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Solving the Range Minimum Query problem

Project description

PyPI version PyPI pyversions PyPI license

This data structure solves the range minimum query problem of finding the minimal value in a sub-array of an array of comparable objects. Different from the original problem, this data structure also supports updating the values.


This package is available on PyPI. Just use pip install -U RangeMinQuery to install it. Then you can import it using from RangeMinQuery import SegmentTree.



Use SegmentTree() to initialize the tree with a set of keys, in comparable and hashable type.

  • func=min specifies how the best value is computed for any range of keys.

  • default=None specifies the default value for each key.

  • maxChildNum=2 specifies the maximum number of children for each node.

tree = SegmentTree(
    {1, 2, 3, 4, 5},
    func=min, default=0, maxChildNum=2

The space complexity should be $O(n)$.


You need to use update() to initialize the values, or update the values if necessary, by specifying a dictionary of key/value pairs. Currently, adding new keys is not supported yet.

tree.update({1: 3, 4: 6})

Given m values updated, the time complexity should be $O(m^2)$.


Use query() to to find the best value of a range of keys. The range is denoted by a tuple (a, b), representing each key x such that a <= x < b. The range here is closed on the left side and open on the right side, consistent with Python tradition.

tree.query((1, 3))

The time complexity should be $O(log n)$.

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