Container With Most Water Solutions in C++

// Source : https://oj.leetcode.com/problems/container-with-most-water/
// Author : Hao Chen
// Date   : 2014-06-22



class Solution {
public:
    int maxArea(vector<int> &height) {

        int maxArea = 0;
        // two pointers scan from two sides to middle
        int left = 0; 
        int right = height.size()-1;

        int area; 
        while ( left < right ){
            // calculate the area
            area = (right - left) * ( height[left] < height[right] ? height[left] : height[right]);
            // tracking the maxium area
            maxArea = area > maxArea ? area : maxArea;
            // because the area is decided by the shorter edge
            // so we increase the area is to increase the shorter edge
            //
            //     height[left] < height[right] ? left++ : right-- ;
            //
            // However, the above code could cause the unnecessary `area` cacluation
            // We can do some improvement as below:
            if (height[left] < height[right]) {
                do {
                    left++;
                } while (left < right && height[left-1] >= height[left]);
            } else {
                do {
                    right--;
                } while (right > left && height[right+1] >= height[right]);
            }
        }
        
        return maxArea;
    }
};

#include <algorithm>
#include <vector>

using std::vector; using std::prev; using std::max; using std::min;

class Solution {
public:
    int maxArea(vector<int> &height) {
        int ret{0};
        for (auto beg = height.begin(), end = prev(height.end()); beg < end; *beg < *end ? ++beg : --end)
            ret = max(ret, static_cast<int>(end - beg)*min(*beg, *end));
        return ret;
    }
};

/// Source : https://leetcode.com/problems/container-with-most-water/
/// Author : liuyubobobo
/// Time   : 2018-08-13

#include <iostream>
#include <vector>
#include <cassert>

using namespace std;

/// Brute Force
/// Time Complexity: O(n^2)
/// Space Complexity: O(1)
class Solution {
public:
    int maxArea(vector<int>& height) {

        assert(height.size() >= 2);

        int area = 0;
        for(int i = 0 ; i < height.size() ; i ++)
            for(int j = i + 1; j < height.size() ; j ++)
                area = max(area , min(height[i], height[j]) * (j - i));
        return area;
    }
};


int main() {

    vector<int> nums1 = {1, 1};
    cout << Solution().maxArea(nums1) << endl;

    return 0;
}

/// Source : https://leetcode.com/problems/container-with-most-water/
/// Author : liuyubobobo
/// Time   : 2016-12-26

#include <iostream>
#include <vector>
#include <cassert>

using namespace std;

/// Two Pointers
/// Time Complexity: O(n)
/// Space Complexity: O(1)
class Solution {
public:
    int maxArea(vector<int>& height) {

        assert(height.size() >= 2);

        int l = 0, r = height.size() - 1;
        int area = 0;
        while(l < r){
            area = max(area , min(height[l], height[r]) * (r - l));
            if(height[l] < height[r])
                l ++;
            else
                r --;
        }
        return area;
    }
};


int main() {

    vector<int> nums1 = {1, 1};
    cout << Solution().maxArea(nums1) << endl;

    return 0;
}