⚡Container With Most Water

#11. Container With Most Water

• Difficulty: Medium
• Company: LeetCode75
• Review: January 14, 2025
• Select: Array, Stack, Two Pointers
• Solved: Yes
• Tried: 2

My Solution: Brute Force - Unsolved

class Solution(object):
    def maxArea(self, height):
        """
        :type height: List[int]
        :rtype: int
        """
        curr = 0
        for i in range(len(height)):
            for j in range(i+1,len(height)):
                curr = max(curr,((j+1)-(i+1))*min(height[i],height[j]))
        return curr

• Description:
  • Tried to calculate all possible cases using a nested loop.
  • Issue: This approach resulted in a time limit exceeded error for large inputs.
• Time: $O(n^2)$
• Space: O(1)

Reflection

• Exploring various solutions and clearly communicating them
• Stating the runtime and space complexity (the pros and cons) of each solution)
• Translating ideas into code
• Debugging your code

Best Answer

Using Two Pointers

class Solution(object):
    def maxArea(self, height):
        """
        :type height: List[int]
        :rtype: int
        """
        max_area = 0 #Saving for answer
        s,e=0,len(height)-1 #Assign the first index and last index
        while s<e:
            area = (e-s) * min(height[s],height[e]) #Current area using e,s
            max_area = max(area, max_area) #Re-assing the max_area comparing with area
            if height[s]< height[e]:
                s+=1
            else:
                e -=1
        return max_area

• Complexity
  • $N$ : The number of elements in the height list.
• Time: $O(N)$
  • A single while loop ensures each pointer moves exactly once.
• Space: O(1)
  • No additional memory is used beyond a few variables.