Two Pointer Technique on Arrays in DSA Go - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is the Two Pointer Technique on Arrays?

It is a method where two pointers move through the array to solve problems efficiently, often from different ends or at different speeds.

beginner

In the Two Pointer Technique, when do pointers usually move towards each other?

When searching for pairs or conditions involving two elements, pointers start at opposite ends and move inward to find matches.

intermediate

How does the Two Pointer Technique help in removing duplicates from a sorted array?

One pointer tracks the position to insert unique elements, while the other scans the array. When a new unique element is found, it is placed at the insert position.

beginner

What is a common use case of the Two Pointer Technique in arrays?

Finding pairs with a given sum, merging sorted arrays, or partitioning arrays based on a condition.

intermediate

Explain why the Two Pointer Technique is efficient compared to nested loops.

It reduces time complexity by avoiding repeated scanning of elements, often achieving linear time instead of quadratic.

What is the initial position of pointers in the Two Pointer Technique when searching for pairs in a sorted array?

A Random positions

B Both at start

C One at start, one at end

D Both at end

Pointers start at the beginning and end to efficiently find pairs by moving inward.

Which problem can be solved using the Two Pointer Technique?

A Merging two sorted arrays

B Finding duplicates in an unsorted array without sorting

C Calculating factorial

D Binary search on a linked list

Merging two sorted arrays is a classic use of two pointers moving through each array.

What is the time complexity of the Two Pointer Technique in most array problems?

A O(n^2)

B O(n log n)

C O(log n)

D O(n)

Two pointers usually scan the array once, resulting in linear time O(n).

When removing duplicates from a sorted array using two pointers, what does the first pointer do?

A Tracks the position to insert unique elements

B Scans the array for duplicates

C Moves randomly

D Deletes elements

The first pointer marks where the next unique element should be placed.

In the Two Pointer Technique, what happens if the sum of elements at pointers is less than the target?

A Move the right pointer backward

B Move the left pointer forward

C Stop the process

D Swap the pointers

Moving the left pointer forward increases the sum to approach the target.

Describe the Two Pointer Technique and give two examples of problems it can solve.

Think about how two pointers move through the array to find answers efficiently.

Explain how the Two Pointer Technique improves efficiency compared to nested loops.

Consider how scanning the array once is better than scanning it multiple times.

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