The Median Maintenance Problem: Difference between revisions
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Every time a new element arrives, it will be inserted in H<sub>low</sub>. | Every time a new element arrives, it will be inserted in H<sub>low</sub>. | ||
While processing an odd element, H<sub>low</sub> contains (k + 1)/2 elements and H<sub>high</sub> contains (k-1)/2 elements. The median, which according to the definition, is the (k + 1)/2 | While processing an odd element, H<sub>low</sub> contains (k + 1)/2 elements and H<sub>high</sub> contains (k-1)/2 elements. The median, which according to the definition, is the (k + 1)/2 element in the sequence of sorted k elements, is at the top of the H<sub>low</sub> heap and can be retrieved with the MAXIMUM() operation. | ||
While processing an even element, H<sub>low</sub> contains immediately after insertion a larger number of elements than it should, so both heaps should be rebalanced by extracting the maximum element from H<sub>low</sub> with REMOVE-MAX() and inserting it in H<sub>high</sub> with INSERT(). After the sequence of operations is complete, H<sub>low</sub> and H<sub>high</sub> will both contain the same number of elements k/2. The current median will still be at the top of H<sub>low</sub> and can be retrieved with the MAXIMUM() operation. | While processing an even element, H<sub>low</sub> contains immediately after insertion a larger number of elements than it should, so both heaps should be rebalanced by extracting the maximum element from H<sub>low</sub> with REMOVE-MAX() and inserting it in H<sub>high</sub> with INSERT(). After the sequence of operations is complete, H<sub>low</sub> and H<sub>high</sub> will both contain the same number of elements k/2. The current median will still be at the top of H<sub>low</sub> and can be retrieved with the MAXIMUM() operation. |
Revision as of 23:08, 13 October 2021
Internal
Problem
Give a sequence of numbers x1, x2, .... xn, report at each step i the median number among the numbers seen so far. For a given k, the and assuming that x1, x2, .... xk are sorted in an ascending order, the median will be on (k + 1)/2 position if k is odd and on k/2 position is k is even.
Discussion
The problem can be resolved by repeatedly solving the selection problem on the set of numbers seen so far, but the selection problem has a running time of O(n), so repeating the selection algorithm for each number will have a running time of O(n2). The median maintenance problem is a canonical use case for a heap.
Solution
Use two heaps: a max heap Hlow that supports the REMOVE-MAX operation with O(1) running time and a min heap Hhigh that supports the REMOVE-MIN operation, also with O(1) running time.
The idea is to maintain the first smallest half elements in Hlow and the largest half of the elements in Hhigh.
Every time a new element arrives, it will be inserted in Hlow.
While processing an odd element, Hlow contains (k + 1)/2 elements and Hhigh contains (k-1)/2 elements. The median, which according to the definition, is the (k + 1)/2 element in the sequence of sorted k elements, is at the top of the Hlow heap and can be retrieved with the MAXIMUM() operation.
While processing an even element, Hlow contains immediately after insertion a larger number of elements than it should, so both heaps should be rebalanced by extracting the maximum element from Hlow with REMOVE-MAX() and inserting it in Hhigh with INSERT(). After the sequence of operations is complete, Hlow and Hhigh will both contain the same number of elements k/2. The current median will still be at the top of Hlow and can be retrieved with the MAXIMUM() operation.
More details: