11. Container With Most Water

• Difficulty: Medium

• Topics: Array, Two Pointers

• Similar Questions:

  • Trapping Rain 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 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.

Example:

Input: [1,8,6,2,5,4,8,3,7]
Output: 49

Solutions:

class Solution {
public:
    int maxArea(vector<int>& height) {
        int left = 0;
        int right = height.size() - 1;
        int ret = 0;
        while (left < right) {
            ret = max(ret, (right - left) * min(height[left], height[right]));
            if (height[left] <= height[right]) {
                ++left;
            } else {
                --right;
            }
        }

        return ret;
    }
};

Proof sketch

Let's dp[i][j] denote the optimal solution if the two axises are between i and j, inclusive. The induction then should be:

dp[i][j] = max((j - i) * min(height[i], height[j]), max(dp[i+1][j], dp[i][j-1]));

The only difference between dp[i+1][j] and dp[i][j-1] is the ability to use which axis as the boundary. Without losing generality, suppose height[i] < height[j]. If the optimal result comes from dp[i][j-1] and ith axis is used as the boundary. However, the result is definitely smaller than (j - i) * min(height[i], height[j]).