Medium
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 the line i
is at (i, ai)
and (i, 0)
. Find two lines, which, together with the x-axis forms a container, such that the container contains the most water.
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
Example 3:
Input: height = [4,3,2,1,4]
Output: 16
Example 4:
Input: height = [1,2,1]
Output: 2
Constraints:
n == height.length
2 <= n <= 105
0 <= height[i] <= 104
To solve the Container With Most Water problem in Swift using a Solution
class, we’ll follow these steps:
Solution
class with a method named maxArea
that takes an array of integers height
as input and returns the maximum area of water that can be contained.left
pointing to the start of the array and right
pointing to the end of the array.maxArea
to store the maximum area encountered so far, initially set to 0.left
is less than right
.(right - left) * min(height[left], height[right])
.maxArea
if the current area is greater than maxArea
.height[left] < height[right]
, increment left
, otherwise decrement right
.left
becomes greater than or equal to right
.maxArea
.Here’s the implementation:
class Solution {
func maxArea(_ height: [Int]) -> Int {
var i = 0
var j = height.count - 1
var mx = Int.min
// Base condition
guard height.count != 2 else {
return min(height[0], height[1])
}
// Hypothesis
// Induction
while (j > i) {
let dis = j - i
var area = height[i] * dis
if height[i] < height[j] {
i += 1
} else {
area = height[j] * dis
j -= 1
}
mx = max(mx, area)
}
return mx
}
}
This implementation provides a solution to the Container With Most Water problem in Swift.