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

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

Problem

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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

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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

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• n == height.length
• 2 <= n <= 105
• 0 <= height[i] <= 104

Solution

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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.

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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;
  }
};