Leetcode 11 - Container with most water

Note:

  • My intuition is to use double pointers but how we should determine to move left or right pointer?
  • No matter which we move, W will be smaller.
  • To at least not make our next container smaller, we have to preserve the higher L. So, we should move the smaller pointer who has a shorter L.
  • This can be proved by math, but we don’t bother to.

Question:

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.

Example:

img

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Input: height = [1,8,6,2,5,4,8,3,7]
Output: 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.

Code:

Double pointers O(n)

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