Container With Most Water

LeetCode 11 • Medium • Two Pointers

Find two lines that form the container holding the most water. Area = width × min(h[left], h[right]). Move the shorter pointer.

TimeO(n)single pass

SpaceO(1)two pointers

Left: — Right: — Max: —

Left

Right

Water

Left

—

Right

—

Max

Speed:

Ready

Press Play. Two pointers at both ends. Area = width × min(h[left], h[right]). Move the shorter pointer inward.

✎ Whiteboard

Size:3

⌨ Type It

Practice until you don't need to look. Use the guide comments below as scaffolding. The green highlights are the nuances to burn into memory.

# ─── CONTAINER MOST WATER (LeetCode 11) ───

# Pattern: Two pointers — move the shorter one (can't improve)

# Time: O(n) Space: O(1)

# 1. Define: maxArea(height)

# 2. Init: left, right = 0, len(height)-1; max_water = 0

# 3. Loop: while left < right:

# 4. Area: w = right - left; h = min(height[left], height[right])

# max_water = max(max_water, w * h)

# 5. Move shorter: if height[left] < height[right]: left += 1 else: right -= 1

# 6. Return: return max_water

# Vars: left, right, max_water, w, h

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