11. Container With Most Water.md

11. Container With Most Water

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 x-axis forms a container, such that the container contains the most water.

Note: You may not slant the container and n is at least 2.

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:

Input: [1,8,6,2,5,4,8,3,7]
Output: 49

Solution

• java

  • mine Runtime: 2 ms, faster than 95.41%, Memory Usage: 40 MB, less than 94.23% of Java online submissions

  //O(n)time O(1)space
  public int maxArea(int[] height) {
      int s = 1,e = height.length;
      int res = 0;
      int maxS = height[s - 1],maxE = height[e - 1];
      int maxSIndex = s,maxEIndex = e;
      while(s < e){
          res = Math.max(res, (maxEIndex - maxSIndex) * Math.min(maxS,maxE));
          if(maxS <= maxE){
              s++;
              if(height[s - 1] > maxS){
                  maxSIndex = s;
                  maxS = height[s - 1];
              }
          }else{
              e--;
              if(height[e - 1] > maxE){
                  maxEIndex = e;
                  maxE = height[e - 1];
              }
          }
      }
      return res;
  }
  

  • the most votes Runtime: 2 ms, faster than 95.41%, Memory Usage: 39.2 MB, less than 98.72% of Java online submissions

  public int maxArea(int[] height) {
      int left = 0, right = height.length - 1;
      int maxArea = 0;
  
      while (left < right) {
          maxArea = Math.max(maxArea, Math.min(height[left], height[right])
                  * (right - left));
          if (height[left] < height[right])
              left++;
          else
              right--;
      }
  
      return maxArea;
  }