Container With Most Water in C++

In this article, we will discuss the problem of finding the container with most water in C++. We will explore different approaches to tackle this problem and implement an optimized solution. So, let’s get started!

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Introduction to the problem

The problem of finding the container with the most water is a classic problem in computer science and is often used to test algorithmic thinking. Given a list of non-negative integers representing the heights of walls, our goal is to find two walls that, together with the x-axis, form a container that holds the maximum amount of water.

Understanding the problem statement

To understand the problem better, let’s consider an example. Suppose we have the following wall heights:

[1, 8, 6, 2, 5, 4, 8, 3, 7]

We can visualize these heights as a set of walls. The container’s height is determined by the shorter wall, and the width is determined by the distance between the walls. Our task is to find two walls that maximize the area of the container.

Exploring the brute force approach

One way to solve this problem is to consider all possible pairs of walls and calculate the area for each pair. We can then return the maximum area found. This approach has a time complexity of O(n^2) as we need to consider all possible combinations.

Implementing the optimized two-pointer approach

To improve the efficiency, we can use a two-pointer approach. We start with two pointers, one at the beginning of the array and the other at the end. We calculate the area between the walls represented by the pointers. Then, we move the pointer that corresponds to the shorter wall towards the center. By doing this, we ensure that we are always considering walls with the potential for a larger area.

We continue this process until the two pointers meet. At each step, we update the maximum area if we find a larger one. This approach reduces the time complexity to O(n) as we only need to traverse the array once.

Let’s implement this optimized approach in C++:

#include <iostream>
#include <vector>
#include <algorithm>

int maxArea(std::vector<int>& height) {
    int left = 0;
    int right = height.size() - 1;
    int maxArea = 0;

    while (left < right) {
        int area = std::min(height[left], height[right]) * (right - left);
        maxArea = std::max(maxArea, area);

        if (height[left] < height[right])

    return maxArea;

int main() {
    std::vector<int> heights = {1, 8, 6, 2, 5, 4, 8, 3, 7};
    int result = maxArea(heights);
    std::cout << "The maximum area is: " << result << std::endl;

    return 0;

Analyzing the time complexity

The optimized two-pointer approach reduces the time complexity to O(n), where n is the number of walls. This improvement is significant compared to the brute force approach, especially for large input sizes.

Testing the solution with examples

Let’s test our solution with a few examples to validate its correctness:

Example 1: Input: [1, 8, 6, 2, 5, 4, 8, 3, 7] Output: The maximum area is: 49

Example 2: Input: [4, 3, 2, 1, 4] Output: The maximum area is: 16

Example 3: Input: [1, 2, 1] Output: The maximum area is: 2

The solution produces the correct output for all the given examples, demonstrating its correctness.


In this article, we explored the problem of finding the container with the most water in C++. We discussed the brute force approach and its limitations. Then, we implemented an optimized solution using the two-pointer approach, which significantly improved the efficiency. We analyzed the time complexity of the solution and tested it with different examples, ensuring its correctness. By employing the optimized approach, we can efficiently solve this problem in C++.

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