Container with most water ( #11 Leetcode)

You are given an integer array height of length n. There are n vertical lines drawn such that the two endpoints of the ith line are (i, 0) and (i, height[i]).

Find two lines that together with the x-axis form a container, such that the container contains the most water.

Return the maximum amount of water a container can store.

Notice that you may not slant the container.

 

Example 1:

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.

Example 2:

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

 

Constraints:

• n == height.length
• 2 <= n <= 105
• 0 <= height[i] <= 104

This can be solved by two pointers.

1. Start with pointer left=0 and pointer right=length-1
2. The max water is limited by the pointer with smaller height
3. When moving a pointer, the width of the area decrease
4. If we move the pointer with higher height, we will never get a
   greater area, the max height will be at most the one of the pointer with smaller height.
5. So we need to move the pointer with smaller height to have a chance to get higher height at the next step.

Things to consider:

1. As we are concerned with maximum area, whenever we see the height is low, we will either increment from start or decrement from the end.

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

Simplified Solution 

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