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↓

Process elements at left and right

↓

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

↩Back to Check condition

Start with two pointers at array ends, check condition, process elements, move pointers inward, repeat until pointers meet.

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 using two pointers moving inward.

Execution Table

Step	Operation	left	right	Sum arr[left]+arr[right]	Action	Visual State
1	Initialize pointers	0	5	arr[0]+arr[5]=1+12=13	13 < 15, move left++	[1, 3, 5, 7, 9, 12] (left=0, right=5)
2	Move left pointer	1	5	arr[1]+arr[5]=3+12=15	Sum == target, found pair	[1, 3, 5, 7, 9, 12] (left=1, right=5)
3	Break loop	1	5	15	Stop searching	[1, 3, 5, 7, 9, 12] (left=1, right=5)

💡 Pointers met condition sum == target, loop stops.

Variable Tracker

Variable	Start	After Step 1	After Step 2	Final
left	0	1	1	1
right	5	5	5	5
sum	N/A	13	15	15

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 left moves to a bigger number, increasing the sum. See execution_table step 1 where sum 13 < 15, so left moves from 0 to 1.

Why do we stop when left pointer meets right pointer?

When left >= right, all pairs have been checked. In execution_table step 3, the loop stops because the pair sum equals target, or pointers crossed.

What if the array is not sorted?

The two pointer technique relies on sorting to decide pointer moves. Without sorting, this logic fails, so the method won't find pairs correctly.

Visual Quiz - 3 Questions

Look at the execution_table, what is the value of 'left' after step 1?

A 1

B 0

C 5

D N/A

💡 Hint

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

At which step does the sum of arr[left] + arr[right] equal the target?

A Step 1

B Step 2

C Step 3

D Never

💡 Hint

Look at the 'Sum arr[left]+arr[right]' and 'Action' columns in execution_table.

If the array was not sorted, what would happen to the pointer movement logic?

A It would still work correctly

B Pointers would move randomly

C The logic to move pointers based on sum comparison would fail

D The pointers would never move

💡 Hint

Refer to key_moments explanation about sorting necessity.

Concept Snapshot

Two Pointer Technique on Arrays:
- Use two pointers at start and end of sorted array
- Check sum of elements at pointers
- If sum < target, move left pointer right
- If sum > target, move right pointer left
- Stop when pointers meet or pair found

Full Transcript

The two pointer technique uses two indexes starting at the ends of a sorted array. We check the sum of the elements at these pointers. If the sum is less than the target, we move the left pointer right to increase the sum. If the sum is greater, we move the right pointer left to decrease the sum. We repeat this until the pointers meet or we find the target sum. This method works only on sorted arrays because the pointer movement depends on the order of elements.

Practice

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

easy

A. Use one pointer to traverse and another to store the sum.

B. Use two arrays to store elements separately.

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

D. Use two pointers to swap elements randomly.

Solution

1. Step 1: Understand the technique

   The two pointer technique involves using two indices or pointers to scan the array, often from the start and end, to solve problems efficiently.

2. Step 2: Identify the correct description

   Use two indices to traverse the array from different ends or positions simultaneously. correctly describes using two pointers to traverse the array from different ends or positions simultaneously.

3. Final Answer:

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

4. Quick Check:

   Two pointers move from different positions = A [OK]

Hint: Two pointers move from different ends or positions [OK]

Common Mistakes:

• Thinking two pointers mean two separate arrays
• Confusing pointers with storing sums
• Assuming pointers swap elements randomly

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 = 0, right = n - 1;

B. int left = 1, right = n;

C. int left = n, right = 0;

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

Solution

1. Step 1: Understand pointer initialization

   For two pointers starting at the beginning and end of an array, left should be 0 and right should be n - 1 because array indices start at 0.

2. Step 2: Check options

   int left = 0, right = n - 1; correctly initializes left = 0 and right = n - 1. Other options have incorrect indices.

3. Final Answer:

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

4. Quick Check:

   Start at 0 and end at n-1 = C [OK]

Hint: Array indices start at 0, so right = n - 1 [OK]

Common Mistakes:

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

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

int arr[] = {1, 3, 5, 7, 9};
int left = 0, right = 4;
int sum = 0;
while (left < right) {
    sum += arr[left] + arr[right];
    left++;
    right--;
}
printf("%d", sum);

medium

A. 25

B. 15

C. 30

D. 20

Solution

1. Step 1: Trace the loop iterations

   Initial pointers: left=0, right=4
   Iteration 1: sum=0+(1+9)=10, left=1, right=3
   Iteration 2: sum=10+(3+7)=20, left=2, right=2
   Loop ends because left is not less than right.

2. Step 2: Calculate final sum

   Sum after loop = 20. Note that the middle element (index 2) is not added because left == right stops the loop.

3. Final Answer:

   20 -> Option D

4. Quick Check:

   Sum pairs (1+9)+(3+7) = 20 [OK]

Hint: Sum pairs while left < right, middle element skipped [OK]

Common Mistakes:

• Including middle element when left == right
• Adding elements twice
• Incorrect loop condition

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

int arr[] = {2, 4, 6, 8, 10};
int left = 0, right = 4;
int target = 14;
while (left <= right) {
    int sum = arr[left] + arr[right];
    if (sum == target) {
        printf("Found");
        break;
    } else if (sum < target) {
        left++;
    } else {
        right--;
    }
}

medium

A. The pointers left and right are initialized incorrectly.

B. The loop condition should be left < right, not left <= right.

C. The sum comparison logic is reversed.

D. The printf statement should be outside the loop.

Solution

1. Step 1: Analyze loop condition

   The loop uses left <= right, which allows pointers to meet or cross. This can cause incorrect checks or double counting.

2. Step 2: Correct loop condition

   Using left < right ensures pointers do not cross and only valid pairs are checked.

3. Final Answer:

   The loop condition should be left < right, not left <= right. -> Option B

4. Quick Check:

   Pointers must not cross: left < right [OK]

Hint: Use left < right to avoid pointer crossing [OK]

Common Mistakes:

• Using <= instead of < in loop condition
• Swapping pointer increment/decrement logic
• Incorrect pointer initialization

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

hard

A. Initialize two pointers at start and end; move pointers inward based on sum comparison; skip duplicates to find unique pairs.

B. Use nested loops to check all pairs and store sums in a hash set.

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

D. Initialize both pointers at start and move both forward to find pairs.

Solution

1. Step 1: Use two pointers at array ends

   Start with left = 0 and right = n - 1. Calculate sum of elements at these pointers.

2. Step 2: Move pointers and skip duplicates

   If sum equals target, record pair and move both pointers skipping duplicates. If sum < target, move left pointer right. If sum > target, move right pointer left.

3. Final Answer:

   Initialize two pointers at start and end; move pointers inward based on sum comparison; skip duplicates to find unique pairs. -> Option A

4. Quick Check:

   Two pointers from ends + skip duplicates = D [OK]

Hint: Move pointers inward and skip duplicates for unique pairs [OK]

Common Mistakes:

• Using nested loops instead of two pointers
• Not skipping duplicates causing repeated pairs
• Moving both pointers forward instead of inward