11. Container With Most Water

11. Container With Most Water

https://leetcode.com/problems/container-with-most-water/

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.

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

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.

Approache 1 : DP - Left And Right Points

Prove：

Suppose height[ i ] <= Height[ j ]

MaxArea[i, j] = MaxArea( MaxArea[i+1, j] , ∀ Area[i, k] i<= k <= j)

MaxArea[i, j] = MaxArea( MaxArea[i+1, j] , Area[i, j] , > ∀ Area[i, k] i+1<= k <= j)

⇒ MaxArea[i, j] = MaxArea( MaxArea[i+1, j] , Area[i, j] )

class Solution {							
public:							
    int maxArea(vector<int>& height) {							
        int ans = 0, cur;							
        int l = 0;							
        int r = height.size() - 1;							
							
        while(l < r) {							
            cur = min(height[l], height[r]) * (r - l);							
            ans = max(ans, cur);										
			height[l] < height[r] ? l++	: r--;						
        }							
        return ans;							
    }							
};							

Runtime: 16 ms, faster than 96.32% of C++ online submissions for Container With Most Water.

Memory Usage: 9.8 MB, less than 72.17% of C++ online submissions for Container With Most Water.

Summary :

1. Consider all combinations of i and j, optimize unnecessary parts (blue).