Google Interview Question for Software Engineers

Country: United States
Interview Type: In-Person

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of 1 vote

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def solve(board):
    m, n = len(board), len(board[0])
    '''create adjacency list
    rows[0] has all the nodes in 1st row
    rows[1] has all the nodes in 2nd row
    cols[0] has all the nodes in 1st column
    cols[1] has all the nodes in 2nd column
    rows, cols = [set() for i in range(m)], [set() for j in range(n)]
    for i in range(m):
        for j in range(n):
            if board[i][j] is 'O':

    #count the degree (number of neighbors) of every orb
    degrees = dict()
    for i in range(m):
        for j in rows[i]:
            degree = len(rows[i]) + len(cols[j]) - 2
            if degree > 0:
                degrees[(i, j)] = degrees

    #erase the node with minimum positive degree until all nodes have 0 degree
    while degrees:
        erase_orb(degrees, board, rows, cols)

    for row in board:
        print row

#erase the orb with the minimum degree
def erase_orb(degrees, board, rows, cols):
    #find the node with minimum positive degree
    i, j = min(degrees.items(), key=lambda x: x[1])[0]
    #erase this node
    degrees.pop((i, j))
    board[i][j] = 'X'
    print 'erase ', (i, j)

    #remove one degree from every neighbor of the node erased
    for y in rows[i]:
        degrees[(i, y)] -= 1
        if degrees[(i, y)] is 0:
            degrees.pop((i, y))

    for x in cols[j]:
        degrees[(x, j)] -= 1
        if degrees[(x, j)] is 0:
            degrees.pop((x, j))

    ['X', 'X', 'X', 'X', 'X', 'O'],
    ['O', 'X', 'X', 'X', 'X', 'X'],
    ['O', 'X', 'X', 'X', 'X', 'O'],
    ['X', 'X', 'O', 'X', 'O', 'X'],
    ['X', 'O', 'X', 'X', 'X', 'O']

- aonecoding4 November 28, 2018 | Flag Reply
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of 0 vote

O(N^2) time, O(N^2) memory

int playOrbGame(vector<vector<bool>>& orbs)
	if (orbs.empty() || orbs[0].empty())
	  return 0;
	auto N = orbs.size();
	auto M = orbs[0].size();
	vector<vector<int>> orb_column_sums(N, vector<int>(M,0));
	for(size_t i = N - 2; i + 1 > 0; --i)
	  for(size_t j = 0; j < M; ++j)
		orb_column_sums[i][j] = orbs[i+1][j] ? orb_column_sums[i+1][j] + 1 : orb_column_sums[i+1][j];
	int orbs_left = 0;
	for(size_t i = 0; i < N; ++i)
	  int min_column_orbs = max(M,N) + 1;
	  int column_to_keep_orb = -1;
	  for (size_t j = 0; j < M; ++j)
	    if (!orbs[i][j])
		if (orb_column_sums[i][j] < min_column_orbs)
		  min_column_orbs = orb_column_sums[i][j];
		  column_to_keep_orb = j;
	  if (column_to_keep_orb != -1)
	    for(size_t k = i + 1; k < N; ++k)
		  orbs[k][column_to_keep_orb] = false;
		for(size_t l = 0; l < M; ++l)
		  if (l != column_to_keep_orb)
		    orbs[i][l] = false;
	return orbs_left;

- Antake December 02, 2018 | Flag Reply
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of 0 vote

just to see all the zeros that are connected will be in a single component.
So from a connected component of Zeros we only require one zero from that component.
So the answers it the no of connected component.
Can be done in O(N^2) time to add everything in union find by path compression.
0(N^2) space

- googleHelper December 03, 2018 | Flag Reply

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