# Leetcode: Container With Most Water

Get the most water from containers Similar Problems:

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.

Note: You may not slant the container and n is at least 2.

Github: code.dennyzhang.com

Credits To: leetcode.com

Leave me comments, if you have better ways to solve.

Key Observations:

```By moving shorter end pointer further doesn't eliminate the final answer
```
• Solution: Two Pointer
```// Blog link: https://code.dennyzhang.com/container-with-most-water
// Basic Ideas: Two pointer
//   Then move the pointer of shorter value
//     Let's check the case that both pointers are the same the value.
//     We won't find a better solution with only one edge of left pointer or right pointer.
//     Thus we can move either pointer
// Complexity: Time O(n), Space O(1)
func maxArea(height []int) int {
l, r := 0, len(height)-1
res, v := 0, 0
for l<r {
if height[l] <= height[r] {
v = height[l]*(r-l)
l++
} else {
v = height[r]*(r-l)
r--
}
if v>res { res = v }
}
return res
}
```

• Solution: Two Pointer + Pruning
```// Blog link: https://code.dennyzhang.com/container-with-most-water
// Basic Ideas: Two pointer
//   Then move the pointer of shorter value
//     Let's check the case that both pointers are the same the value.
//     We won't find a better solution with only one edge of left pointer or right pointer.
//     Thus we can move either pointer
// Complexity: Time O(n), Space O(1)
func maxArea(height []int) int {
l, r := 0, len(height)-1
res, v := 0, 0
for l<r {
lmax, rmax := height[l], height[r]
if height[l] <= height[r] {
v = height[l]*(r-l)
} else {
v = height[r]*(r-l)
}
if v>res { res = v }
if height[l] <= height[r] {
for l<r && height[l]<=lmax { l++ }
} else {
for l<r && height[r]<=rmax { r-- }
}
}
return res
}
```

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