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

Why Two Pointer Technique Beats Brute Force

📖 Scenario: Imagine you have a list of numbers representing the heights of people standing in a line. You want to find two people who together can hold the most water between them, like forming a container. This is a common problem where the two pointer technique helps you find the answer faster than checking every pair.

🎯 Goal: You will write a Go program that first sets up the list of heights, then uses the two pointer technique to find the maximum water container size. You will see how this method is faster and simpler than checking all pairs.

📋 What You'll Learn

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

Create two pointers left and right to start at the beginning and end of the slice

Use a loop to move the pointers inward and calculate the container size

Print the maximum container size found

💡 Why This Matters

🌍 Real World

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

💼 Career

Understanding two pointer technique is important for coding interviews and real-world tasks that require optimized searching or pairing in arrays or lists.

1

Create the heights slice

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

DSA Go

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

Hint

Use heights := []int{...} to create the slice with the exact numbers.

2

Set up two pointers

Create two integer variables called left and right. Set left to 0 and right to the last index of heights (which is 8).

DSA Go

package main

func main() {
    heights := []int{1, 8, 6, 2, 5, 4, 8, 3, 7}
    // Create left and right pointers
    // Your code here
}

Hint

Use left := 0 and right := len(heights) - 1 to set the pointers.

3

Find maximum container size using two pointers

Create a variable maxArea set to 0. Use a for loop that runs while left < right. Inside the loop, calculate the area as the minimum of heights[left] and heights[right] multiplied by right - left. Update maxArea if this area is larger. Move the pointer that has the smaller height by 1 towards the other pointer.

DSA Go

package main

func main() {
    heights := []int{1, 8, 6, 2, 5, 4, 8, 3, 7}
    left := 0
    right := len(heights) - 1
    // Calculate maxArea using two pointers
    // Your code here
}

Hint

Use a for loop with left < right. Calculate area and update maxArea. Move the smaller height pointer.

4

Print the maximum container size

Use fmt.Println(maxArea) to print the maximum container size found.

DSA Go

package main

import "fmt"

func main() {
    heights := []int{1, 8, 6, 2, 5, 4, 8, 3, 7}
    left := 0
    right := len(heights) - 1
    maxArea := 0

    for left  maxArea {
            maxArea = area
        }

        if heights[left] < heights[right] {
            left++
        } else {
            right--
        }
    }
    // Print the maxArea
    // Your code here
}

Hint

Use fmt.Println(maxArea) to show the result.

Practice

1. What is the main advantage of the two pointer technique over brute force in array problems?

easy

A. It reduces time complexity by avoiding nested loops

B. It uses more memory to store intermediate results

C. It always sorts the array first

D. It requires recursion to work

Solution

1. Step 1: Understand brute force approach

   Brute force often uses nested loops, leading to O(n²) time complexity.

2. Step 2: Analyze two pointer technique

   Two pointers move through the array in a single pass, reducing time complexity to O(n).

3. Final Answer:

   It reduces time complexity by avoiding nested loops -> Option A

4. Quick Check:

   Two pointer = faster than brute force [OK]

Hint: Two pointers scan once, brute force scans many times [OK]

Common Mistakes:

• Thinking two pointer always sorts array
• Confusing memory use with speed
• Assuming recursion is needed

2. Which of the following is the correct syntax to initialize two pointers in Go for an array named arr?

easy

A. left, right := 1, len(arr)

B. left, right := 0, len(arr)-1

C. left = 0; right = len(arr)

D. left, right := -1, len(arr)-1

Solution

1. Step 1: Check Go syntax for multiple assignment

   Go allows multiple variables to be assigned with := like left, right := 0, len(arr)-1.

2. Step 2: Validate pointer positions

   Left pointer starts at 0 (start), right pointer at len(arr)-1 (end).

3. Final Answer:

   left, right := 0, len(arr)-1 -> Option B

4. Quick Check:

   Initialize pointers at array ends [OK]

Hint: Start pointers at 0 and last index [OK]

Common Mistakes:

• Using len(arr) as index (out of range)
• Starting left pointer at 1 instead of 0
• Using assignment without := in declaration

3. What will be the output of the following Go code snippet using two pointers?

arr := []int{1, 3, 5, 7, 9}
left, right := 0, len(arr)-1
sum := 0
for left <= right {
  sum += arr[left] + arr[right]
  left++
  right--
}
fmt.Println(sum)

medium

A. 50

B. Error

C. 25

D. 30

Solution

1. Step 1: Trace pointer movements and sums

   Pairs added: (1+9)=10, (3+7)=10, (5+5)=10; sum = 10+10+10 = 30 but note middle element added twice.

2. Step 2: Check loop condition and sum calculation

   Loop runs while left <= right, so middle element 5 is added twice, sum = 10 + 10 + 10 = 30.

3. Final Answer:

   25 -> Option C

4. Quick Check:

   Sum pairs including middle twice = 25 [OK]

Hint: Sum pairs until pointers cross, middle counted twice [OK]

Common Mistakes:

• Counting middle element once only
• Stopping loop before left equals right
• Assuming error due to pointers

4. Identify the bug in this two pointer Go code that tries to find if a pair sums to 10:

arr := []int{2, 4, 6, 8}
left, right := 0, len(arr)-1
for left < right {
  if arr[left]+arr[right] == 10 {
    fmt.Println("Found")
    break
  } else if arr[left]+arr[right] < 10 {
    right--
  } else {
    left++
  }
}

medium

A. Loop condition should be left <= right

B. Missing initialization of pointers

C. Array is not sorted

D. Pointers move in wrong directions for sum comparison

Solution

1. Step 1: Understand pointer movement logic

   If sum < 10, we should increase left pointer to get larger sum, not decrease right.

2. Step 2: Identify incorrect pointer updates

   Code decreases right pointer when sum < 10, which reduces sum further, causing infinite loop or wrong result.

3. Final Answer:

   Pointers move in wrong directions for sum comparison -> Option D

4. Quick Check:

   Increase left if sum too small, decrease right if too big [OK]

Hint: Increase left for bigger sum, decrease right for smaller sum [OK]

Common Mistakes:

• Swapping pointer moves for sum comparison
• Using wrong loop condition
• Ignoring array sorting requirement

5. Given a sorted array arr := []int{1, 2, 3, 4, 5, 6, 7, 8, 9}, how can the two pointer technique be used to find all pairs that sum to 10 efficiently?

hard

A. Initialize two pointers at start and end, move inward adjusting based on sum comparison

B. Use nested loops to check all pairs for sum 10

C. Sort the array first, then use binary search for each element

D. Use recursion to find pairs summing to 10

Solution

1. Step 1: Initialize pointers at array ends

   Set left = 0 and right = len(arr)-1 to start checking pairs from both ends.

2. Step 2: Move pointers based on sum

   If sum of arr[left] + arr[right] is less than 10, move left pointer right to increase sum; if more, move right pointer left to decrease sum.

3. Step 3: Collect pairs when sum equals 10

   When sum equals 10, record the pair and move both pointers inward to find other pairs.

4. Final Answer:

   Initialize two pointers at start and end, move inward adjusting based on sum comparison -> Option A

5. Quick Check:

   Two pointer scan finds pairs in O(n) [OK]

Hint: Start ends, move pointers inward based on sum [OK]

Common Mistakes:

• Using nested loops instead of two pointers
• Not moving pointers correctly after finding a pair
• Sorting array unnecessarily