Container With Most Water – LeetCode Solution Java , Python 3, Python 2 , C , C++, Best and Optimal Solutions , All you need.

You are given an integer array height of length n. There are n vertical lines are drawn such that the two endpoints of the ith line is (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

C++ Container With Most Water LeetCode Solution

Start by evaluating the widest container, using the first and the last line. All other possible containers are less wide, so to hold more water, they need to be higher. Thus, after evaluating that widest container, skip lines at both ends that don’t support a higher height. Then evaluate that new container we arrived at. Repeat until there are no more possible containers left.
int maxArea(vector<int>& height) {
    int water = 0;
    int i = 0, j = height.size() - 1;
    while (i < j) {
        int h = min(height[i], height[j]);
        water = max(water, (j - i) * h);
        while (height[i] <= h && i < j) i++;
        while (height[j] <= h && i < j) j--;
    }
    return water;
}

Java Container With Most Water LeetCode Solution

class Solution {
    public int maxArea(int[] H) {
        int ans = 0, i = 0, j = H.length-1, res = 0;
        while (i < j) {
            if (H[i] <= H[j]) {
                res = H[i] * (j - i);
                i++;
            } else {
                res = H[j] * (j - i);
                j--;
            }
            if (res > ans) ans = res;
        }
        return ans;
    }
}
 

Python 3 Container With Most Water LeetCode Solution

class Solution:
    def maxArea(self, H: List[int]) -> int:
        ans, i, j = 0, 0, len(H)-1
        while (i < j):
            if H[i] <= H[j]:
                res = H[i] * (j - i)
                i += 1
            else:
                res = H[j] * (j - i)
                j -= 1
            if res > ans: ans = res
        return ans

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