54 lines
1.3 KiB
Ruby
54 lines
1.3 KiB
Ruby
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# Given a sorted array and a value x, Find Ceil(ceiling of x is the smallest element in array greater than or equal to x)
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# and Floor(the greatest element smaller than or equal to x) of x.
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#Algorithm: Binary-search
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#Time-complexity: O(logn),Auxiliary-space:O(1)
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def find_floor_ceil(a,key)
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len=a.length
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floor=find_floor(a,0,len,key)
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ceil=find_ceil(a,-1,len-1,key)
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if floor==-1
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puts "floor doesn't exist"
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if ceil==-1
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puts "Ceil doesn't exist"
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return -1
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else
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return "Ceil is #{a[ceil]}"
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end
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else
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if ceil==-1
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puts "Ceil doesn't exist"
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return "Floor is #{a[floor]}"
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else
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return "Floor is #{a[floor]} and Ceil is #{a[ceil]}"
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end
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end
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end
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#Method to find floor
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def find_floor(a,lo,hi,key)
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return -1 if key<a[lo]
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while hi-lo>1
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mid=lo+(hi-lo)/2
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if a[mid]<=key
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lo=mid
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else
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hi=mid
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end
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end
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return lo
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end
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#Method to find Ceil
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def find_ceil(a,lo,hi,key)
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return -1 if key>a[hi]
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while hi-lo>1
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mid=lo+(hi-lo)/2
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if a[mid]>=key
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hi=mid
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else
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lo=mid
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end
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end
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return hi
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end
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find_floor_ceil([1, 2, 8, 10, 10, 12, 19],19) # => Floor is 19 and Ceil is 19
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