Facebook Rotary Lock Practice Puzzle
Anonymous User
6893
MetaInterview

image
image

This is the solution for the single lock which is in another set of practice questions is the easy version of the above.
image

This is the table for explanation:
D0,1: goes down from 0 to 1
R0,1: goes right from 0 to 1
image

This is the solution:

def getMinCodeEntryTime(N: int, M: int, C: List[int]) -> int:
  # Write your code here
  row = list()
  col = list()
  C.append(1)
  
  for i in range(M):
    min_it = getMTFromAToB(N, C[i-1], C[i])
    min_row = float('inf')
    min_col = float('inf')
    
    for j in range(i):
      min_jt = getMTFromAToB(N, C[j-1], C[i])
      
      min_row = min(min_row, row[j] + min_jt)
      row[j] += min_it
      
      min_col = min(min_col, col[j] + min_jt)
      col[j] += min_it
    
    if not row:
      min_row = min_it
      min_col = min_it
      
    row.append(min_row)
    col.append(min_col)
  
  return min(*row, *col)
Comments (28)