A dense, fast Int map implemented as a 64-ary tree that covers an interval $[0, n)$.

Since: 1.1.0.0

$O(n)$ Creates an IntMap for an interval $[0, n)$.

Since: 1.1.0.0

$O(n + m \log n)$ Creates an IntMap for an interval $[0, n)$ with initial values.

Since: 1.1.0.0

$O(1)$ Returns the capacity $n$, where the interval $[0, n)$ is covered by the map.

Since: 1.1.0.0

$O(1)$ Returns the number of elements in the map.

Since: 1.1.0.0

$O(1)$ Returns whether the map is empty.

Since: 1.1.0.0

$O(\log n)$ Looks up the value for a key.

Since: 1.1.0.0

$O(\log n)$ Tests whether a key $k$ is in the map.

Since: 1.1.0.0

$O(\log n)$ Tests whether a key $k$ is not in the map.

Since: 1.1.0.0

$O(\log n)$ Looks up the $(k, v)$ pair with the smallest key $k$ such that $k \ge k_0$.

Since: 1.1.0.0

$O(\log n)$ Looks up the $(k, v)$ pair with the smallest $k$ such that $k \gt k_0$.

Since: 1.1.0.0

$O(\log n)$ Looks up the $(k, v)$ pair with the largest key $k$ such that $k \le k_0$.

Since: 1.1.0.0

$O(\log n)$ Looks up the $(k, v)$ pair with the largest key $k$ such that $k \lt k_0$.

Since: 1.1.0.0

$O(\log n)$ Looks up the $(k, v)$ pair with the minimum key $k$.

Since: 1.1.0.0

$O(\log n)$ Looks up the $(k, v)$ pair with the maximum key $k$.

Since: 1.1.0.0

$O(\log n)$ Inserts a $(k, v)$ pair into the map. If an entry with the same key already exists, it is overwritten.

Since: 1.1.0.0

$O(\log n)$ Inserts a $(k, v)$ pair into the map. If an entry with the same key already exists, it overwritten with $f(v_{\mathrm{new}}, v_{\mathrm{old}})$.

Since: 1.1.0.0

$O(\log n)$ Modifies the value associated with a key. If an entry with the same key already does not exist, nothing is performed.

Since: 1.1.0.0

$O(\log n)$ Modifies the value associated with a key. If an entry with the same key already does not exist, nothing is performed.

Since: 1.1.0.0

$O(\log n)$ Deletes the $(k, v)$ pair with the key $k$ from the map. Does nothing if no such key exists. Returns whether the key existed.

Since: 1.1.0.0

$O(\log n)$ Deletes the $(k, v)$ pair with the key $k$ from the map. Does nothing if no such key exists.

Since: 1.1.0.0

$O(\log n)$ Deletes the $(k, v)$ pair with the minimum key $k$ in the map.

Since: 1.1.0.0

$O(\log n)$ Deletes the $(k, v)$ pair with maximum key $k$ in the map.

Since: 1.1.0.0

$O(n \log n)$ Enumerates the keys in the map.

Since: 1.1.0.0

$O(n \log n)$ Enumerates the elements (values) in the map.

Since: 1.1.0.0

$O(n \log n)$ Enumerates the key-value pairs in the map.

Since: 1.1.0.0