Container With Most Water

Why it belongs on this sheet

It is a classic two-pointer problem where brute force is easy to think of, but the greedy elimination is the real interview value.

Pattern

Move the limiting side

Approach

Start with the widest container. The height is limited by the shorter wall, so moving the taller wall cannot help if the shorter wall stays the bottleneck. Move the shorter side inward.

Java solution

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

    while (left < right) {
      int width = right - left;
      int area = Math.min(height[left], height[right]) * width;
      best = Math.max(best, area);

      if (height[left] < height[right]) {
        left++;
      } else {
        right--;
      }
    }

    return best;
  }
}

Complexity

• Time: O(n)
• Space: O(1)

Interview note

What matters here is the proof idea for why moving the shorter side is safe.