2. Container With Most Water.md

Question #2. Container With Most Water

Given n non-negative integers a1, a2, ..., an , where each represents a point at coordinate (i, ai). n vertical lines are drawn such that the two endpoints of the line i is at (i, ai) and (i, 0). Find two lines, which, together with the x-axis forms a container, such that the container contains the most water.

Notice that you may not slant the container.

Examples

• 

  maxArea([1, 8, 6, 2, 5, 4, 8, 3, 7]) // should return 49

  Explanation: The above vertical lines are represented by array [1,8,6,2,5,4,8,3,7]. In this case, the max area of water (blue section) the container can contain is 49.

• maxArea([1, 1]) // should return 1

• maxArea([4, 3, 2, 1, 4]) // should return 16

• maxArea([1, 2, 1]) // should return 2

Solution

Brute-force Solution

• Time Complexity: O(N²)
• Space Complexity: O(1)

/**
 * @param {number[]} height
 * @return {number}
 */
function maxArea(height) {
  //  Initialize a variable to track the largest area
  let largest = 0;
  //  Loop over the height array (a)
  for (let a = 0; a < height.length; a++) {
    //  Loop over the succeeding elements (b)
    for (let b = a + 1; b < height.length; b++) {
      //  Using the two current elements,
      //  calculate the area of the container
      //? Area = Length * Width
      //? Length = min(height[a], height[b])
      //? Width = b - a
      const area = Math.min(height[a], height[b]) * (b - a);
      //  Update the largest area as needed
      largest = Math.max(largest, area);
    }
  }
  //  Return the largest area
  return largest;
}

Optimal Solution

• Time Complexity: O(N)
• Space Complexity: O(1)

/**
 * @param {number[]} height
 * @return {number}
 */
function maxArea(height) {
  //  Initialize two pointers (a and b)
  //? a points to start, b points to end
  let a = 0;
  let b = height.length - 1;
  //  Initialize a variable to track the largest area
  let largest = 0;
  //  Loop while a is behind b:
  while (a < b) {
    //  Using the two current elements,
    //  calculate the area of the container
    //? Area = Length * Width
    //? Length = min(height[a], height[b])
    //? Width = b - a
    const area = Math.min(height[a], height[b]) * (b - a);
    //  Update the largest area as needed
    largest = Math.max(largest, area);
    //  If height[a] is less than height[b]:
    if (height[a] < height[b]) {
      //  Move a forward
      a++;
    }
    //  Otherwise:
    else {
      //  Move b backward
      b--;
    }
  }
  //  Return the largest area
  return largest;
}