description

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

submission

impl Solution {
    // area = W x H, if we start at the leftmost and rightmost lines,
    // and keep shrinking W = (R - L) until it becomes 0, the area can
    // only increase if H increase, since H = min(H[l], H[r]), we must
    // always choose the shorter side to shrink at each step.
    // When H[l] == H[r], choose either side is OK, because if there
    // is a larger rectangle between l and r, it must have two side
    // l' and r' satisifing l < l' && H[l'] > H[l] and
    // r' < r && H[r'] > H[r], meaning eventually we have to move
    // both sides, so we can't miss it with these steps.
    // By following these steps, we are searching for the maxium area
    // within the leftmost and rightmost lines with the best strategy
    // we can possibly formulate.
    pub fn max_area(height: Vec<i32>) -> i32 {
        let mut ans = 0;
        // with Vec we get a DoubleEndedIterator
        let mut iter = height.iter().enumerate();
        let mut l = iter.next();
        let mut r = iter.next_back();
        while let (Some((i, h1)), Some((j, h2))) = (l, r) {
            // record the maxium area
            ans = ans.max(h1.min(h2) * (j - i) as i32);
            // push inward on the shorter side
            if h1 < h2 {
                l = iter.next();
            } else {
                r = iter.next_back();
            }
        }
        ans
    }
}