maxWaterTrappedBetweenWall

/**
 * Interview Question:
 * Given an array of non-negative integers where each value represents the
 * height of a vertical wall, find the maximum amount of water that can be held
 * between any two walls and the x-axis.
 *
 * Requirements:
 * - Choose two walls from the array
 * - The water height is limited by the shorter wall
 * - The water width is the distance between the two chosen walls
 * - Return the maximum possible area: width * min(leftHeight, rightHeight)
 *
 * Approach:
 * Use two pointers, one at the start and one at the end. Calculate the current
 * area, update the maximum, then move the pointer at the shorter wall inward.
 * Moving the taller wall cannot improve the area because the width shrinks and
 * the shorter wall is still the limiting height.
 *
 * Complexity:
 * O(n) time and O(1) space.
 */

/**
 * @param {number[]} walls
 * @return {number}
 */
export default function maximumWaterBetweenWalls(walls) {
  let left = 0;
  let right = walls.length - 1;
  let maxVolume = 0;

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

    maxVolume = Math.max(maxVolume, volumne);
    // Move the shorter wall inward.
    // Why? The current volume is limited by the shorter wall.
    // Moving the taller wall cannot help because width gets smaller
    // while the limiting height stays the same or lower.
    if (walls[left] < walls[right]) {
      left++;
    } else {
      right--;
    }
  }
  return maxVolume;
}