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.
Solution
The first though when I saw this question is use iteration from the first index to the last index.
And use a variable to store the max Area
C# Solution
❌Solution1 ➡ Too slow
public class Solution {
public int MaxArea(int[] height) {
int l = height.Length;
int max = 0;
int maxH = -1;
for(int i = 0; i<l-1;i++)
{
if(height[i] < maxH)
{
continue;
}
for(int j =i+1; j<l;j++)
{
int cal = Math.Min(height[i],height[j]) * (j-i);
//max = cal>max? cal : max;
if(cal>max)
{
maxH = height[i];
max = cal;
}
}
}
return max;
}
}
With solution1, it was to slow for this question, so let us try to use two pointer to convergence the array.
Solution2
public class Solution {
public int MaxArea(int[] height) {
int i = 0;
int j = height.Length-1;
int maxAmt = 0;
int iH;
int jH;
int cal;
while(j>i)
{
iH = height[i];
jH = height[j];
//Convergence back and forth
//which is higher,than will convergence from the other side
if(iH>jH)
{
cal = jH*(j-i);
j--;
}
else
{
cal = iH*(j-i);
i++;
}
if(cal>maxAmt)
maxAmt = cal;
}
return maxAmt;
}
}
Java Solution
Solution1
class Solution {
public int maxArea(int[] height) {
int i = 0;
int j = height.length-1;
int maxAmt = 0;
int iH;
int jH;
int cal;
while(j>i)
{
iH = height[i];
jH = height[j];
//Convergence back and forth
//which is higher,than will convergence from the other side
if(iH>jH)
{
cal = jH*(j-i);
j--;
}
else
{
cal = iH*(j-i);
i++;
}
if(cal>maxAmt)
maxAmt = cal;
}
return maxAmt;
}
}
Runtime : 4ms、3ms、2ms
Python3 Solution
Solution1
class Solution:
def maxArea(self, height: List[int]) -> int:
i = 0;
j = len(height)-1;
maxAmt = 0;
while(j>i):
iH = height[i];
jH = height[j];
if(iH>jH):
cal = jH*(j-i);
j-=1;
else:
cal = iH*(j-i);
i+=1;
if(cal>maxAmt):
maxAmt = cal;
return maxAmt;
JavaScript Solution
Solution1
/**
* @param {number[]} height
* @return {number}
*/
var maxArea = function(height) {
var i = 0;
var j = height.length-1;
var maxAmt = 0;
var iH;
var jH;
var cal;
while(j>i)
{
iH = height[i];
jH = height[j];
if(iH>jH)
{
cal = jH*(j-i);
j--;
}
else
{
cal = iH*(j-i);
i++;
}
if(cal>maxAmt)
{
maxAmt = cal;
}
}
return maxAmt;
};
Submission Detail
C#
Java
Python3
JavaScript
Conclusion
🧡If my solution helps, that is my honor!
"Human nature is like water. It takes the shape of its container."
Wallace Stevens
ai_front = {"insertion_before":"BEFORE","insertion_after":"AFTER","insertion_prepend":"PREPEND CONTENT","insertion_append":"APPEND CONTENT","insertion_replace_content":"REPLACE CONTENT","insertion_replace_element":"REPLACE ELEMENT","visible":"VISIBLE","hidden":"HIDDEN","fallback":"FALLBACK","automatically_placed":"Automatically placed by AdSense Auto ads code","cancel":"Cancel","use":"Use","add":"Add","parent":"Parent","cancel_element_selection":"Cancel element selection","select_parent_element":"Select parent element","css_selector":"CSS selector","use_current_selector":"Use current selector","element":"ELEMENT","path":"PATH","selector":"SELECTOR"};