|
| 1 | +from typing import List |
| 2 | + |
| 3 | + |
| 4 | +class Solution: |
| 5 | + def maxArea(self, heights: List[int]) -> int: |
| 6 | + """ |
| 7 | + Find the maximum area that can form a container with the most water. |
| 8 | +
|
| 9 | + Problem: Given n non-negative integers representing heights of vertical lines, |
| 10 | + find two lines that together with the x-axis forms a container that holds the most water. |
| 11 | +
|
| 12 | + Approach: |
| 13 | + - Use two pointers technique starting from both ends of the array |
| 14 | + - Calculate area at each step and keep track of maximum |
| 15 | + - Move the pointer with smaller height inward (since it limits the container height) |
| 16 | +
|
| 17 | + Proof of correctness: |
| 18 | + - Every time we move a pointer, the width decreases by 1. |
| 19 | + - Why we move the pointer with smaller height: |
| 20 | + 1) If we keep the smaller height and move the taller pointer instead, |
| 21 | + the width decreases but the height remains limited by the smaller value, |
| 22 | + so the area will always decrease. |
| 23 | + 2) However, if we move the pointer with smaller height, we might find a taller line. |
| 24 | + In this case, even though width decreases, height might increase enough |
| 25 | + to create a larger area. |
| 26 | + - This way, we examine all possible combinations that could give us the maximum area. |
| 27 | + - We only explore combinations that have the potential to create an area larger |
| 28 | + than the maximum we've found so far. |
| 29 | +
|
| 30 | + Time Complexity: O(n) where n is the length of heights array |
| 31 | + Space Complexity: O(1) using constant extra space |
| 32 | +
|
| 33 | + Args: |
| 34 | + heights: List of heights of the vertical lines |
| 35 | +
|
| 36 | + Returns: |
| 37 | + Maximum water area that can be contained |
| 38 | + """ |
| 39 | + max_area = 0 # Initialize the maximum area to 0 |
| 40 | + left, right = 0, len(heights) - 1 # Start with leftmost and rightmost positions |
| 41 | + |
| 42 | + # Continue until the pointers meet |
| 43 | + while left < right: |
| 44 | + # Calculate width between current lines (difference in positions) |
| 45 | + width = right - left |
| 46 | + |
| 47 | + # Height is limited by the shorter line |
| 48 | + current_height = min(heights[left], heights[right]) |
| 49 | + |
| 50 | + # Calculate current area and update max_area if larger |
| 51 | + current_area = width * current_height |
| 52 | + max_area = max(max_area, current_area) |
| 53 | + |
| 54 | + # Move the pointer with smaller height inward |
| 55 | + # (Moving the smaller one gives potential for larger area since |
| 56 | + # width will decrease but height might increase) |
| 57 | + if heights[left] < heights[right]: |
| 58 | + left += 1 |
| 59 | + else: |
| 60 | + right -= 1 |
| 61 | + |
| 62 | + return max_area |
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