Container With Most Water

Problem Statement

Given an array of heights representing vertical lines, find two lines that, together with the x-axis, form a container that holds the most water. This is a medium-level problem solved using two pointers.

Example

Input: height = [1, 8, 6, 2, 5, 4, 8, 3, 7]

Output: 49 (lines at indices 1 and 8 with height 7 and width 7 form area 49)

Code

public class Solution {
    public int maxArea(int[] height) {
        int left = 0, right = height.length - 1;
        int maxWater = 0;
        
        while (left < right) {
            int width = right - left;
            int minHeight = Math.min(height[left], height[right]);
            maxWater = Math.max(maxWater, width * minHeight);
            
            if (height[left] < height[right]) {
                left++;
            } else {
                right--;
            }
        }
        return maxWater;
    }
}
            

def max_area(height):
    left, right = 0, len(height) - 1
    max_water = 0
    
    while left < right:
        width = right - left
        min_height = min(height[left], height[right])
        max_water = max(max_water, width * min_height)
        
        if height[left] < height[right]:
            left += 1
        else:
            right -= 1
    return max_water

# Example usage
print(max_area([1, 8, 6, 2, 5, 4, 8, 3, 7]))  # 49
            

function maxArea(height) {
    let left = 0, right = height.length - 1;
    let maxWater = 0;
    
    while (left < right) {
        let width = right - left;
        let minHeight = Math.min(height[left], height[right]);
        maxWater = Math.max(maxWater, width * minHeight);
        
        if (height[left] < height[right]) {
            left++;
        } else {
            right--;
        }
    }
    return maxWater;
}

// Example usage
console.log(maxArea([1, 8, 6, 2, 5, 4, 8, 3, 7])); // 49
            

Explanation

• Use two pointers: left at start, right at end.
• Calculate area as width (distance between pointers) times the minimum height.
• Update max area if current area is larger.
• Move the pointer with the shorter height inward to potentially increase area.

Note