -1)): # return the minimum value from the last row return min(dp[i]) else: # move to the next row return matrix(board, dp, i+1, 0) else: # move to the next column return matrix(board, dp, i, j+1)