Container With Most Water

1. Define the problem

You are given an integer array height of length n. There are n vertical lines drawn such that the two endpoints of the ith line are (i, 0) and (i, height_i). Find two lines that together with the x-axis form a container, such that the container contains the most water . Return the maximum amount of water a container can store. Notice that you may not slant the container. Start pointers at both ends and always move the pointer at the shorter line inward — moving the taller line can only shrink the width without any chance of gaining height.

Constraints

n == height.length
2 ≤ n ≤ 10⁵
0 ≤ height_i ≤ 10⁴

Example

Inputheight = [1, 8, 6, 2, 5, 4, 8, 3, 7]

Output49

Explanation Lines at index 1 (height 8) and index 8 (height 7) form the container: width 7 x height min(8, 7) = 7 gives area 49, the maximum possible.

2. Visualize the solution

Converge two pointers, always moving the shorter side

Statusinit

left=1 (idx 0), right=7 (idx 8): width 8 x min(1, 7) = 8. best = 8.

What happens in this step

left = 0 (height 1), right = 8 (height 7)
width = 8 - 0 = 8, area = 8 x min(1, 7) = 8

height[left]=1 is the shorter side, so left moves inward from index 0 to index 1. Best area so far: 8.

Step 1 of 4

Steps to visualize

Point left at the first line and right at the last line.
Compute the area: width between the pointers times the shorter of the two heights.
Update the best area seen so far.
Move whichever pointer is at the shorter line one step inward.
Stop when the pointers meet.

3. Walk through the code

Walk through the code

Same walkthrough, now with the code. Press Next to move one step and watch which lines run.

Converge two pointers, always moving the shorter side

Statusinit

left=1 (idx 0), right=7 (idx 8): width 8 x min(1, 7) = 8. best = 8.

What happens in this step

left = 0 (height 1), right = 8 (height 7)
width = 8 - 0 = 8, area = 8 x min(1, 7) = 8

height[left]=1 is the shorter side, so left moves inward from index 0 to index 1. Best area so far: 8.

Step 1 of 4

4. Solution

Solution

solution.tsTypeScript

function maxArea(height) {
  let left = 0;
  let right = height.length - 1;
  let best = 0;

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

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

  return best;
}

Time: O(n)
Space: O(1)

5. Test cases

Test cases

Input	Expected	Covers
`height = [1, 8, 6, 2, 5, 4, 8, 3, 7]`	`49`	example from the docstring
`height = [1, 1]`	`1`	smallest valid input: only two lines
`height = [4, 4, 4, 4]`	`12`	every line the same height, so width alone decides the winner
`height = [1, 2, 3, 4, 5, 6]`	`9`	strictly increasing heights favor a middle pair, not the ends
`height = [0, 3, 0, 5, 0]`	`6`	a zero-height line at the boundary holds no water

Container With Most Water