Two Pointer Technique on Arrays in DSA C++ - Execution Trace

Concept Flow - Two Pointer Technique on Arrays

Initialize two pointers: left=0, right=end

↓

Check condition: left < right?

No→Stop

Yes↓

Compare or process elements at left and right

↓

Move pointers: left++ or right-- or both

↩Back to condition check

Start with two pointers at array ends, compare or process elements, move pointers inward until they meet or cross.

Execution Sample

DSA C++

int left = 0, right = n - 1;
while (left < right) {
  if (arr[left] + arr[right] == target) {
    // found pair
    break;
  } else if (arr[left] + arr[right] < target) {
    left++;
  } else {
    right--;
  }
}

Find two numbers in a sorted array that sum to a target by moving two pointers inward.

Execution Table

Step	Operation	left	right	arr[left]	arr[right]	Sum	Action	Visual State
1	Initialize pointers	0	5	1	12	13	Check sum vs target=9	[1, 3, 4, 6, 8, 12]
2	Sum=13 > 9	0	5	1	12	13	right--	[1, 3, 4, 6, 8, 12]
3	Move right to 4	0	4	1	8	9	Sum == target, found pair	[1, 3, 4, 6, 8, 12]
4	Stop	0	4	1	8	9	Break loop	[1, 3, 4, 6, 8, 12]

💡 Pointers met or crossed or pair found with sum equal to target.

Variable Tracker

Variable	Start	After Step 1	After Step 2	After Step 3	Final
left	0	0	0	0	0
right	5	5	4	4	4
sum	N/A	13	13	9	9

Key Moments - 3 Insights

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

Because the array is sorted, decreasing right pointer moves to a smaller number, reducing the sum (see execution_table row 2).

Why stop when left meets right?

Because pointers crossing means all pairs checked; no new pairs exist beyond this (see execution_table row 4).

Why do we not move both pointers at the same time?

Moving both could skip valid pairs; we only move one pointer based on sum comparison to find the correct pair (see execution_table rows 2 and 3).

Visual Quiz - 3 Questions

Look at the execution_table, what is the value of 'right' after step 2?

A 3

B 4

C 5

D 0

💡 Hint

Check the 'right' column in execution_table row 3.

At which step does the sum equal the target?

A Step 1

B Step 2

C Step 3

D Step 4

💡 Hint

Look at the 'Sum' and 'Action' columns in execution_table.

If the array was not sorted, how would the two pointer technique change?

A We would need to sort the array first.

B It would still work the same way.

C We would move both pointers simultaneously.

D We would only use one pointer.

💡 Hint

Two pointer technique relies on sorted arrays to decide pointer movement (see key_moments).

Concept Snapshot

Two Pointer Technique on Arrays:
- Use two pointers at array ends.
- Move pointers inward based on condition.
- Works best on sorted arrays.
- Efficient for pair sum and similar problems.
- Stops when pointers meet or condition met.

Full Transcript

The two pointer technique uses two indexes starting at opposite ends of a sorted array. We check the sum of elements at these pointers. If the sum is equal to the target, we found the pair. 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 pair is found. This method is efficient because it avoids checking all pairs by leveraging the sorted order.

Practice

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

easy

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

B. Use two nested loops to compare every pair of elements

C. Sort the array first and then use binary search

D. Use a single pointer to iterate through the array

Solution

1. Step 1: Understand the technique

   The Two Pointer Technique uses two indices to traverse the array, often from opposite ends or different positions, to solve problems efficiently.

2. Step 2: Compare with other options

   Options B and C describe different methods, and D uses only one pointer, so they do not match the Two Pointer Technique.

3. Final Answer:

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

4. Quick Check:

   Two pointers = two indices scanning array [OK]

Hint: Two pointers move independently to cover array efficiently [OK]

Common Mistakes:

• Thinking it always requires sorting first
• Confusing with nested loops approach
• Using only one pointer instead of two

2. Which of the following is the correct syntax to initialize two pointers left and right for an array arr of size n in C++?

easy

A. int left = n - 1, right = 0;

B. int left = 1, right = n;

C. int left = n, right = 0;

D. int left = 0, right = n - 1;

Solution

1. Step 1: Understand pointer initialization

   For two pointers scanning from start and end, left starts at 0 and right at n-1 (last index).

