Introduction to Algorithms
3rd Edition
ISBN: 9780262033848
Author: Thomas H. Cormen, Ronald L. Rivest, Charles E. Leiserson, Clifford Stein
Publisher: MIT Press
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Chapter 14.3, Problem 6E
Program Plan Intro
To maintain a dynamic set Q of numbers that supports the operation MIN-GAP that gives the magnitude of difference of the two closest numbers in Q.
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If we can assume that the keys in the list have been arranged in order (forexample, numerical or alphabetical order), then we can terminate unsuccessfulsearches more quickly. If the smallest keys come first, then we can terminatethe search as soon as a key greater than or equal to the target key has beenfound. If we assume that it is equally likely that a target key not in the list is inany one of the n + 1 intervals (before the first key, between a pair of successivekeys, or after the last key), then what is the average number of comparisonsfor unsuccessful search in this version?
Consider an initially empty hash table, with an initial size of two buckets, and a load factor threshold of 0.725. After inserting the values 27, 13, 11, 47, 35, and 19 in that order, using big-oh notation, give the average-case time to search for a number between 10 and 50, and the worst-case for the same search. You may assume that the size of the hash table is doubled when necessary.
public int hashCode( int k ) {
int h = k * 3 - 2;
return h % this.size();
}
Problem 2: In this problem we assume that h: U → {0, 1, . . . , m − 1} is a good hashfunction, that is, every key k has the same probability 1/m to map to any place in the tableT of length m.(i) What is the probability that three pairwise distinct elements u1, u2, u3 ∈ U aremapped by the function h to the same place in the table (that is, h(u1) = h(u2) =h(u3))?(ii) We insert three elements into an empty hash table T using the hash function h. Ifcollision is solved by chaining, what is the probability that T[0] and T[1] are empty?
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- Suppose we have a hash map that uses the standard “mod” hash function shown in the chapter and uses linear probing for collision resolution. The starting hash table length is 5, and the table chooses to rehash to twice its former size once the load factor reaches or exceeds 0.5. If we begin with an empty map, what will be the final state of the hash table after the following key/value pairs are added and removed? Draw the entire array and the contents of each index, including any resizing and rehashing necessary. Write “X” in any index in which an element is removed and not replaced by another element. Also write the size, capacity, and load factor of the final hash table. HashMap<Integer, String> map = new HashMap<Integer, String>(); map.put(7, "Jessica"); map.put(34, "Tyler"); map.put(17, "Ryan"); map.put(15, "Tina"); map.put(84, "Saptarshi"); map.remove("Tyler"); map.put(7, "Meghan"); map.put(33, "Kona"); map.remove(17); map.put(6, "Tina"); map.remove(84); map.put(15,…arrow_forwardLet S be the set consisting of all 10 digits and all upper- and lowercase letters. Let U be the set of all (ordered) strings consisting of exactly four elements from S. Construct an injective hash function h: U -> N with the smallest possible value of maxUh(u). Modify your hash function to give a simply uniform map h' to the set {0, ..., 30}. In other words, if U is a probability space with P(u) = P(u') for all u, u' in U, then the random variable h': U -> {0, ..., 30} should have a uniform distribution..arrow_forwardIt is straightforward and efficient to compute the union of two sets using Boolean values. We may create a new union set by Oring the matching items of the two BitArrays since the union of two sets is a combination of the members of both sets. At other words, if the value in the corresponding place of either BitArray is True, a member is added to the new set.Computing the intersection of two sets is analogous to computing the union, except that the And operator is used instead of the Or operator Using the same technique we used to detect the difference, we can determine if one set is a subset of another. For example, if:setA(index) && !(setB(index))evaluates to False then setA is not a subset of setB.The BitArray Set ImplementationWrite The code for a CSet class based on a BitArray.arrow_forward
- What is the worst-case performance of a lookup operation in a hashmap and why? Group of answer choices A- O(1), hashmap always has a constant time lookup, and that is why we like using this associative data structure. B- O(lg(n)) hashmap has a log(n) lookup because we are able to perform a binary search on the keys because our hashmap always maintains a sorted order of entries added. C- O(n) because we can have a bad hash function that puts all of our items in the same bucket, thus we would have to iterate through all n items.arrow_forward#. Consider a hash table of size 11 with hash function h(x) = x mod 11.Draw both hash tables (Closed-Bucket and Open-Bucket) that results after inserting, in the given order, the following values: 45, 12, 35, 67, 25, 52, 88, 70, 68, 53, 55arrow_forwardNote: Your solution should have O(n) time complexity, where n is the number of elements in l, and O(1) additional space complexity, since this is what you would be asked to accomplish in an interview. Given a linked list l, reverse its nodes k at a time and return the modified list. k is a positive integer that is less than or equal to the length of l. If the number of nodes in the linked list is not a multiple of k, then the nodes that are left out at the end should remain as-is. You may not alter the values in the nodes - only the nodes themselves can be changed.arrow_forward
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