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Ruby

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# A sorted array is rotated at some unknown point, find the minimum element in it
# Time-complexity: O(logn),Auxiliary-space:O(1)
# Algorithm: Binary-Search
#This method cannot handle duplicates
def search_minimum(a)
lo=0
hi=a.length-1
while(lo<=hi)
mid= lo+(hi-lo)/2
if (mid>0 && a[mid-1]>a[mid])
return "Minimum element is #{a[mid]} at #{mid}"
elsif a[mid]>a[hi]
lo=mid+1
else
hi=mid-1
end
end
return "Minimum element is #{a[lo]} at #{lo}"
end
search_minimum([5, 6, 1, 2, 3, 4]) #=> Minimum element is 1 at 2
# It doesnt look possible to search minimum in O(Logn) time in all cases when duplicates are allowed.
# Consider {2, 2, 2, 2, 2, 2, 2, 2, 0, 2} and {2, 0, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2}.