CS223 Advanced Data Structures and Algorithms 1

Priority Queue and Binary HeapPriority Queue and Binary Heap

Neil TangNeil Tang02/09/201002/09/2010

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Class OverviewClass Overview

Priority queue

Binary heap

Heap operations: insert, deleteMin, de/increaseKey, delete, buildHeap

Application

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Priority QueuePriority Queue

A priority queue is a queue in which each element has a

priority and elements with higher priorities are supposed to

be removed before the elements with lower priorities.

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Possible SolutionsPossible Solutions

Linked list: Insert at the front (O(1)) and traverse the list to delete (O(N)).

Linked list: Keep it always sorted. traverse the list to insert (O(N)) and delete the first element (O(1)).

Binary search tree

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Binary HeapBinary Heap

A binary heap is a binary tree that is completely filled, with possible exception of the bottom level and in which for every node X, the key in the parent of X is smaller than (or equal to) the key in X.

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Binary HeapBinary Heap

A complete binary tree of height h has between 2h and 2h+1 -1 nodes. So h = logN.

For any element in array position i, its left child in position 2i and the right child is in position (2i+1), and the parent is in i/2.

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Insert 14Insert 14

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Insert (Percolate Up)Insert (Percolate Up)

Time complexity: O(logN)

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deleteMindeleteMin

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deleteMin deleteMin (Percolate Down)(Percolate Down)

Time complexity: O(logN)

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Other OperationsOther Operations

decreaseKey(p,)

increaseKey(p, )

delete(p)?

delete(p)=decreaseKey(p,)+deleteMin()

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buildHeapbuildHeap

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buildHeapbuildHeap

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buildHeapbuildHeap

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buildHeapbuildHeap

Theorem: For the perfect binary tree of height h with (2h+1-1) nodes the sum of the heights of the nodes is (2h+1-1-(h+1)).

Time complexity: 2h+1-1-(h+1) = O(N).

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ApplicationsApplications

Problem: find the kth smallest element.

Algorithm: buildHeap, then deleteMin k times.

Time complexity: O(N+klogN) = O(NlogN).

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ApplicationsApplications

Problem: find the kth largest element.

Algorithm: buildHeap with the first k elements, check the rest one by one. In each step, if the new element is larger than the element in the root node, deleteMin and insert the new one.

Time complexity: O(k+(N-k)logk) = O(NlogN).