Two-pointer technique in Data Structures Theory - Time & Space Complexity

Time Complexity: Two-pointer technique

O(n)

Understanding Time Complexity

The two-pointer technique helps solve problems by moving two markers through data.

We want to know how the time needed grows as the data size grows.

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

function twoPointerSum(arr, target) {
  let left = 0;
  let right = arr.length - 1;
  while (left < right) {
    let sum = arr[left] + arr[right];
    if (sum === target) return true;
    else if (sum < target) left++;
    else right--;
  }
  return false;
}
    

This code checks if two numbers in a sorted list add up to a target by moving two pointers.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: The while loop moves pointers through the array.
• How many times: Each pointer moves at most once through the array, so up to n steps total.

How Execution Grows With Input

As the list gets bigger, the pointers move more steps but only once each.

Input Size (n)	Approx. Operations
10	Up to 10 steps
100	Up to 100 steps
1000	Up to 1000 steps

Pattern observation: The steps grow roughly in a straight line with input size.

Final Time Complexity

Time Complexity: O(n)

This means the time needed grows directly with the size of the input list.

Common Mistake

[X] Wrong: "Since there are two pointers, the time is doubled or squared."

[OK] Correct: The pointers move in one pass without nested loops, so steps add up linearly, not multiply.

Interview Connect

Understanding how two pointers move through data helps solve many real problems efficiently and shows clear thinking about performance.

Self-Check

"What if the array was not sorted? How would the time complexity change?"

Practice

1. What is the main purpose of the two-pointer technique in data structures?

easy

A. To store data in two separate arrays

B. To use two markers to scan data efficiently

C. To sort data using two different algorithms

D. To create two copies of the data for backup

Solution

1. Step 1: Recall the main purpose of the two-pointer technique

   The two-pointer technique uses two markers or pointers to scan through data efficiently, reducing unnecessary checks.

2. Step 2: Compare with other options

   The other options describe unrelated actions like storing data in separate arrays, sorting data, or creating backups, which are not the main purpose of this technique.

3. Final Answer:

   To use two markers to scan data efficiently -> Option B

4. Quick Check:

   Two-pointer technique = efficient scanning [OK]

Hint: Two pointers scan data fast by moving through it together [OK]

Common Mistakes:

• Confusing two-pointer with sorting algorithms
• Thinking it copies data instead of scanning
• Assuming it stores data separately

2. Which of the following is the correct way to initialize two pointers left and right for scanning an array arr from both ends?

easy

A. left = 0; right = len(arr) - 1

B. left = 1; right = len(arr)

C. left = len(arr); right = 0

D. left = -1; right = len(arr) + 1

Solution

1. Step 1: Identify valid pointer positions

   For scanning from both ends, left starts at 0 (first index) and right at len(arr) - 1 (last index).

2. Step 2: Check other options for errors

   The other options use invalid indices like starting at 1 or beyond array length, which cause errors or miss elements.

3. Final Answer:

   left = 0; right = len(arr) - 1 -> Option A

4. Quick Check:

   Start pointers at array ends = left = 0; right = len(arr) - 1 [OK]

Hint: Start left at 0 and right at last index for two-pointer scan [OK]

Common Mistakes:

• Starting pointers outside array bounds
• Using length instead of length minus one
• Swapping left and right initial values

3. Consider the following code using the two-pointer technique to check if a string is a palindrome:

def is_palindrome(s):
    left, right = 0, len(s) - 1
    while left < right:
        if s[left] != s[right]:
            return False
        left += 1
        right -= 1
    return True

print(is_palindrome('radar'))

What will be the output?

medium

A. True

B. None

C. Error

D. False

Solution

1. Step 1: Understand the palindrome check logic

   The code compares characters from the start and end moving inward. If any pair differs, it returns False; otherwise, True.

2. Step 2: Apply to 'radar'

   Characters match at all pairs: 'r'=='r', 'a'=='a', middle 'd' ignored. So, it returns True.

3. Final Answer:

   True -> Option A

4. Quick Check:

   Palindrome check for 'radar' = True [OK]

Hint: Check characters from ends moving inward for palindrome [OK]

Common Mistakes:

• Confusing return values for palindrome
• Not updating pointers correctly
• Ignoring middle character in odd length strings

4. The following code attempts to find if there exists a pair in a sorted array arr that sums to target using the two-pointer technique:

def has_pair(arr, target):
    left, right = 0, len(arr)
    while left < right:
        current_sum = arr[left] + arr[right]
        if current_sum == target:
            return True
        elif current_sum < target:
            left += 1
        else:
            right -= 1
    return False

What is the error in this code?

medium

A. The function should return True at the end

B. The loop condition should be while left <= right

C. The current_sum calculation is wrong

D. The right pointer is initialized incorrectly

Solution

1. Step 1: Check pointer initialization

   The right pointer is set to len(arr), which is outside valid index range (max index is len(arr)-1).

2. Step 2: Identify correct initialization

   Right pointer should be len(arr) - 1 to point to the last element.

3. Final Answer:

   The right pointer is initialized incorrectly -> Option D

4. Quick Check:

   Right pointer must be last index, not length [OK]

Hint: Right pointer = last index, not array length [OK]

Common Mistakes:

• Setting right pointer to array length
• Using wrong loop condition
• Miscomputing sum inside loop

5. You have two sorted arrays A and B. Using the two-pointer technique, how can you efficiently find all pairs (a, b) such that a + b = target? Choose the best approach.

hard

A. Use nested loops to check every pair from A and B

B. Sort both arrays again and then check pairs

C. Use one pointer on A starting at 0 and one on B starting at end, move pointers based on sum comparison

D. Use a hash map to store sums of all pairs

Solution

1. Step 1: Understand two-pointer approach for two sorted arrays

   Start pointer i at beginning of A and pointer j at end of B. Calculate sum A[i]+B[j].

2. Step 2: Move pointers based on sum

   If sum equals target, record pair and move pointers. If sum less than target, move i forward; if greater, move j backward. This avoids nested loops and is efficient.

3. Final Answer:

   Use one pointer on A starting at 0 and one on B starting at end, move pointers based on sum comparison -> Option C

4. Quick Check:

   Two-pointer on sorted arrays finds pairs efficiently [OK]

Hint: One pointer start, one pointer end; move based on sum [OK]

Common Mistakes:

• Using nested loops causing slow performance
• Resorting already sorted arrays unnecessarily
• Using hash map instead of two-pointer for sorted arrays