## Problem Statement

Given n non-negative integers a1, a2, …, an, where each represents a point at coordinate (i, ai). ‘n’ vertical lines are drawn such that the two endpoints of line i is at (i, ai) and (i, 0).

Find two lines, which together with the x-axis form a container, such that the container contains the most water.

Your program should return an integer that corresponds to the maximum area of water that can be contained ( Yes, we know maximum area instead of maximum volume sounds weird.

But this is a 2D plane we are working with for simplicity ).

Note: You may not slant the container.

### Confused about your next job?

**Examples:Input: **height[] = {1,8,6,2,5,4,8,3,7}** ****Output:** 49**Explanation: **The maximum area is the area denoted with the blue section. The total area is 49 units.

**Input: **height[] = {1, 1}** ****Output:** 1

## Approach 1: Brute Force

The most basic approach to solve this problem is to simplify all possible areas for every pair of line segments and maximize it.

**Algorithm**

- Initialise
**maxarea = 0**. - Run a nested loop i = 0 till i = N and j = i + 1 till N.
- The area would be the
**min(height[i], height[j]) * (j – i)** - Maximize the area for every iteration.

- The area would be the
- Return
**maxarea.**

### C++ Code

int maxArea(vector<int> height) { int maxarea = 0; for (int i = 0; i < height.size(); i++) for (int j = i + 1; j < height.size(); j++) maxarea = max(maxarea, min(height[i], height[j]) * (j - i)); return maxarea; }

### Java Code

public class Solution { public int maxArea(int[] height) { int maxarea = 0; for (int i = 0; i < height.length; i++) for (int j = i + 1; j < height.length; j++) maxarea = Math.max(maxarea, Math.min(height[i], height[j]) * (j - i)); return maxarea; } }

### Python Code

def maxArea(heights): maxarea = 0 for i in range(len(heights)): for j in range(i + 1, len(heights)): maxarea = max(min(heights[i], heights[j]) * (j - i) return maxarea

**Time Complexity:**O(N^2), where N is the size of the array**Space Complexity:**O(1) since no extra space is used.

## Approach 2: Two Pointers

The most basic observation is that the area formed between the lines will always be limited by the height of the minimum. So, the larger the distance, the greater the area.

Check the below illustration to solve the problem:

**Algorithm:**

- Take two pointers, one at
**i = 0**and another pointer**j = N – 1**. - Consider a variable
**maxarea = 0**, for calculating**maxarea**till index**i.** - At every step, calculate the
**maxarea**between them the two lines pointed**i.e. height[i]**and**height[j]**and update**maxarea**. - Keep incrementing the pointer pointed at smaller height towards each other.

### C++ Implementation

int maxArea(vector<int> height) { int maxarea = 0, l = 0, r = height.size() - 1; while (l < r) { maxarea = max(maxarea, min(height[l], height[r]) * (r - l)); if (height[l] < height[r]) l++; else r--; } return maxarea; }

### Java Implementation

public class Solution { public int maxArea(int[] height) { int maxarea = 0, l = 0, r = height.length - 1; while (l < r) { maxarea = Math.max(maxarea, Math.min(height[l], height[r]) * (r - l)); if (height[l] < height[r]) l++; else r--; } return maxarea; } }

### Python Implementation

def maxArea(height): L, R, width, res = 0, len(height) - 1, len(height) - 1, 0 for w in range(width, 0, -1): if height[L] < height[R]: res, L = max(res, height[L] * w), L + 1 else: res, R = max(res, height[R] * w), R - 1 return res

**Time Complexity:**O(N), where N is the size of the array**Space Complexity:**O(1) since no extra space is used.

**Practice Question**

## FAQs

**How is the area between two points found out?**Since the width of each consecutive block is considered to be 1, the length is simply the distance between the two points and the height is the minimum between the two heights.

**What is the most efficient approach to solving** this problem?

The two-pointer method is the most efficient way to solve this problem. The time complexity is O(N) and space complexity is O(1).