Two Pointer Technique on Arrays in DSA Go - Execution Trace

Concept Flow - Two Pointer Technique on Arrays

Initialize left pointer at start

↓

Initialize right pointer at end

↓

Check condition on elements at left and right

↓

If condition met

→Move left pointer right

↓

Else

→Move right pointer left

↓

Repeat until left >= right

↓

Done

Start with two pointers at opposite ends of the array, move them towards each other based on a condition until they meet or cross.

Execution Sample

DSA Go

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

Find two numbers in a sorted array that add up to 10 using two pointers moving inward.

Execution Table

Step	Operation	left	right	arr[left]	arr[right]	Sum	Action	Visual State
1	Initialize pointers	0	4	1	9	10	Check sum == 10	[1, 3, 5, 7, 9]
2	Sum == 10	0	4	1	9	10	Break loop	[1, 3, 5, 7, 9]
3	Loop ends	0	4	1	9	10	left >= right or break	[1, 3, 5, 7, 9]

💡 Sum of arr[left] + arr[right] equals 10, loop breaks.

Variable Tracker

Variable	Start	After Step 1	After Step 2	Final
left	0	0	0	0
right	4	4	4	4
arr[left]	1	1	1	1
arr[right]	9	9	9	9
Sum	N/A	10	10	10

Key Moments - 3 Insights

Why do we move the left pointer when the sum is less than the target?

Because the array is sorted, increasing the left pointer moves to a larger number, increasing the sum.

Why does the loop stop when left >= right?

When left meets or passes right, all pairs have been checked.

What happens if no pair sums to the target?

The pointers move inward until they cross without breaking early. The loop ends naturally when left >= right, as explained in the exit_note.

Visual Quiz - 3 Questions

Look at the execution_table, what is the sum of arr[left] + arr[right] at step 1?

A 9

B 8

C 10

D 11

💡 Hint

Check the 'Sum' column in execution_table row for step 1.

At which step does the loop break because the sum equals the target?

A Step 2

B Step 3

C Step 1

D Loop never breaks

💡 Hint

Look at the 'Action' column where it says 'Break loop'.

If the array was [1, 2, 3, 4, 5] and target was 10, what would happen to the pointers?

A They would find a pair at first step

B They would cross without finding a pair

C Only right pointer moves

D Only left pointer moves

💡 Hint

Refer to key_moments about what happens if no pair sums to target.

Concept Snapshot

Two Pointer Technique on Arrays:
- Use two pointers at start and end
- Move pointers inward based on condition
- Stop when pointers meet or condition met
- Efficient for sorted arrays
- Commonly used for pair sum problems

Full Transcript

The two pointer technique uses two indexes starting at opposite ends of an array. We check the sum of elements at these pointers. If the sum matches the target, we stop. If the sum is less, we move the left pointer right to increase sum. If the sum is more, we move the right pointer left to decrease sum. This continues until pointers meet or cross. This method is efficient for sorted arrays and helps find pairs quickly.

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