[Leetcode 43] 11 Container With Most Water

Problem：

Given n non-negative integers a₁, a₂, ..., a_n, where each represents a point at coordinate (i, a_i). n vertical lines are drawn such that the two endpoints of line i is at (i, a_i) 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.

Analysis:

Naive O(n^2) solution is too expensive.

After searching on the web, find a O(n) solution. Similar to 2-sum, use two pointers points to the start and end of the given array, compute the area they can form and compare this result with the current max area. Update the max area if it's less than the new result. Then always choose the pointer which points to the shorter one, move it toward the center of the array and repeat step 1. 

This solution is true because only two edges that both larger or equal to the current two edges can generate area that migth greater than the current area, which has a larger bottom. 

Code:

Naive solution

class Solution {
public:
    int maxArea(vector<int> &height) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        int marea = 0;
        
        for (int i=0; i<height.size(); i++) {
            for (int j=i+1; j<height.size(); j++) {
                int h = height[i] < height[j] ? height[i] : height[j];
                int area = (j-i) * h;
                if (area > marea) 
                    marea = area;
            }
        }
        
        return marea;
    }
};

O(N) solution

class Solution {
public:
    int maxArea(vector<int> &height) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        int maxArea = 0, i = 0, j = height.size()-1;
        
        while (i < j) {
            int area = min(height[i], height[j]) * (j-i);
            if (area > maxArea)
                maxArea = area;
                
            if (height[i] < height[j])
                i++;
            else
                j--;
        }
        
        return maxArea;
    }
    
    int min(int a, int b) {
        return a<b?a:b;
    }
};

Attention:

O(n^2) solution can not be adopted if the data is too large.