Why Two Pointer Technique on Arrays in DSA Go?

The Big Idea

What if you could find pairs in a list without checking every single combination?

The Scenario

Imagine you have a long list of numbers and you want to find two numbers that add up to a specific target. Doing this by checking every possible pair one by one feels like searching for a needle in a haystack.

The Problem

Checking every pair means a lot of repeated work and takes a long time, especially if the list is very long. It's easy to make mistakes or miss pairs because you have to remember which pairs you already checked.

The Solution

The two pointer technique uses two markers that move through the list from different ends. This way, you can find pairs or solve problems by moving pointers smartly without checking every pair, saving time and effort.

Before vs After

✗ Before

for i := 0; i < len(arr); i++ {
    for j := i + 1; j < len(arr); j++ {
        if arr[i] + arr[j] == target {
            return i, j
        }
    }
}

✓ After

left, right := 0, len(arr)-1
for left < right {
    sum := arr[left] + arr[right]
    if sum == target {
        return left, right
    } else if sum < target {
        left++
    } else {
        right--
    }
}

What It Enables

This technique makes it easy and fast to solve problems involving pairs or ranges in sorted arrays without checking every possibility.

Real Life Example

Think of two people starting at opposite ends of a hallway trying to find a pair of shoes that together match a certain size. Instead of checking every pair, they move closer or farther based on the sizes they find, quickly finding the right match.

Key Takeaways

Manual pair checking is slow and error-prone.

Two pointers move from both ends to find solutions efficiently.

Works best on sorted arrays for problems involving pairs or sums.

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