Container With Most Water
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# LeetCode #11: Container With Most Water (C/C++).

medium

source: https://leetcode.com/problems/container-with-most-water
C/C++ Solution to LeetCode problem 11. Container With Most Water.

## Problem

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.

## Examples

### 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

## Solution

We will use two pointers (indexes), l (left) and r (right):

• l starts at index 0.
• r starts at the last index.
• While l < r:
• Calculate area (shortest_bar * distance_between_bars)
• If the area is bigger, then we update the area.
• Move the pointers:
• If l bar shorter, then move l to the right.
• Else move r to the left.
• Return the max area.

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 class Solution { public: int maxArea(vector<int>& height) { int l = 0; int r = height.size() - 1; int max = 0; while (l < r) { int area = (r - l) * min(height[r], height[l]); if (area > max) max = area; if (height[l] < height[r]) l += 1; else r -= 1; } return max; } };