Interview Question

Interview Los Angeles, CA

Third person: Given a 2-d array, write code to print it out

  in a snake pattern. For example, if the array is this: 1, 2, 3 4, 5, 6 7, 8, 9 the routine prints this: 1,2,3,6,9,8,7,4,5 The array is an NxN array. The final question was just how to write a connection pool (i.e, a class that returns connections to the user, and if the user is done, returns them back to the pool)
Answer

Interview Answer

4 Answers

1

#!/usr/bin/python import math; # test several small sized matrices def main(): for N in range(0, 8): A = []; count = 1; for row in range(0, N): A.append([]); for col in range(0, N): A[row].append(count); count += 1; printMat(A); print snake(A); # prints the matrix in standard form def printMat(A): if len(A) == 0: print "[\t[\t]\t]"; return; print "[\t", for row in range(0, len(A)): if row == 0: print "[\t", else: print "\t[\t", for col in range(0, len(A[row])): if col == len(A[row]) - 1: if row == len(A) - 1: print str(A[row][col]) + "]\t]"; else: print str(A[row][col]) + "]"; else: print str(A[row][col]) + ",\t", # return array reading the matrix in snake fashion def snake(A): N = len(A); array = []; # outermost layer is 0 for layer in range(0, int(math.floor((N + 1) / 2))): d_u_count_range = max(4 * N - 8 * layer - 4, 0); l_r_count_range = d_u_count_range / 4; u_d_count_range = d_u_count_range / 2; r_l_count_range = 3 * l_r_count_range; for count in range(0, d_u_count_range): # append left to right values if count < l_r_count_range: array.append(A[layer][layer + count]); # append up to down values elif count < u_d_count_range: array.append(A[layer + count - l_r_count_range] [N - 1 - layer]); # append right to left values elif count < r_l_count_range: array.append(A[layer + l_r_count_range] [N - 1 - layer - (count - u_d_count_range)]); # append down to up values else: array.append(A[layer + l_r_count_range - (count - r_l_count_ran\ ge)] [layer]); # append center value if matrix had odd dimensions if N % 2 == 1: array.append(A[(N - 1) / 2][(N - 1) / 2]); return array; if __name__ == '__main__': main();

random_guy on Mar 24, 2013
0

arr=[[1, 2, 3, 4], [5, 6, 7, 8], [9, 10,11,12], [13,14,15,16]] # arr=[[1,2,3],[4,5,6],[7,8,9]] rows=len(arr) cols=len(arr[0]) LEFT, RIGHT, UP, DOWN=0, 1, 2, 3 delta=[(-1,0), # LEFT (1,0), # RIGHT (0,-1), # UP (0,1)] # DOWN change_direction=(UP, # LEFT now goes UP DOWN, # RIGHT now goes DOWN RIGHT, # UP now goes RIGHT LEFT) # DOWN now goes LEFT def snake(x, y): visited=[] # this matrix holds True for every point we had visited for i in range(rows): visited.append([False]*cols) # initially False - never visited direction=RIGHT # initial direction route=[(x, y)] # we are located at x,y on map visited[y][x] = True while True: nx,ny = delta[direction] nx += x ny += y # change direction if new coordinate is not valid if nx >= cols or nx < 0 or ny>=rows or ny < 0 or visited[ny][nx]: direction=change_direction[direction] nx,ny = delta[direction] nx += x ny += y # if still not valid - break if nx>= cols or nx < 0 or ny>=rows or ny < 0 or visited[ny][nx]: break # advance to new coordinates x, y = nx, ny route.append((x, y)) # add it to route visited[y][x] = True # mark this place as visited print(",".join([str(arr[y][x]) for x,y in route])) if __name__ == '__main__': snake(0,0)

idispatch on Apr 17, 2013
0

public static void printSpiral(Array a) { ps(a, 0, a.size()-1); } public static void ps(Array a, int s, int f) { if (f < s) return; for (int i = s; i <= f; i++) System.out.println(a.get(i,s)); for (int j = s+1; j < f; j++) System.out.println(a.get(f, j)); for (int i = f; i >= s; i--) System.out.println(a.get(i, f)); for (int j = f-1; j > s; j--) System.out.println(a.get(s,j)); ps(a, s+1, f-1); }

Rahul on May 4, 2013
0

public static void print2DArraySnakePattern(int[][] arr) { if (arr == null) { return; } for (int i=0; i < arr.length; i++) { if (i % 2 == 0) { for (int j=0; j < arr[0].length;j++) { System.out.print(arr[i][j]+","); } } else { for (int j=arr[0].length -1; j >= 0;j--) { System.out.print(arr[i][j]+","); } } } }

Ben H on Jul 27, 2013

Add Answers or Comments

To comment on this, Sign In or Sign Up.