Container With Most Water Leetcode Solution | Easy Approach

Minimum Cost to Merge Stones
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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

Container With Most Water Solutions

Time: O(1)
Space: O(1)

C++

class Solution {
 public:
  int maxArea(vector<int>& 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;
  }
};

Java

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;
  }
}

Python

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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