Why Two Pointer Technique Beats Brute Force in DSA Python - See It Work

Why Two Pointer Technique Beats Brute Force

📖 Scenario: Imagine you have a list of numbers representing the heights of walls. You want to find two walls that together can hold the most water between them. This is like finding the biggest container formed by two walls.

🎯 Goal: You will first try the simple but slow way (brute force) to find the maximum water container. Then, you will use the faster two pointer technique to do the same task efficiently.

📋 What You'll Learn

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

Create a variable called max_area_brute and set it to 0

Use nested for loops with variables i and j to check all pairs of walls in heights

Calculate area as (j - i) * min(heights[i], heights[j]) inside the loops and update max_area_brute if this area is bigger

Create two pointers left and right starting at the ends of heights

Create a variable called max_area_two_pointer and set it to 0

Use a while loop to move left and right pointers towards each other, calculating area and updating max_area_two_pointer

Print both max_area_brute and max_area_two_pointer to compare results

💡 Why This Matters

🌍 Real World

This technique helps in problems where you need to find pairs or ranges efficiently, like in water container problems, string matching, or array processing.

💼 Career

Understanding two pointer technique is important for coding interviews and real-world tasks that require optimized solutions over brute force.

1

Create the list of wall heights

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

DSA Python

# Create the list heights with the given values
# Your code here

Hint

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

2

Set up variables for brute force and two pointer methods

Create a variable called max_area_brute and set it to 0. Also create variables left and right with values 0 and len(heights) - 1 respectively. Finally, create a variable called max_area_two_pointer and set it to 0.

DSA Python

heights = [1, 8, 6, 2, 5, 4, 8, 3, 7]
# Create max_area_brute = 0
# Create left = 0 and right = len(heights) - 1
# Create max_area_two_pointer = 0
# Your code here

Hint

Use simple assignment statements to create these variables.

3

Calculate max area using brute force and two pointer technique

Use nested for loops with variables i and j to check all pairs of walls in heights. Calculate area as (j - i) * min(heights[i], heights[j]) and update max_area_brute if this area is bigger. Then use a while loop with left and right pointers moving towards each other. Calculate area inside the loop and update max_area_two_pointer. Move left pointer forward if heights[left] is less than heights[right], else move right pointer backward.

DSA Python

heights = [1, 8, 6, 2, 5, 4, 8, 3, 7]
max_area_brute = 0
left = 0
right = len(heights) - 1
max_area_two_pointer = 0
# Add nested for loops for brute force
# Add while loop for two pointer technique
# Your code here

Hint

Use two nested loops for brute force and a while loop with two pointers for the efficient method.

4

Print the results to compare both methods

Print max_area_brute and max_area_two_pointer on separate lines to compare the maximum water container areas found by both methods.

DSA Python

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

for i in range(len(heights)):
    for j in range(i + 1, len(heights)):
        area = (j - i) * min(heights[i], heights[j])
        if area > max_area_brute:
            max_area_brute = area

while left  max_area_two_pointer:
        max_area_two_pointer = area
    if heights[left] < heights[right]:
        left += 1
    else:
        right -= 1
# Print max_area_brute and max_area_two_pointer
# Your code here

Hint

Use two print statements, one for each max area variable.