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feat(patterns): pointer/array tutorials

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Kai Chappell
2025-08-18 22:15:43 +01:00
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backend/data/patterns/fast-slow-pointers.yaml Normal file
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name: Fast & Slow Pointers
slug: fast-slow-pointers
difficulty_level: 2
description: >
Use two pointers moving at different speeds to detect cycles, find midpoints,
or identify patterns in sequences. The fast pointer advances twice as quickly,
allowing detection of structural properties without extra space.
when_to_use: |
- Detecting cycles in linked lists or sequences
- Finding the middle of a linked list
- Finding the start of a cycle
- Happy number problem
- Palindrome linked list verification
metaphor: |
Imagine two runners on a circular track. If one runs twice as fast as the other,
the fast runner will eventually lap the slow runner—they'll meet at some point
on the track. This proves the track is circular (has a cycle).
Another analogy: finding the middle of a line of people. Have two people start
at the front—one takes one step at a time, the other takes two. When the fast
person reaches the end, the slow person is at the middle.
core_concept: |
The **fast & slow pointers** technique (also called Floyd's cycle detection or
"tortoise and hare") uses two pointers moving at different speeds:
- **Slow pointer**: Moves 1 step at a time
- **Fast pointer**: Moves 2 steps at a time
Key insights:
1. **Cycle detection**: In a cyclic structure, fast will eventually catch up to
slow (they'll meet inside the cycle). In a non-cyclic structure, fast will
reach the end.
2. **Finding middle**: When fast reaches the end, slow is at the middle (fast
traveled 2x the distance).
3. **Finding cycle start**: After detecting a cycle, reset one pointer to start.
Move both at the same speed—they meet at the cycle start. (Mathematical proof:
the distances work out perfectly.)
visualization: |
**Cycle Detection:**
```
List: 1 → 2 → 3 → 4 → 5
↑ ↓
7 ← 6
Step 1: slow=1, fast=1
Step 2: slow=2, fast=3
Step 3: slow=3, fast=5
Step 4: slow=4, fast=7
Step 5: slow=5, fast=4
Step 6: slow=6, fast=6 ← They meet! Cycle exists.
```
**Finding Middle:**
```
List: 1 → 2 → 3 → 4 → 5 → null
Step 1: slow=1, fast=1
Step 2: slow=2, fast=3
Step 3: slow=3, fast=5
Step 4: fast reaches null
slow is at middle (3)
```
**Finding Cycle Start:**
```
After detecting cycle at node X:
1. Reset slow to head, keep fast at meeting point
2. Move both at same speed (1 step each)
3. They meet at cycle start
Why? Math: Let's say:
- Distance from head to cycle start = A
- Distance from cycle start to meeting point = B
- Cycle length = C
At meeting: slow traveled A + B
fast traveled A + B + nC (some complete cycles)
Since fast travels 2x: 2(A + B) = A + B + nC
Therefore: A + B = nC, so A = nC - B = (n-1)C + (C-B)
This means: distance from head to cycle start
= distance from meeting point to cycle start (going forward)
```
code_template: |
class ListNode:
def __init__(self, val=0, next=None):
self.val = val
self.next = next
def has_cycle(head: ListNode) -> bool:
"""Detect if linked list has a cycle."""
if not head or not head.next:
return False
slow = head
fast = head
while fast and fast.next:
slow = slow.next
fast = fast.next.next
if slow == fast:
return True
return False
def find_cycle_start(head: ListNode) -> ListNode:
"""Find the node where the cycle begins."""
if not head or not head.next:
return None
# Phase 1: Detect cycle
slow = fast = head
while fast and fast.next:
slow = slow.next
fast = fast.next.next
if slow == fast:
break
else:
return None # No cycle
# Phase 2: Find cycle start
slow = head
while slow != fast:
slow = slow.next
fast = fast.next
return slow
def find_middle(head: ListNode) -> ListNode:
"""Find the middle node of linked list."""
if not head:
return None
slow = fast = head
while fast and fast.next:
slow = slow.next
fast = fast.next.next
return slow # Middle (or second middle if even length)
def is_happy_number(n: int) -> bool:
"""Check if number is happy (sum of squared digits eventually = 1)."""
def get_next(num: int) -> int:
total = 0
while num > 0:
digit = num % 10
total += digit * digit
num //= 10
return total
slow = n
fast = get_next(n)
while fast != 1 and slow != fast:
slow = get_next(slow)
fast = get_next(get_next(fast))
return fast == 1
def is_palindrome_linked_list(head: ListNode) -> bool:
"""Check if linked list is a palindrome."""
