11. Container With Most Water

Start with the widest container (left=0, right=n-1). Move the shorter side inward since it limits the height and we're reducing width.

Approach

1. Two pointers at both ends.
2. Compute area = min(left_h, right_h) * width.
3. Move the pointer with the shorter height inward.
4. Track maximum area.

Complexity

• Time: O(n)
• Space: O(1)

class Solution {
    public int maxArea(int[] height) {
        int left = 0, right = height.length - 1;
        int maxWater = 0;

        while (left < right) {
            int w = right - left;
            int h = Math.min(height[left], height[right]);
            maxWater = Math.max(maxWater, w * h);
            if (height[left] < height[right]) left++;
            else right--;
        }

        return maxWater;
    }
}