Fixes issue #12268: Largest rectangle histogram

data_structures/stacks/largest_rectangle_histogram.py

@@ -0,0 +1,39 @@
+def largest_rectangle_area(heights: list[int]) -> int:
+    """
+    Inputs an array of integers representing the heights of bars,
+    and returns the area of the largest rectangle that can be formed
+
+    >>> largest_rectangle_area([2, 1, 5, 6, 2, 3])
+    10
+
+    >>> largest_rectangle_area([2, 4])
+    4
+
+    >>> largest_rectangle_area([6, 2, 5, 4, 5, 1, 6])
+    12
+
+    >>> largest_rectangle_area([1])
+    1
+    """
+    stack: list[int] = []
+    max_area = 0
+    heights = [*heights, 0]  # make a new list by appending the sentinel 0
+    n = len(heights)
+
+    for i in range(n):
+        # make sure the stack remains in increasing order
+        while stack and heights[i] < heights[stack[-1]]:
+            h = heights[stack.pop()]  # height of the bar
+            # if stack is empty, it means entire width can be taken from index 0 to i-1
+            w = i if not stack else i - stack[-1] - 1  # calculate width
+            max_area = max(max_area, h * w)
+
+        stack.append(i)
+
+    return max_area
+
+
+if __name__ == "__main__":
+    import doctest
+
+    doctest.testmod()

machine_learning/frequent_pattern_growth.py

@@ -240,7 +240,7 @@ def ascend_tree(leaf_node: TreeNode, prefix_path: list[str]) -> None:
         ascend_tree(leaf_node.parent, prefix_path)
 
 
-def find_prefix_path(base_pat: frozenset, tree_node: TreeNode | None) -> dict:  # noqa: ARG001
+def find_prefix_path(_: frozenset, tree_node: TreeNode | None) -> dict:
     """
     Find the conditional pattern base for a given base pattern.
 

machine_learning/loss_functions.py

@@ -629,13 +629,15 @@ def smooth_l1_loss(y_true: np.ndarray, y_pred: np.ndarray, beta: float = 1.0) ->
     return np.mean(loss)
 
 
-def kullback_leibler_divergence(y_true: np.ndarray, y_pred: np.ndarray) -> float:
+def kullback_leibler_divergence(
+    y_true: np.ndarray, y_pred: np.ndarray, epsilon: float = 1e-10
+) -> float:
     """
     Calculate the Kullback-Leibler divergence (KL divergence) loss between true labels
     and predicted probabilities.
 
-    KL divergence loss quantifies dissimilarity between true labels and predicted
-    probabilities. It's often used in training generative models.
+    KL divergence loss quantifies the dissimilarity between true labels and predicted
+    probabilities. It is often used in training generative models.
 
     KL = Σ(y_true * ln(y_true / y_pred))
 
@@ -649,6 +651,7 @@ def kullback_leibler_divergence(y_true: np.ndarray, y_pred: np.ndarray) -> float
     >>> predicted_probs = np.array([0.3, 0.3, 0.4])
     >>> float(kullback_leibler_divergence(true_labels, predicted_probs))
     0.030478754035472025
+
     >>> true_labels = np.array([0.2, 0.3, 0.5])
     >>> predicted_probs = np.array([0.3, 0.3, 0.4, 0.5])
     >>> kullback_leibler_divergence(true_labels, predicted_probs)
@@ -659,7 +662,13 @@ def kullback_leibler_divergence(y_true: np.ndarray, y_pred: np.ndarray) -> float
     if len(y_true) != len(y_pred):
         raise ValueError("Input arrays must have the same length.")
 
-    kl_loss = y_true * np.log(y_true / y_pred)
+    # negligible epsilon to avoid issues with log(0) or division by zero
+    epsilon = 1e-10
+    y_pred = np.clip(y_pred, epsilon, None)
+
+    # calculate KL divergence only where y_true is not zero
+    kl_loss = np.where(y_true != 0, y_true * np.log(y_true / y_pred), 0.0)
+
     return np.sum(kl_loss)