Array_Leetcode_Max_Water_Container_11.cpp

/*11. Container With Most Water
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*/

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

        int sum = 0;
        int n = height.size();
        //Brute force but TLE
        // for (int i = 0; i < n; i++) {
        //     for (int j = 0; j < n; j++) {

        //         int maxwater = (i - j) * min(height[i], height[j]);
        //         sum = max(sum, maxwater);
        //     }
        // }

        //Optimal O(N) & O(1)

        int l=0;
        int r= n-1;

        while(l<r)
        {
          int hight = r-l;
          int base = min(height[l],height[r]);
          int maxwater = hight * base;
          sum = max(sum,maxwater);
           
      if(height[l] > height[r]) //two pointer me kahi na kahi to shrink karega
           {                  // to usse width kam hogi to apne ko chote wale 
               r--;          //side ko age ya biche badhana hai taki area badhe
           }
           else l++;
        }

        return sum;
    }
};