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

Linked List.

Master pointer manipulation in linked structures

complexity

time

O(n)

space

O(1)

practice questions

browse and solve all 10 pattern questions →

02' — visual

interactive

Sliding window — k = 3

Maximum sum subarray

Window

[0, 2]

Sum

Step

0 / 3

03'

mental
model.

Use slow and fast pointers to traverse the list at different speeds. Fast pointer moves 2x the speed of slow pointer.

Linked List patterns focus on traversing, reversing, and manipulating linked structures using pointer manipulation.

when to use ✓

01Reverse a linked list
02Detect cycle (Floyd's algorithm)
03Find middle of linked list
04Merge two sorted lists
05Remove nth node from end

when not ✗

01Random access is needed
02Cache locality matters

04'

recognition signals

Linked list traversal or manipulation

Need O(1) insertions/deletions

Cycle detection

Find middle, merge, or reverse

Reversal (full or partial)Fast and slow pointerCycle detectionMerge operationsIntersection of two lists

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

common mistakes

01Losing references to next nodes during reversal
02Not handling null pointers
03Confusing recursion with iteration

interview tips

01Draw the linked list on paper
02Use dummy nodes to simplify edge cases
03Practice both iterative and recursive approaches

06'

practice this pattern

10 curated problems from LeetCode

See all

Intersection of Two Linked Lists# · Linked List

EASY

Design Linked List# · Linked List

MEDIUM

Remove Nth Node From End of List# · Linked List

MEDIUM

Remove Duplicates from Sorted List II# · Linked List

MEDIUM

Swap Nodes in Pairs# · Linked List

MEDIUM

Copy List with Random Pointer# · Linked List

MEDIUM

Partition List# · Linked List

MEDIUM

Rotate List# · Linked List

MEDIUM

07'

faq

← back to patterns

pattern / EASY

Linked List.

Master pointer manipulation in linked structures

complexity

time

O(n)

space

O(1)

practice questions

browse and solve all 10 pattern questions →

02' — visual

interactive

Sliding window — k = 3

Maximum sum subarray

Window

[0, 2]

Sum

Step

0 / 3

03'

mental
model.

Use slow and fast pointers to traverse the list at different speeds. Fast pointer moves 2x the speed of slow pointer.

Linked List patterns focus on traversing, reversing, and manipulating linked structures using pointer manipulation.

when to use ✓

01Reverse a linked list
02Detect cycle (Floyd's algorithm)
03Find middle of linked list
04Merge two sorted lists
05Remove nth node from end

when not ✗

01Random access is needed
02Cache locality matters

04'

recognition signals

Linked list traversal or manipulation

Need O(1) insertions/deletions

Cycle detection

Find middle, merge, or reverse

Reversal (full or partial)Fast and slow pointerCycle detectionMerge operationsIntersection of two lists

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