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