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33 lines
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Ruby

#Given an array A[] consisting 0s, 1s and 2s, put all 0s first, then all 1s and all 2s in last.
##Brute-force: Count 0s and 1s (using 2 varibales)and rewrite array first with counted zeros then by counted 1s and then by 2s
#Time-complexity:O(n),Auxiliary-space:O(1)
##Algorithm: Dutch National Flag Algorithm
#Time-complexity:O(n),Auxiliary-space:O(1)
def segregate(a)
right=a.length-1
left=0
current=0
while(current<=right)
if a[current]==0
swap(a,left,current)
left+=1
current+=1
elsif a[current]==2
swap(a,right,current)
right-=1
else
current+=1
end
end
print a
end
def swap(a,i,j)
a[i],a[j]=a[j],a[i]
end
segregate([0, 1, 1, 0, 1, 2, 1, 2, 0, 0, 0, 1])