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Container With Most Water LC #11 Medium Two Pointers
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.

Example 1
Input: height = [1,8,6,2,5,4,8,3,7]
Output: 49
Explanation: Lines at index 1 and 8 form the best container: min(8,7)×7 = 49.
Example 2
Input: height = [1,1]
Output: 1
Constraints: n == height.length; 2 ≤ n ≤ 105; 0 ≤ height[i] ≤ 104
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Approach
Greedy Two Pointer
L at left, R at right. Compute area = min(h[L],h[R])×(R-L). Move the shorter wall inward — the taller wall can never help with a smaller base. O(n) time, O(1) space.
Height Bars
area = ?
Variables
L
—
R
—
Area
—
Max Area
0
Step Logic
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Algorithm
1
L=0, R=n-1, maxArea=0
2
area = min(h[L], h[R]) × (R-L)
3
maxArea = max(maxArea, area)
4
Move shorter wall: h[L]<h[R] → L++ else R--
Time
O(n)
Space
O(1)
Why It Works

Width decreases as pointers move inward. To possibly increase area, need greater height. The shorter wall limits height — moving it may find a taller wall. Moving the taller wall can only decrease or maintain height with a smaller width — it never helps. So we always move the shorter wall inward, guaranteeing we never miss the optimal pair.