What is the Sliding Window pattern?

A technique using two pointers to maintain a window over a linear data structure, processing elements efficiently in O(n).

When should I use Sliding Window?

When the problem asks about contiguous subarrays or substrings, longest/shortest sequences, or when brute force is O(n²) and can be optimized.

Is Sliding Window important for interviews?

Yes — it appears in 20-30% of coding interview questions at FAANG companies, especially for array and string problems.

What is the time complexity?

O(n) time, since each element is processed at most twice. Space is O(1) for fixed windows or O(k) for variable windows with a hashmap.

← back to patterns

pattern / EASY

Two Pointers.

Use two references to traverse a data structure

complexity

time

O(n)

space

O(1)

practice questions

browse and solve all 5 pattern questions →

02' — visual

interactive

Sliding window — k = 3

Maximum sum subarray

Window

[0, 2]

Sum

Step

0 / 3

03'

mental
model.

Two pointers move through the data structure based on a condition. Often one starts at the beginning and one at the end, moving toward each other.

Two Pointers is a technique where two pointers iterate through the data structure in tandem, often used to find pairs or compare elements efficiently.

when to use ✓

01Finding pair with target sum in sorted array
02Removing duplicates from sorted array
03Container with most water
043Sum, 4Sum problems

when not ✗

01Unsorted data without preprocessing
02When you need to maintain original order and pointers would break it

04'

recognition signals

Sorted array or linked list

Need to find pairs/triplets with a target sum

Need to compare elements from both ends

In-place operations on array

Opposite ends pointersFast and slow pointersSame direction pointersThree pointers for triplet problems

05'

templates

python

1# Beginner: Fixed-size sliding window
2def max_sum_k(nums, k):
3   n = len(nums)
4   if n < k:
5      return 0
6   window_sum = sum(nums[:k])
7   max_sum = window_sum
8   for i in range(k, n):
9      window_sum += nums[i] - nums[i - k]
10      max_sum = max(max_sum, window_sum)
11   return max_sum