Container With Most Water - Array - Medium - LeetCode
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Container With Most Water - Array - Medium - LeetCode

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  • 1The problem involves finding two vertical lines that form a container with the maximum water capacity based on given heights.
  • 2The maximum area of water is calculated using a two-pointer approach, optimizing for time complexity at O(n).
  • 3Example inputs demonstrate varying heights, with the maximum area achieved being 49 for the array [1,8,6,2,5,4,8,3,7].

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"The problem involves finding two vertical lines that form a container with the maximum water capacity based on given heights."

Container With Most Water - Array - Medium - LeetCode

Given n non-negative integers a1, a2, ..., an , where each represents a point at coordinate (i, ai). n vertical lines are drawn such that the two endpoints of line i is at (i, ai) and (i, 0). Find two lines, which, together with the x-axis forms a container, such that the container contains the most water.

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 Example 3:

Input: height = [4,3,2,1,4] Output: 16 Example 4:

Input: height = [1,2,1] Output: 2

Constraints:

2 <= height.length <= 3 * 104 0 <= height[i] <= 3 * 104

public class Solution {
 public int MaxArea(int[] height) {
 
 int max = 0;
 int i=0;
 int j = height.Length-1;
 while(i<j){
 max = Math.Max(max, Math.Min(height[i],height[j])*(j-i));
 if(height[i]<height[j]){
 i++;
 }
 else{
 j--;
 }
 }
 
 return max;
 }
}

Time Complexity: O(n)

Space Complexity: O(1)

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sumitc91

Published on 18 October 2020 · 1 min read · 179 words

Part of AskGif Blog · coding

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