2. Step 2: Check each option

   int left = 0, right = n - 1; correctly sets left=0 and right=n-1. Options A, B, and C have incorrect indices that may cause out-of-bound errors.

3. Final Answer:

   int left = 0, right = n - 1; -> Option D

4. Quick Check:

   Start = 0, End = n-1 [OK]

Hint: Start left at 0 and right at n-1 for full array scan [OK]

Common Mistakes:

• Setting right pointer to n (out of bounds)
• Swapping left and right initial values
• Starting left pointer at 1 instead of 0

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

int arr[] = {1, 2, 3, 4, 5, 6};
int left = 0, right = 5;
int sum = 0;
while (left < right) {
    sum += arr[left] + arr[right];
    left++;
    right--;
}
std::cout << sum << std::endl;

medium

A. 27

B. 18

C. 21

D. 15

Solution

1. Step 1: Trace the loop iterations

   Initial pointers: left=0, right=5
   Iteration 1: sum=0+arr[0]+arr[5]=0+1+6=7, left=1, right=4
   Iteration 2: sum=7+arr[1]+arr[4]=7+2+5=14, left=2, right=3
   Iteration 3: sum=14+arr[2]+arr[3]=14+3+4=21, left=3, right=2 (loop ends)

2. Step 2: Final sum value

   Sum after loop is 21.

3. Final Answer:

   21 -> Option C

4. Quick Check:

   Sum pairs (1+6)+(2+5)+(3+4) = 21 [OK]

Hint: Add pairs from ends moving inward until pointers cross [OK]

Common Mistakes:

• Including middle element when pointers cross
• Adding elements twice
• Incorrect loop condition (<= instead of <)

4. Identify the error in the following code that uses the two pointer technique to find if a sorted array contains two numbers that sum to a target value.

bool hasPairSum(int arr[], int n, int target) {
    int left = 0, right = n - 1;
    while (left <= right) {
        int sum = arr[left] + arr[right];
        if (sum == target) return true;
        else if (sum < target) left++;
        else right--;
    }
    return false;
}

medium

A. The loop condition should be left < right instead of left <= right

B. The pointers should be initialized as left = 1 and right = n

C. The sum calculation should be arr[left] - arr[right]

D. The function should return false inside the loop

Solution

1. Step 1: Analyze loop condition

   The loop runs while left <= right, which allows pointers to be equal, causing sum of same element twice.

2. Step 2: Correct loop condition

   Changing condition to left < right ensures pointers do not overlap, avoiding incorrect sum calculation.

3. Final Answer:

   The loop condition should be left < right instead of left <= right -> Option A

4. Quick Check:

   Pointers must not cross or overlap [OK]

Hint: Use left < right to avoid using same element twice [OK]

Common Mistakes:

• Allowing pointers to be equal causing wrong sums
• Incorrect pointer initialization
• Returning false too early inside loop

5. Given a sorted array arr of size n, how can you use the two pointer technique to find all unique pairs that sum up to a target value t? Which approach is correct?

hard

A. Use nested loops to check every pair and store unique pairs in a set

B. Initialize left = 0 and right = n - 1, move pointers inward, skip duplicates after finding a pair

C. Sort the array, then use binary search for each element to find complement

D. Initialize left = 1 and right = n, move pointers outward

Solution

1. Step 1: Set pointers and move inward

   Start with left at 0 and right at n-1. Move pointers inward based on sum compared to target.

2. Step 2: Skip duplicates to find unique pairs

   After finding a pair, increment left and decrement right while skipping duplicate values to avoid repeated pairs.

3. Final Answer:

   Initialize left = 0 and right = n - 1, move pointers inward, skip duplicates after finding a pair -> Option B

4. Quick Check:

   Two pointers inward + skip duplicates = unique pairs [OK]

Hint: Skip duplicates after finding pairs to avoid repeats [OK]

Common Mistakes:

• Not skipping duplicates causing repeated pairs
• Using nested loops causing higher time complexity
• Incorrect pointer initialization or movement