Array rotation-IV (Block Swap Algorithm)
Given an zero indexed array and positive integer d,rotate the array by d-steps to the left.
Block Swap Algorithm:
Block swap algorithm for array rotation
Initialize A=[0..d-1], B=[d..size-1]
until size of A is equal to size of B
a) If A is shorter
1. Divide B into Bl and Br such that Br is of same length as A.
2. Swap A and Br to change ABlBr into BrBlA.
3. Now A is at its final place, so recur on pieces of B.
b) If A is longer
1. Divide A into Al and Ar such that Al is of same length as B.
2. Swap Al and B to change AlArB into BArAl.
3. Now B is at its final place, so recur on pieces of A.
Finally when A and B are of equal size, block swap them.
Time-complexity:O(n)
Auxiliary-space:O(1)
Implementation
RECURSIVE
#Driver function
def rotate_array(a, d) #Input array "a" and rotation by "d" elemets
finish =a.length-1
block_swap(a,0,finish,d)
end
def block_swap(a,start,finish,d)
n=finish-start+1
if n>0
if d>n
d%=n
end
if d==0
return a
end
if d==n-d
swap(a,start,start+d,d)
return a
elsif d<n-d
swap(a,start,finish-d+1,d)
block_swap(a,start,finish-d,d)
else
swap(a,start,d,n-d)
block_swap(a,n-d,n-1,(2*d)-n)
end
end
return a
end
#Utility function for swapping
def swap(a,start1,start2,d)
for i in 0...d
temp = a[start1+i]
a[start1+i] = a[start2+i]
a[start2+i] = temp
end
end
rotate_array([1,2,3,4,5,6,7,8,9,10],6)
Iterative
def block_swap(a,d)
n=a.length
if n>0
if d>=n
d%=n
end
if d==0
return a
end
if d==n-d
swap(a,0,d,d)
end
i= d
j= n-d
while(i!=j)
if i<j
swap(a,d-i,d+j-i,i)
j-=i
else
swap(a,d-i,d,j)
i-=j
end
end
swap(a,d-i,d,i)
end
return a
end
def swap(a,start1,start2,d)
if (start1 != start2)
for i in 0...d
temp = a[start1+i]
a[start1+i] = a[start2+i]
a[start2+i] = temp
end
end
end
block_swap([1,2,3,4,5,6,7,8,9],5)