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Unformatted text preview: time. A flip indicates a merge operation. So the total running time for m insertions is ∑-= + = ≤ ≤ 1 1 ) (lg 2 2 2 k i i i n mO mk m T So the amortized running time for each operation is ) (lg / ) (lg n O m n mO = . c. To delete an element, say x , in i A , first we find the smallest array that is not empty. Assume it is j A . Delete x from i A . If j i ≠ , take an element out of j A and insert it into i A . Whether j i ≠ or not, now the length of j A is 1 2-j . Then we divide array j A into j arrays, 1 1 , , ,-j A A A L , which are consistent with this data structure. COT 6401 Analysis of Algorithms Homework #4 by Changfu Wu P.2/2 5.22 [Solution] Algorithm: SELECTION( k n , ) 1 ] 1 [ Key candidate ← 2 for 2 ← i to 1 +-k n 3 if ] [ i Key candidate 4 ] [ i Key candidate ← 5 return candidate Number of comparisons: k n-Lower bound of comparisons: k n-...
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