Container With Most Water — Leetcode

Topic: Array
Difficulty: Medium
Question Number: 11

Problem statement:
You are given an integer array height of length n. There are n vertical lines drawn such that the two endpoints of the ith line are (i, 0) and (i, height[i]).

Find two lines that together with the x-axis form a container, such that the container contains the most water.

Return the maximum amount of water a container can store.

Notice that you may not slant the container.

Example 1:

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.

Example 2:

Input: height = [1,1]
Output: 1

Explanation:

This is a quite confusing question but easy too. Let's first understand this question with a brute-force approach.

Checking for height
First, from the left [0] and right[1] find the area of water that can be filled.
height[0] = 1
height[1] = 8
Water can be filled up to height ‘1’ only. so, Height= 1

Checking for width
distance between left[0] and right[1] is 1
right-left => 1–0 = 1
so, Width = 1

Checking for area
h x w
1 x 1 = 1

Similarly, the left[0] will be checked with the right[2,3,4,5,6,7,8].
Then, left[1,2,3,4,5,6,7,8] will be checked with right[0,1,2,3,4,5,6,7,8]

After, calculating the areas from each possible bar. Writing them down in a matrix form table to find the maximum area.
MAX AREA = 49

but, the time complexity is O(n²)

Not a good approach :(

Let’s try the same question with the optimized approach.

• First mark the left-right pointers where we can get the maximum width. Which always going to be the first number of the array and the last number of the array.
• Find the area of the water and store the value in the area variable
• Follow
  if (height[left] <= height[right])
  left++;
  else
  right — ;
  and move the left-right pointer accordingly.
• from all the areas calculated, find the max area

As you can notice from the above approach, we minimized and optimized the way of getting the answer.

Time complexity- O(n)

Code:

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

Thank you

Oshi Raghav