Two Pointer Technique on Arrays in DSA Go - Build from Scratch

Two Pointer Technique on Arrays

📖 Scenario: You are working with a list of numbers representing daily temperatures. You want to find pairs of days where the temperature difference is exactly a certain value.

🎯 Goal: Build a program that uses the two pointer technique to find all pairs of numbers in a sorted array whose difference equals a given target.

📋 What You'll Learn

Create a sorted array of integers called temps with exact values

Create an integer variable targetDiff for the difference to find

Use the two pointer technique with variables left and right to find pairs with difference equal to targetDiff

Print all pairs found in the format (left_value, right_value)

💡 Why This Matters

🌍 Real World

Finding pairs of values with a specific difference is useful in temperature analysis, stock price changes, or matching time intervals.

💼 Career

Two pointer technique is a common pattern in coding interviews and helps solve array problems efficiently.

1

Create the sorted array of temperatures

Create a sorted array of integers called temps with these exact values: 1, 3, 5, 7, 9, 11

DSA Go

# Create the sorted array temps with values 1, 3, 5, 7, 9, 11
# Your code here

Hint

Use temps := []int{1, 3, 5, 7, 9, 11} to create the array.

2

Set the target difference

Create an integer variable called targetDiff and set it to 4

DSA Go

package main

func main() {
    temps := []int{1, 3, 5, 7, 9, 11}
    // Create targetDiff variable and set to 4
    // Your code here
}

Hint

Use targetDiff := 4 to set the difference to find.

3

Use two pointers to find pairs with the target difference

Use two integer variables left and right initialized to 0 and 1 respectively. Use a for loop to move the pointers through temps and find all pairs where temps[right] - temps[left] == targetDiff. Print each pair as fmt.Printf("(%d, %d)\n", temps[left], temps[right]). Increment left or right appropriately to find all pairs without duplicates.

DSA Go

package main

import "fmt"

func main() {
    temps := []int{1, 3, 5, 7, 9, 11}
    targetDiff := 4

    // Use two pointers left and right to find pairs with difference targetDiff
    // Your code here
}

Hint

Use a for loop with left and right pointers. Compare temps[right] - temps[left] with targetDiff and move pointers accordingly.

4

Print all pairs found

Run the program to print all pairs in temps whose difference equals targetDiff. The output should show pairs in the format (left_value, right_value) on separate lines.

DSA Go

package main

import "fmt"

func main() {
    temps := []int{1, 3, 5, 7, 9, 11}
    targetDiff := 4

    left, right := 0, 1
    for right < len(temps) {
        diff := temps[right] - temps[left]
        if diff == targetDiff {
            fmt.Printf("(%d, %d)\n", temps[left], temps[right])
            left++
            right++
        } else if diff < targetDiff {
            right++
        } else {
            left++
            if left == right {
                right++
            }
        }
    }

    // Output is printed inside the loop
    // Your code here
}

Hint

The pairs printed should be (1, 5), (3, 7), (5, 9), and (7, 11) each on a new line.

Practice

1. What is the main idea behind the Two Pointer Technique when used on arrays?

easy

A. Use two arrays to store elements separately.

B. Use one pointer to traverse and another to store values.

C. Use two indices to scan the array from different ends or positions simultaneously.

D. Use two nested loops to compare all pairs of elements.

Solution

1. Step 1: Understand the technique

   The two pointer technique uses two indices or pointers to traverse the array, often starting from different ends or positions.

2. Step 2: Compare with other options

   The other options describe different methods, but only using two indices to scan the array from different ends or positions simultaneously correctly describes the two pointer approach.

3. Final Answer:

   Use two indices to scan the array from different ends or positions simultaneously. -> Option C

4. Quick Check:

   Two pointers move independently to solve problems efficiently [OK]

Hint: Two pointers move from ends or positions to reduce time [OK]

Common Mistakes:

• Thinking it uses two separate arrays
• Confusing with nested loops approach
• Assuming only one pointer moves

2. Which of the following is the correct way to initialize two pointers left and right for an array arr in Go?

easy

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

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

C. left := -1; right := len(arr)

D. left := 0; right := len(arr)

Solution

1. Step 1: Understand array indexing in Go

   Arrays and slices in Go are zero-indexed, so the first element is at index 0 and the last at len(arr)-1.

2. Step 2: Check pointer initialization

   Setting left to 0 and right to len(arr)-1 correctly points to the first and last elements respectively.

