Container with the most water.

Leetcode link

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Problem Statement

There is an array of heights of a set of containers of water. We need to find the water containers that can hold the highest area of water.

Here, the output is 49 because, 7 is the minimum height, and there are 7 spaces in the x axis. 7*7=49

Algorithm

• First, we define a left and a right pointer. We also define the maxresult and area variables.
• while the left pointer is less than the right pointer,
  • find the area of the left and right pointer. (r-l)x(min(height(r,l))
  • if greater than max then update it.
  • is the left pointer smaller? inc left.
  • else, dec right
• return the max area.

Code

class Solution {
    public int maxArea(int[] height) {
       int left = 0; 
       int right = height.length-1;
       int area = 0;
       int maxarea=0;
       while (left<right){
           area = (right-left) *(Math.min(height[left],height[right]));
           maxarea = Math.max(maxarea, area);
           if (height[left]<height[right]){
               left = left+1;
           }
           else{
               right = right-1;
           }
       }
 
       return maxarea;
    }
}