How do you solve Container With Most Water on LeetCode?

Container With Most Water (LeetCode #11) can be solved using Brute Force or Optimal Solution (Two Pointers). The optimal approach is Optimal Solution (Two Pointers) with O(n) time complexity and O(1) space complexity. Key insight: The area of water that can be contained is determined by the shorter of the two lines and the distance between them. This means we should focus on maximizing the height and the distance.

What is the brute force solution for Container With Most Water?

The brute force approach for Container With Most Water is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking every possible pair of lines to calculate the area they can contain. This is straightforward but inefficient because it requires checking all combinations. The optimal Optimal Solution (Two Pointers) approach improves this to O(n).

What algorithm or data structure is used to solve Container With Most Water?

Container With Most Water uses the following concepts: Array, Two Pointers, Greedy. The recommended patterns to study are: Hash Map, Array, Two Pointers.

What are the interview tips for Container With Most Water?

When you first see this problem, mention that you're considering a brute force approach and then discuss how you might optimize it. Communicate your thought process clearly, especially when transitioning from brute force to the two-pointer approach, explaining why it is more efficient. Mention edge cases such as the smallest possible input (two lines) and cases where all heights are the same.

What are common mistakes when solving Container With Most Water?

Students often forget to account for the shorter line when calculating the area, which can lead to incorrect results. Another common mistake is to not update the pointers correctly, which can cause the algorithm to miss potential larger areas.

#11

Container With Most Water

Medium

ArrayTwo PointersGreedyHash MapArrayTwo Pointers

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution (Two Pointers)★
Time	O(n²)	O(n)
Space	O(1)	O(1)

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Intuition

Time O(n)Space O(1)

The two-pointer approach is efficient because it reduces the number of comparisons by leveraging the properties of the container. By moving the pointer pointing to the shorter line, we can potentially find a taller line that may yield a larger area.

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Algorithm

3 steps

1Step 1: Initialize two pointers, one at the start and one at the end of the height array.
2Step 2: Calculate the area formed by the lines at the two pointers and update the maximum area if this area is larger.
3Step 3: Move the pointer that points to the shorter line inward, as this may lead to a larger area with a taller line.

solution.py15 lines

1# Full working Python code with comments
2
3def maxArea(height):
4    max_area = 0  # Initialize max area
5    left, right = 0, len(height) - 1  # Two pointers
6    while left < right:
7        # Calculate area
8        area = min(height[left], height[right]) * (right - left)
9        max_area = max(max_area, area)  # Update max area
10        # Move the pointer pointing to the shorter line
11        if height[left] < height[right]:
12            left += 1
13        else:
14            right -= 1
15    return max_area

ℹ

Complexity note: The time complexity is O(n) because we only traverse the height array once with two pointers. The space complexity is O(1) since we use a constant amount of extra space.

1The area of water that can be contained is determined by the shorter of the two lines and the distance between them. This means we should focus on maximizing the height and the distance.
2Using two pointers allows us to efficiently explore potential containers without needing to check every pair, leading to a significant reduction in time complexity.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.