@@ -925,17 +925,18 @@ trees. Building a traditional decision tree (as in the other GBDTs
925925requires sorting the samples at each node (for
926926each feature). Sorting is needed so that the potential gain of a split point
927927can be computed efficiently. Splitting a single node has thus a complexity
928- of :math: `\mathcal {O}(\text {n_features} * n \log (n))` where :math: `n` is the
929- number of samples at the node.
928+ of :math: `\mathcal {O}(n_ \text {features} \times n \log (n))` where :math: `n`
929+ is the number of samples at the node.
930930
931931:class: `HistGradientBoostingClassifier ` and
932932:class: `HistGradientBoostingRegressor `, in contrast, do not require sorting the
933933feature values and instead use a data-structure called a histogram, where the
934934samples are implicitly ordered. Building a histogram has a
935935:math: `\mathcal {O}(n)` complexity, so the node splitting procedure has a
936- :math: `\mathcal {O}(\text {n_features} * n)` complexity, much smaller than the
937- previous one. In addition, instead of considering :math: `n` split points, we
938- here consider only ``max_bins `` split points, which is much smaller.
936+ :math: `\mathcal {O}(n_\text {features} \times n)` complexity, much smaller
937+ than the previous one. In addition, instead of considering :math: `n` split
938+ points, we here consider only ``max_bins `` split points, which is much
939+ smaller.
939940
940941In order to build histograms, the input data `X ` needs to be binned into
941942integer-valued bins. This binning procedure does require sorting the feature
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