if not head or not head.next:
return True
# Find middle
slow = fast = head
while fast and fast.next:
slow = slow.next
fast = fast.next.next
# Reverse second half
prev = None
while slow:
next_node = slow.next
slow.next = prev
prev = slow
slow = next_node
# Compare halves
left, right = head, prev
while right:
if left.val != right.val:
return False
left = left.next
right = right.next
return True
recognition_signals:
- "linked list cycle"
- "detect cycle"
- "find middle"
- "happy number"
- "palindrome linked list"
- "circular array"
- "Floyd's"
- "tortoise and hare"
- "meeting point"
- "cycle start"
common_mistakes:
- title: Not checking fast.next before advancing
description: |
Accessing `fast.next.next` without first checking `fast.next` causes
null pointer errors when the list has even length.
fix: |
Always check both `fast` and `fast.next`:
```python
while fast and fast.next:
fast = fast.next.next
```
- title: Wrong initialization for cycle detection
description: |
Starting slow and fast at different positions (e.g., slow=head, fast=head.next)
changes the math for finding the cycle start.
fix: |
Start both at head for consistency. The algorithms are designed assuming
both start at the same position.
- title: Forgetting to handle empty or single-node lists
description: |
Accessing head.next or head.next.next on empty or single-node lists
causes errors.
fix: |
Add early returns:
```python
if not head or not head.next:
return False # or appropriate value
```
- title: Confusing meeting point with cycle start
description: |
Returning the meeting point instead of finding the actual cycle start
gives the wrong answer.
fix: |
After detecting a cycle (meeting point), reset one pointer to head and
advance both at the same speed to find the cycle start.
variations:
- name: Cycle detection
description: |
Determine if a linked list or sequence has a cycle. If fast catches slow,
there's a cycle.
example: "Linked List Cycle, Circular Array Loop"
- name: Finding cycle start
description: |
After detecting a cycle, find the node where the cycle begins using the
two-phase approach.
example: "Linked List Cycle II"
- name: Finding middle
description: |
When fast reaches the end, slow is at the middle. Useful for divide and
conquer on linked lists.
example: "Middle of Linked List, Sort List (merge sort needs middle)"
- name: Happy number
description: |
Treat the sequence of digit-square sums as a linked list. Either reaches 1
(happy) or cycles (unhappy).
example: "Happy Number"
- name: Palindrome check
description: |
Find middle, reverse second half, compare. Combines finding middle with
linked list reversal.
example: "Palindrome Linked List"
related_patterns:
- two-pointers
- linkedlist-reversal
prerequisite_patterns:
- two-pointers

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name: Overlapping Intervals
slug: intervals
difficulty_level: 2
description: >
Process and manipulate intervals (ranges) that may share common regions.
The key insight is that sorting intervals by start time allows efficient
detection and handling of overlaps through a single linear pass.
when_to_use: |
- Merging overlapping intervals
- Inserting an interval into a sorted list
- Finding gaps between intervals
- Meeting room scheduling
- Finding interval intersections
metaphor: |
Imagine scheduling meeting rooms. Each meeting is an interval of time. When
two meetings overlap, you need either two rooms or to merge them into one
longer booking. By sorting meetings by start time, you can easily spot
overlaps—if the next meeting starts before the current one ends, they overlap.
Another analogy: merging overlapping highlighter marks on a page. Sort them
left to right, and if one mark starts before the previous ends, combine them
into one continuous highlight.
core_concept: |
The **interval pattern** relies on sorting intervals by start time. Once sorted,
overlapping detection becomes simple:
**Two intervals [a, b] and [c, d] overlap if c <= b** (assuming a <= c after sorting)
Key operations:
1. **Merge**: Extend the current interval's end to include overlapping intervals
2. **Insert**: Find where the new interval overlaps and merge as needed
3. **Count overlaps**: Track how many intervals are "active" at any point
For problems needing simultaneous tracking (like minimum meeting rooms), use
a **sweep line** approach: sort all start and end points together, then sweep
through counting active intervals.
visualization: |
**Merging overlapping intervals:**
```
Input: [[1,3], [2,6], [8,10], [15,18]]
(sorted by start)
Process [1,3]: result = [[1,3]]
Process [2,6]: 2 <= 3? Yes, overlap!
