M.T.Ramkrushna

Posted on Sep 14

DSA: Topic 3: Two Pointers

#interview #python #dsa

Interview frequency: ⭐⭐⭐⭐⭐ (Top 5 pattern)

Many candidates fail interviews not because they don't know algorithms, but because they don't recognise when to use Two Pointers.

What is the Two Pointers technique?

Instead of using one index:

i = 0

you use two indices.

left = 0
right = len(arr) - 1

These pointers move based on the problem.

Example:

Index : 0  1  2  3  4

Array : 2  4  7  9  15
         ↑         ↑
       left      right

The pointers may:

Move towards each other
Move in the same direction
Move at different speeds

When should you think of Two Pointers?

This is the most important part.

Whenever you see:

Sorted array
Pair problems
Remove duplicates
Move elements
Reverse array/string
Palindrome
Merge sorted arrays

Your brain should immediately ask:

"Can I solve this using two pointers?"

Pattern 1: Opposite Direction

Pointers start from both ends.

L ------------->

<------------- R

Used for:

Pair Sum
Reverse Array
Palindrome
Container With Most Water

Example 1: Reverse an Array

Input

nums = [1,2,3,4,5]

Instead of creating another array, swap from both ends.

left = 0
right = len(nums)-1

while left < right:
    nums[left], nums[right] = nums[right], nums[left]

    left += 1
    right -= 1

Output

[5,4,3,2,1]

Complexity

Time : O(n)

Space: O(1)

Interviewers love in-place solutions.

Example 2: Valid Palindrome

madam

m == m

a == a

d

Algorithm

left = 0

right = len(s)-1

while left < right:

    if s[left] != s[right]:
        return False

    left += 1

    right -= 1

return True

Time

O(n)

Very common interview problem.

Example 3: Two Sum II (Sorted Array)

Input

[2,4,7,11,15]

target = 18

Instead of checking every pair:

2 + 15 = 17

Need larger

Move left

4 + 15 = 19

Too large

Move right

4 + 11 = 15

Need larger

Move left

7 + 11 = 18

Found

Code

left = 0
right = len(nums)-1

while left < right:

    total = nums[left] + nums[right]

    if total == target:
        return [left,right]

    elif total < target:
        left += 1

    else:
        right -= 1

Notice how the sorted property lets us decide which pointer to move.

Pattern 2: Same Direction

Used for:

Remove duplicates
Move zeroes
Partition arrays

Example:

[0,1,0,3,12]

Expected

[1,3,12,0,0]

Use

slow

fast

Fast explores.

Slow remembers where the next valid element should go.

Move Zeroes

slow = 0

for fast in range(len(nums)):

    if nums[fast] != 0:

        nums[slow], nums[fast] = nums[fast], nums[slow]

        slow += 1

Output

[1,3,12,0,0]

Interview frequency:

⭐⭐⭐⭐⭐

Pattern 3: Fast and Slow Pointer

Mostly used in Linked Lists.

slow

fast

slow -> 1 step

fast -> 2 steps

Applications

Detect cycle
Find middle node
Happy Number

We'll revisit this when we study linked lists.

How to Identify Two Pointer Problems

Ask yourself:

Is the array sorted?

Example

Pair Sum

Use opposite pointers.

Do I need to compare two ends?

Palindrome

Reverse String

Use opposite pointers.

Am I moving/removing elements?

Move Zeroes

Remove Duplicates

Use slow and fast pointers.

Am I merging two sorted arrays?

Use one pointer for each array.

Common Interview Mistakes

Mistake 1

Using nested loops.

for i:

    for j:

Many pair problems can be solved with two pointers in O(n) instead of O(n²).

Mistake 2

Using Two Pointers on an unsorted array when the problem depends on order.

For example:

Two Sum

If the array isn't sorted and you can't sort it (because you need original indices), use a Hash Map instead.

Mistake 3

Moving the wrong pointer.

Remember:

If the array is sorted:

Sum too small

Move left

Sum too big

Move right

Real Interview Questions

Master these:

Valid Palindrome ⭐⭐⭐⭐⭐
Reverse String ⭐⭐⭐⭐⭐
Two Sum II ⭐⭐⭐⭐
Move Zeroes ⭐⭐⭐⭐⭐
Remove Duplicates from Sorted Array ⭐⭐⭐⭐⭐
Squares of a Sorted Array ⭐⭐⭐⭐
Container With Most Water ⭐⭐⭐⭐
3Sum ⭐⭐⭐⭐⭐ (advanced extension)

Interview Thinking Process

Suppose you're asked:

"Given a sorted array, find two numbers whose sum is target."

A strong interview approach is:

Notice the array is sorted.
Recognise that brute force (O(n²)) is unnecessary.
Suggest using two pointers (left, right).
Explain why moving left increases the sum and moving right decreases it.
Code the O(n) solution.

This demonstrates both algorithmic knowledge and problem-solving, which is exactly what interviewers evaluate.

Next Topic

The next topic is Sliding Window, which is often confused with Two Pointers but solves a different class of problems such as:

Longest substring without repeating characters
Maximum sum subarray of size k
Minimum window substring
Longest repeating character replacement

Sliding Window is another ⭐⭐⭐⭐⭐ interview pattern and builds directly on the pointer concepts you've just learned.

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