MoigeMatino · MoigeMatino · Oct 27, 2024 · Oct 27, 2024 · Oct 27, 2024 · Oct 27, 2024
diff --git a/arrays_and_strings/container_with_most_water/README.md b/arrays_and_strings/container_with_most_water/README.md
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+## **Problem Statement**
+
+You are given an integer array `height` of length `n`. There are `n` vertical lines drawn such that the two endpoints of the `ith` line are `(i, 0)` and `(i, height[i])`.
+
+Find two lines that together with the x-axis form a container, such that the container contains the most water.
+
+Return the maximum amount of water a container can store.
+
+Notice that you may not slant the container.
+
+### Example 1
+    Input: height = [1,8,6,2,5,4,8,3,7]
+    Output: 49
+
+    Explanation: The vertical lines are represented by array [1,8,6,2,5,4,8,3,7]. 
+    In this case, the max area of water the container can contain is 49. Derived from (1, height[1]) and (8, height[8]) which gives us the heights; h1 = 8, h2 = 7. 
+    To get the width: 8 - 1 = 7
+    To get height = min(8, 7) = 7
+    To get area = height * width = 7 * 7 = 49
+
+### Example 2
+    Input: height = [1,1]
+    Output: 1
diff --git a/arrays_and_strings/container_with_most_water/__init__.py b/arrays_and_strings/container_with_most_water/__init__.py
diff --git a/arrays_and_strings/container_with_most_water/container_with_most_water.py b/arrays_and_strings/container_with_most_water/container_with_most_water.py
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+def max_area(height: list[int]) -> int:
+    """
+    Calculate the maximum area of water a container can store, formed by two lines
+    from the given list of heights and the x-axis.
+
+    Parameters:
+    height (list[int]): A list of integers representing the heights of vertical lines.
+
+    Returns:
+    int: The maximum area of water that can be contained.
+
+    """
+    left = 0
+    right = len(height) - 1
+    max_area = 0
+
+    while left < right:
+        # Calculate the height and width of the current container
+        curr_height = min(height[left], height[right])
+        curr_width = right - left
+        curr_area = curr_height * curr_width
+
+        # Update max_area if the current area is larger
+        max_area = max(curr_area, max_area)
+
+        # Move the pointer pointing to the shorter line
+        if height[left] < height[right]:
+            left += 1
+        else:
+            right -= 1
+
+    return max_area
+
+# Approach and Reasoning:
+#     -----------------------
+#     - We use the two-pointer technique to solve this problem efficiently.
+#     - Initialize two pointers, `left` at the start and `right` at the end of the list.
+#     - The width of the container is the distance between the two pointers.
+#     - The height of the container is determined by the shorter of the two lines at the pointers.
+#     - Calculate the area for the current pair of lines and update `max_area` if this area is larger.
+#     - Move the pointer pointing to the shorter line inward to potentially find a taller line,
+#         which might result in a larger area.
+#     - Repeat the process until the two pointers meet.
+
+#     Time Complexity:
+#     ----------------
+#     - The time complexity of this approach is O(n), where n is the number of elements in the `height` list.
+#     - This is because each element is processed at most once as the pointers move towards each other.
+
+#     Space Complexity:
+#     -----------------
+#     - The space complexity is O(1) because we are using a constant amount of extra space,
+#       regardless of the input size.
+