LeetCode #11 Container With Most Water

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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.

Example

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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.

Note

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

Solution1

Method: Brute Force

Time Complexity: O(n ^2)

Space Complexity: O(1)

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class Solution:
    def maxArea(self, height):
        """
        :type height: List[int]
        :rtype: int
        """
        res = float('-inf')
        length = len(height)
        for l_idx in range(length - 1):
            for r_idx in range(l_idx, length):
                h = min(height[l_idx], height[r_idx])
                w = r_idx - l_idx
                res = max(res, h*w)
        
        return res

Solution2

Method: Greedy

Time Complexity: O(n)

Space Complexity: O(1)

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class Solution:
    def maxArea(self, height):
        """
        :type height: List[int]
        :rtype: int
        """
        #            / max{v(i, j), S(i...j-1)};  height(i) >= height(j)
        # S(i..j) = |
        #            \ max{v(i, j), S(i+1...j)};  height(i) < height(j)
        left = 0
        right = len(height) - 1
        res = float('-inf')

        while left < right:
            min_height = min(height[left], height[right])
            res = max(res, min_height * (right - left))
            
            while left <= right and height[left] <= min_height:
                left += 1

            while left <= right and height[right] <= min_height:
                right -= 1
        
        return res

Solution3

Method: Greedy

Time Complexity: O(n)

Space Complexity: O(1)

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class Solution:
    def maxArea(self, height):
        i, j = 0, len(height) - 1
        water = 0
        while i < j:
            water = max(water, (j - i) * min(height[i], height[j]))
            if height[i] < height[j]:
                i += 1
            else:
                j -= 1
        return water