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solutions/testtt.py

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+#!/usr/bin/env python3
+# -*- coding: utf-8 -*-
+
+"""
+Created on 12.01.2025
+@author: Mushtary Alam
+"""
+
+
+def trap(height):
+    """
+    Calculates the amount of rainwater that can be trapped given an elevation map.
+
+    Parameters:
+        height (list[int]): A list of non-negative integers representing the elevation map, where the width of each bar is 1.
+    Returns:
+        int: The total amount of trapped rainwater.
+
+    Logic:
+        - Uses two pointers (left and right) to traverse the elevation map.
+        - Tracks the maximum height to the left and right of each bar.
+        - Accumulates water trapped based on the minimum of these maximum heights.
+
+        Examples:
+        >>> trap([0,1,0,2,1,0,1,3,2,1,2,1])
+        6
+        >>> trap([4,2,0,3,2,5])
+        9
+        >>> trap([1,0,2,1,0,1,3,2,1,2,1])
+        6
+
+    """
+
+    # Defensive assertions for input validation
+    assert isinstance(height, list), "Input must be a list."
+    assert all(isinstance(h, int) and h >= 0 for h in height), (
+        "All elements must be non-negative integers."
+    )
+
+    # If there are fewer than 3 elements, no water can be trapped.
+    if not height or len(height) < 3:
+        return 0
+
+    # Initialize two pointers and variables to track the max heights
+    left, right = 0, len(height) - 1
+    left_max, right_max = height[left], height[right]
+    water = 0
+
+    # Traverse the list with two pointers to calculate trapped water
+    while left < right:
+        if height[left] < height[right]:
+            # Move the left pointer and update the left_max
+            left += 1
+            left_max = max(left_max, height[left])
+            # Calculate the trapped water at the current left position
+            water += max(0, left_max - height[left])
+        else:
+            # Move the right pointer and update the right_max
+            right -= 1
+            right_max = max(right_max, height[right])
+            # Calculate the trapped water at the current right position
+            water += max(0, right_max - height[right])
+
+    return water