3. Final Answer:

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

4. Quick Check:

   Start pointers at 0 and last index [OK]

Hint: Start left at 0 and right at last index (len-1) [OK]

Common Mistakes:

• Setting right to len(arr) causes out-of-range error
• Starting left at 1 misses first element
• Using negative indices which Go does not support

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

arr := []int{1, 3, 5, 7, 9}
left, right := 0, len(arr)-1
for left < right {
    if arr[left]+arr[right] == 10 {
        fmt.Println(left, right)
        break
    } else if arr[left]+arr[right] < 10 {
        left++
    } else {
        right--
    }
}

medium

A. 0 4

B. 1 3

C. 2 3

D. No output

Solution

1. Step 1: Trace the pointers and sums

   Initial pointers: left=0 (arr[0]=1), right=4 (arr[4]=9), sum=10 ==10, so prints 0 4 and breaks.

2. Step 2: Verify the condition and loop

   Actually, sum=1+9=10, so it prints 0 4 and breaks immediately.

3. Final Answer:

   0 4 -> Option A

4. Quick Check:

   Sum 1+9=10 at indices 0 and 4 [OK]

Hint: Check sum at pointers; print indices when sum matches target [OK]

Common Mistakes:

• Not breaking after first match
• Confusing indices with values
• Incorrect pointer updates

4. Find the error in the following Go code using two pointers to find if any two numbers sum to 8:

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

medium

A. The loop condition should be left <= right

B. left pointer should start at 1

C. sum calculation is incorrect

D. right is initialized to len(arr), which is out of range

Solution

1. Step 1: Check pointer initialization

   right is set to len(arr), but valid indices go from 0 to len(arr)-1. This causes out-of-range error when accessing arr[right].

2. Step 2: Verify other parts

   left starting at 0 is correct, loop condition left < right is valid, sum calculation is correct.

3. Final Answer:

   right is initialized to len(arr), which is out of range -> Option D

4. Quick Check:

   Array indices go 0 to len-1, not len [OK]

Hint: Initialize right pointer to len(arr)-1 to avoid out-of-range [OK]

Common Mistakes:

• Setting right to len(arr) causes runtime error
• Changing loop condition unnecessarily
• Misunderstanding array indexing

5. Given a sorted array arr := []int{1, 2, 3, 4, 5, 6, 7, 8, 9}, use the two pointer technique to find all pairs that sum to 10. Which of the following Go code snippets correctly prints all such pairs?

hard

A. left, right := 0, len(arr)-1
for left < right {
sum := arr[left] + arr[right]
if sum == 10 {
fmt.Println(arr[left], arr[right])
left++
right--
} else if sum < 10 {
left++
} else {
right--
}
}

B. left, right := 0, len(arr)-1
for left <= right {
sum := arr[left] + arr[right]
if sum == 10 {
fmt.Println(arr[left], arr[right])
left++
right--
} else if sum < 10 {
left++
} else {
right--
}
}

C. left, right := 0, len(arr)
for left < right {
sum := arr[left] + arr[right]
if sum == 10 {
fmt.Println(arr[left], arr[right])
left++
right--
} else if sum < 10 {
left++
} else {
right--
}
}

D. left, right := 1, len(arr)-1
for left < right {
sum := arr[left] + arr[right]
if sum == 10 {
fmt.Println(arr[left], arr[right])
left++
right--
} else if sum < 10 {
left++
} else {
right--
}
}

Solution

1. Step 1: Check pointer initialization and loop condition

   left, right := 0, len(arr)-1 for left < right { sum := arr[left] + arr[right] if sum == 10 { fmt.Println(arr[left], arr[right]) left++ right-- } else if sum < 10 { left++ } else { right-- } } initializes left=0 and right=len(arr)-1, which are valid indices. Loop runs while left < right to avoid duplicates and out-of-range.

2. Step 2: Verify pointer updates on sum match

   When sum == 10, both pointers move inward to find other pairs. This correctly finds all pairs without repeats.

3. Final Answer:

   left, right := 0, len(arr)-1 for left < right { sum := arr[left] + arr[right] if sum == 10 { fmt.Println(arr[left], arr[right]) left++ right-- } else if sum < 10 { left++ } else { right-- } } -> Option A

4. Quick Check:

   Two pointers move inward, printing pairs when sum matches [OK]

Hint: Move both pointers inward on match to find all pairs [OK]

Common Mistakes:

• Using right = len(arr) causes index error
• Loop condition left <= right prints duplicates
• Starting left at 1 misses pairs with first element