# LeetCode – Container With Most Water (Java)

Problem

Given n non-negative integers a1, a2, ..., an, where each represents a point at coordinate (i, ai). n vertical lines are drawn such that the two endpoints of line i is at (i, ai) and (i, 0). Find two lines, which together with x-axis forms a container, such that the container contains the most water.

Analysis

Initially we can assume the result is 0. Then we scan from both sides. If leftHeight < rightHeight, move right and find a value that is greater than leftHeight. Similarily, if leftHeight > rightHeight, move left and find a value that is greater than rightHeight. Additionally, keep tracking the max value. Java Solution

```public int maxArea(int[] height) { if (height == null || height.length < 2) { return 0; }   int max = 0; int left = 0; int right = height.length - 1;   while (left < right) { max = Math.max(max, (right - left) * Math.min(height[left], height[right])); if (height[left] < height[right]) left++; else right--; }   return max; }```
Category >> Algorithms >> Interview >> Java
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• Rox

lo
I m a little confused about this approch
We have to maximize the Area that can be formed between the vertical lines using the shorter line as length.
If we want to get the maximum Area, why we should use the shorter line as length?
And why the area is “seen” as the area of a trapez and not of a rectangle?
I ask this because there is the possibility to have

height of a3=8
height of a5=2
width=2

Area=10

height of a1=5
height of a7=1
width=6
Area=18

I do not understand why the container is not seen as a trapezoid and as a continer.
I can not understand how this solution solve the situation when, the width between two lines can be short but if the two lines are the higest,the Area will be grather than a continer with a larger width but with shorter lenght
Thanks!

• its not same as largest histogram problem

• Jason Zhu

It’s the same as largest histogram problem [http://www.programcreek.com/2014/05/leetcode-largest-rectangle-in-histogram-java/ ] with the same O(N) solution.

• appow

Oops, should have been <= width

• appow

Javascript solution. Returns an object with fields (i,height).
//heights: int[] array

function maxArea( heights ) {
var width = heights.length-1;
var objects = [];
for(var i = 0; i < width; i++){
var obj = {i:i,height:heights[i]};
objects.push(obj);
}
var sorted = objects.sort(function(a,b){
return (b.height*b.i – a.height*a.i)
});

return sorted
}

• ygw

proof: if Am<An,F(m,n)=(n-m)*Min(Am,An), then F(i,n) for all m<=i<=n. Done.