Q12: How do you solve Container With Most Water using Two Pointers?

Given $n$ vertical lines, find two lines that form a container storing the most water. Initialize pointers at both ends (left = 0, right = len - 1). Calculate area: $text{width} times min(text{height}[left], text{height}[right])$. Move the pointer pointing to the shorter line inward, as moving the taller line can never yield a larger area.

Container With Most Water Solution

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

💡 Shrinks container width while attempting to discover larger height boundaries.

Key Takeaways

Two pointers start at array boundaries.
Move the pointer with the smaller height inward to seek a larger container area.

Knowledge Check

Why is moving the pointer pointing to the taller line guaranteed NOT to increase the container area?

Q12: How do you solve Container With Most Water using Two Pointers?

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