Difference between revisions of "TADM2E 4.23"
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− | + | == Problem == | |
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+ | 4-23. We seek to sort a sequence ''S'' of ''n'' integers with many duplications, such that the number of distinct integers in ''S'' is O(''log n''). Give an O(''n log log n'') worst-case time algorithm to sort such sequences. | ||
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+ | == Solution == | ||
+ | |||
+ | With a self-balancing binary search tree, we can achieve O(''log k'') search and insertion (where ''k'' is the number of elements in the tree). | ||
+ | |||
+ | |||
+ | 1. For each element in the sequence ''S'', try and insert it into the tree. | ||
+ | |||
+ | - If the element is present, we increment a ''count'' variable stored at a node. | ||
+ | |||
+ | - If the element is not present, we insert the node and set the ''count'' = 1. | ||
+ | |||
+ | |||
+ | 2. Traverse the tree in order and add elements according to the ''count''. | ||
+ | |||
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+ | Since the tree is guaranteed not to exceed ''log n'' elements, search and insertion take O(''log log n''). | ||
+ | |||
+ | Thus, this entire algorithm takes O(''n * log log n'' + ''log n'') = O(''n * log log n'') as required. |
Latest revision as of 00:44, 1 August 2020
Problem
4-23. We seek to sort a sequence S of n integers with many duplications, such that the number of distinct integers in S is O(log n). Give an O(n log log n) worst-case time algorithm to sort such sequences.
Solution
With a self-balancing binary search tree, we can achieve O(log k) search and insertion (where k is the number of elements in the tree).
1. For each element in the sequence S, try and insert it into the tree.
- If the element is present, we increment a count variable stored at a node.
- If the element is not present, we insert the node and set the count = 1.
2. Traverse the tree in order and add elements according to the count.
Since the tree is guaranteed not to exceed log n elements, search and insertion take O(log log n).
Thus, this entire algorithm takes O(n * log log n + log n) = O(n * log log n) as required.