unknown Interview Question for Randoms

Country: India
Interview Type: In-Person

Comment hidden because of low score. Click to expand.
of 1 vote

BFS with multiple sources. Something like that:

M = ['110',
C,R = len(M[0]), len(M)
def isB(i,j):
    return i in [0,R+1] or j in [0,C+1] or M[i-1][j-1] == '0'
# add boundary for simpler checks
m = [[[1,0][isB(i,j)] for j in range(C+2)] for i in range(R+2)]
q = [(1,i,1) for i in range(1,R+1) if m[i][1]]
v = set()
while q:
    s,i,j = q.pop()
    if (i,j) in v or not m[i][j]:
    if j == C:
    q = [(s+1,i-1,j),
         (s+1,i,j+1)] + q

- adr September 30, 2019 | Flag Reply

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