LeetCode - 11. Container With Most Water

Thoughts:

I choose two pointer to solve this problem, although this problem is not sorted array, but still need to compare the two values.

Thoughts visuzlization:

height = [1,8,6,2,5,4,8,3,7]
max = 0
l is left pointer, and r is right pointer
I can calculate the area by narrowing the width
and compare the heights

if height[l] is smaller than height[r], move left pointer to right
if height[r] is smaller than height[l], move right pointer to left
if same, narrow the width by moving both
why move both
if I move to the lower one, the area will be smaller
if I move to the higher one, I still need to use the other one that is equal as before, the area will be smaller too

result = (r - l) * Math.min(height[l], height[r])
and compare result with max value

[1,8,6,2,5,4,8,3,7]
 l               r
[1,8,6,2,5,4,8,3,7]
   l             r
[1,8,6,2,5,4,8,3,7]
   l           r
[1,8,6,2,5,4,8,3,7]
   l         r
[1,8,6,2,5,4,8,3,7]
     l     r
[1,8,6,2,5,4,8,3,7]
     l   r
[1,8,6,2,5,4,8,3,7]
     l r

Code:

/**
 * @param {number[]} height
 * @return {number}
 */
var maxArea = function (height) {
  let l = 0;
  let r = height.length - 1;
  let max = (r - l) * Math.min(height[l], height[r]);
  while (l < r) {
    if (height[l] < height[r]) l++;
    else if (height[l] > height[r]) r--;
    else {
      l++;
      r--;
    }
    max = Math.max(max, (r - l) * Math.min(height[l], height[r]));
  }
  return max;
};

Reference

Original link