Container With Most Water (Two Pointer Approach)

Problem Statement

Given heights height[i] of vertical lines, pick two lines that, with the x-axis, form a container holding the most water. Return that maximum area.

Example

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

Two Pointer Approach

Start with the widest container (both ends). The area is min(height[left], height[right]) * (right - left). Moving the taller line can only shrink the area, so always move the shorter line inward, hoping for a taller one.

Complexity

• Time: O(n)
• Space: O(1)

Solution Code

Python

def maxArea(height):
    left, right = 0, len(height) - 1
    best = 0
    while left < right:
        area = min(height[left], height[right]) * (right - left)
        best = max(best, area)
        if height[left] < height[right]:
            left += 1
        else:
            right -= 1
    return best

Java

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

C++

int maxArea(vector<int>& height) {
    int left = 0, right = height.size() - 1, best = 0;
    while (left < right) {
        int area = min(height[left], height[right]) * (right - left);
        best = max(best, area);
        if (height[left] < height[right]) left++;
        else right--;
    }
    return best;
}

Key Takeaways

• Width shrinks every step, so only a taller line can improve the area.
• Always discard the shorter line — it is the limiting factor.
• Greedy two pointers turn O(n^2) brute force into O(n).