Maximum Water Trapped Between Walls

mediumcodingAlgorithmsTwo Pointers

Overview

Given n non-negative integers representing wall heights where the width of each wall is 1, find two walls that together with the x-axis form a container that holds the most water.

Return the maximum amount of water the container can store.

Constraints

  • 2 <= height.length <= 100,000
  • 0 <= height[i] <= 10,000
  • Must run in O(n) time.
  • You may not slant the container.

Examples

maxArea([1, 8, 6, 2, 5, 4, 8, 3, 7]);
// => 49  (walls at index 1 and 8: min(8,7) * 7 = 49)
maxArea([1, 1]);
// => 1

maxArea([4, 3, 2, 1, 4]);
// => 16  (walls at index 0 and 4: min(4,4) * 4 = 16)

Solution

Reveal solution
function maxArea(height) {
  let left = 0, right = height.length - 1;
  let max = 0;

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

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

  return max;
}

Two pointers start at opposite ends. The area is limited by the shorter wall, so move the shorter pointer inward. O(n) time, O(1) space.

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