0011. Container With Most Water

Medium | Two Pointer | 648 ms (94.80%), 27.1 MB (88.44%)

❓ Problem Statement

Source: LeetCode - Container With Most Water GitHub: Solution / Performance

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.

✍🏻 Constraints & Example

Constraints:

• n == height.length

• 2 <= n <= 10^5

• 0 <= height[i] <= 10^4

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.

Input: height = [1,1]
Output: 1

Input: height = [4,3,2,1,4]
Output: 16

Input: height = [1,2,1]
Output: 2

💡 Ideas

To begin with the potential maximum, we use two pointers from left- and right-most elements (due to the LARGEST gap in between).

Calculate the size of the container in each iteration, and move the pointer with SHORTER height until l >= r.

🤖 Python3

class Solution:
    def maxArea(self, height: List[int]) -> int:
        # (base case)
        if len( height ) == 2: return min( height )
        
        # ==================================================
        #  Array + Two Pointer                             =
        # ==================================================
        # time  : O(n)
        # space : O(1)
        
        area = 0
        l, r = 0, len(height) - 1
        
        # start from both-side (due to the LARGEST gap)
        # move the pointer with SHORTER height
        while r > l:
            tmp = min(height[l], height[r]) * (r - l)
            if tmp > area: area = tmp
            
            if height[r] > height[l]: l += 1
            else: r -= 1
                
        return area

🤖 Java

class Solution {
    /**
     * @time  : O(n)
     * @space : O(1)
     */
     
    public int maxArea(int[] height) {
        if( height.length == 2 ) return Math.min( height[0], height[1] );
        
        int l = 0, r = height.length - 1;
        int area = 0;
            
        while( l < r ){
            int tmp = 0;
            
            if( height[l] < height[r] ){
                if( height[l] == 0 ){
                    l += 1;
                    continue;
                }
                
                tmp = height[l] * ( r-l );
                l += 1;
                
            } else{
                if( height[r] == 0 ){
                    r -= 1;
                    continue;
                }
                
                tmp = height[r] * ( r-l );
                r -= 1;
            }
            
            if( tmp > area ) area = tmp;
        }
        
        return area;
    }
}