 # LeetCode #11 Container With Most Water Solution & Explanation

Content

## LeetCode Problem

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;
};``````

## Conclusion

🧡If my solution helps, that is my honor!

```"Human nature is like water. It takes the shape of its container."
Wallace Stevens```

The problem link : Container With Most Water – LeetCode

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