Harsh Rajpal

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# 11. Container With Most Water

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

Constraints:

• n == height.length
• 2 <= n <= 105
• 0 <= height[i] <= 104

Solution:
Algorithm:

1. Initialize the maxArea to 0
2. Initialize the left pointer to 0 and the right pointer to the length of the array - 1
3. Loop through the array while the left pointer is less than the right pointer
4. Calculate the area by multiplying the minimum of the height of the left and right pointer by the difference of the right and left pointer
5. Update the maxArea by comparing the maxArea and the area
6. If the height of the left pointer is less than the height of the right pointer, increment the left pointer, else decrement the right pointer
7. Return the maxArea

Code:

``````class Solution {
public int maxArea(int[] height) {

int maxArea = 0;
int left = 0;
int right = height.length - 1;
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;

}
}
``````

Time Complexity:
O(n)
Space Complexity:
O(1)