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Hello Programmers/Coders, Today we are going to share solutions to the Programming problems of LeetCode Solutions in C++, Java, & Python. At Each Problem with Successful submission with all Test Cases Passed, you will get a score or marks and LeetCode Coins. And after solving maximum problems, you will be getting stars. This will highlight your profile to the recruiters.

In this post, you will find the solution for the Container With Most Water in C++, Java & Python-LeetCode problem. We are providing the correct and tested solutions to coding problems present on LeetCode. If you are not able to solve any problem, then you can take help from our Blog/website.

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Link for the ProblemContainer With Most Water– LeetCode Problem

Container With Most Water– LeetCode Problem

Problem:

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:

question 11
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
Container With Most Water– LeetCode Solutions
class Solution {
 public:
  int maxArea(vector& height) {
    int ans = 0;
    int l = 0;
    int r = height.size() - 1;

    while (l < r) {
      const int minHeight = min(height[l], height[r]);
      ans = max(ans, minHeight * (r - l));
      if (height[l] < height[r])
        ++l;
      else
        --r;
    }

    return ans;
  }
};
class Solution {
  public int maxArea(int[] height) {
    int ans = 0;
    int l = 0;
    int r = height.length - 1;

    while (l < r) {
      final int minHeight = Math.min(height[l], height[r]);
      ans = Math.max(ans, minHeight * (r - l));
      if (height[l] < height[r])
        ++l;
      else
        --r;
    }

    return ans;
  }
}
class Solution:
  def maxArea(self, height: List[int]) -> int:
    ans = 0
    l = 0
    r = len(height) - 1

    while l < r:
      minHeight = min(height[l], height[r])
      ans = max(ans, minHeight * (r - l))
      if height[l] < height[r]:
        l += 1
      else:
        r -= 1

    return ans

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