Container With Most Water in DSA Python - Build from Scratch

Container With Most Water

📖 Scenario: Imagine you have a row of vertical lines drawn on a flat surface. Each line has a height. You want to find two lines that, together with the surface, form a container that holds the most water.This is like choosing two walls to hold water between them. The water amount depends on the shorter wall and the distance between the walls.

🎯 Goal: You will write a program to find the maximum amount of water that can be held between two lines from a list of heights.

📋 What You'll Learn

Create a list called heights with the exact values: [1, 8, 6, 2, 5, 4, 8, 3, 7]

Create a variable called max_water and set it to 0

Use a while loop with variables left and right to find the maximum water container

Print the value of max_water

💡 Why This Matters

🌍 Real World

This problem models situations where you want to maximize capacity between boundaries, like designing containers, dams, or storage tanks.

💼 Career

Understanding two-pointer techniques and optimization is useful for coding interviews and real-world problems involving arrays and searching.

1

Create the list of heights

Create a list called heights with these exact values: [1, 8, 6, 2, 5, 4, 8, 3, 7]

DSA Python

# Create the list of heights
# Your code here

Hint

Use square brackets [] to create the list and separate numbers with commas.

2

Initialize the maximum water variable

Create a variable called max_water and set it to 0

DSA Python

heights = [1, 8, 6, 2, 5, 4, 8, 3, 7]
# Create max_water variable and set it to 0
# Your code here

Hint

Use = to assign the value 0 to max_water.

3

Find the maximum water container

Use a while loop with variables left starting at 0 and right starting at len(heights) - 1. Inside the loop, calculate the water between heights[left] and heights[right]. Update max_water if the current water is greater. Move left forward if heights[left] is less than heights[right], else move right backward.

DSA Python

heights = [1, 8, 6, 2, 5, 4, 8, 3, 7]
max_water = 0
# Use while loop with left and right pointers to find max water
# Your code here

Hint

Use two pointers starting at the ends of the list and move them towards each other.

4

Print the maximum water

Print the value of max_water

DSA Python

heights = [1, 8, 6, 2, 5, 4, 8, 3, 7]
max_water = 0
left = 0
right = len(heights) - 1

while left  max_water:
        max_water = current_water
    if heights[left] < heights[right]:
        left += 1
    else:
        right -= 1

# Print the max_water value
# Your code here

Hint

Use print(max_water) to show the result.