Merge: [1, max(3,6)] = [1,6]
result = [[1,6]]
Process [8,10]: 8 <= 6? No, no overlap
result = [[1,6], [8,10]]
Process [15,18]: 15 <= 10? No, no overlap
result = [[1,6], [8,10], [15,18]]
```
**Visualized on number line:**
```
Before: [1---3]
[2------6]
[8--10]
[15--18]
After: [1--------6] [8--10] [15--18]
```
**Meeting rooms (sweep line):**
```
Meetings: [[0,30], [5,10], [15,20]]
Events (sorted):
time=0: +1 (start) active=1
time=5: +1 (start) active=2 ← max
time=10: -1 (end) active=1
time=15: +1 (start) active=2 ← max
time=20: -1 (end) active=1
time=30: -1 (end) active=0
Max concurrent = 2 → need 2 meeting rooms
```
code_template: |
def merge_intervals(intervals: list[list[int]]) -> list[list[int]]:
"""Merge overlapping intervals."""
if not intervals:
return []
# Sort by start time
intervals.sort(key=lambda x: x[0])
result = [intervals[0]]
for start, end in intervals[1:]:
last_end = result[-1][1]
if start <= last_end: # Overlap
result[-1][1] = max(last_end, end) # Extend
else:
result.append([start, end])
return result
def insert_interval(intervals: list[list[int]],
new: list[int]) -> list[list[int]]:
"""Insert and merge a new interval into sorted list."""
result = []
i = 0
n = len(intervals)
# Add all intervals before new interval
while i < n and intervals[i][1] < new[0]:
result.append(intervals[i])
i += 1
# Merge overlapping intervals with new
while i < n and intervals[i][0] <= new[1]:
new[0] = min(new[0], intervals[i][0])
new[1] = max(new[1], intervals[i][1])
i += 1
result.append(new)
# Add remaining intervals
while i < n:
result.append(intervals[i])
i += 1
return result
def min_meeting_rooms(intervals: list[list[int]]) -> int:
"""Find minimum meeting rooms needed (sweep line)."""
events = []
for start, end in intervals:
events.append((start, 1)) # +1 for start
events.append((end, -1)) # -1 for end
# Sort by time, with ends before starts at same time
events.sort(key=lambda x: (x[0], x[1]))
max_rooms = 0
current_rooms = 0
for _, delta in events:
current_rooms += delta
max_rooms = max(max_rooms, current_rooms)
return max_rooms
def interval_intersection(A: list[list[int]],
B: list[list[int]]) -> list[list[int]]:
"""Find intersection of two sorted interval lists."""
result = []
i = j = 0
while i < len(A) and j < len(B):
# Find overlap
start = max(A[i][0], B[j][0])
end = min(A[i][1], B[j][1])
if start <= end:
result.append([start, end])
# Advance the interval that ends first
if A[i][1] < B[j][1]:
i += 1
else:
j += 1
return result
def can_attend_all(intervals: list[list[int]]) -> bool:
"""Check if a person can attend all meetings."""
intervals.sort(key=lambda x: x[0])
for i in range(1, len(intervals)):
if intervals[i][0] < intervals[i-1][1]:
return False # Overlap found
return True
recognition_signals:
- "intervals"
- "merge"
- "overlapping"
- "meeting rooms"
- "schedule"
- "time slots"
- "range"
- "insert interval"
- "non-overlapping"
- "intersection"
common_mistakes:
- title: Not sorting first
description: |
Trying to process unsorted intervals leads to incorrect results because
overlaps aren't detected properly.
fix: |
Always sort intervals by start time before processing:
```python
intervals.sort(key=lambda x: x[0])
```
- title: Wrong overlap condition
description: |
Using `start < last_end` instead of `start <= last_end` misses adjacent
intervals that should be merged (like [1,2] and [2,3]).
fix: |
Use `<=` for touching intervals, `<` for strict overlap only. Check problem
requirements for whether touching counts as overlapping.
- title: Not updating end correctly when merging
description: |
Setting `end = new_end` instead of `end = max(old_end, new_end)` fails when
a smaller interval is contained within a larger one.
fix: |
Always take the maximum:
```python
result[-1][1] = max(result[-1][1], end)
```
- title: Off-by-one with closed vs open intervals
description: |
Confusion about whether interval endpoints are inclusive `[a, b]` or
exclusive `[a, b)` causes incorrect overlap detection.
fix: |
Clarify the convention from the problem. Most problems use closed intervals
where both endpoints are included.
variations:
- name: Merge intervals
description: |
Combine all overlapping intervals into non-overlapping intervals.
example: "Merge Intervals"
- name: Insert interval
description: |
Insert a new interval into a sorted list, merging with any overlapping
intervals.
example: "Insert Interval"
- name: Meeting rooms
description: |
Find minimum resources needed for concurrent intervals using sweep line
or min-heap.
example: "Meeting Rooms II, Car Pooling"
- name: Interval intersection
description: |
Find common regions between two lists of intervals using two pointers.
example: "Interval List Intersections"
- name: Non-overlapping intervals
description: |
Find minimum removals to make all intervals non-overlapping. Greedy
approach: keep intervals that end earliest.
example: "Non-overlapping Intervals, Erase Overlap Intervals"
related_patterns:
- greedy
- two-pointers
prerequisite_patterns: []

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name: LinkedList In-Place Reversal
slug: linkedlist-reversal
difficulty_level: 2
description: >
Reverse linked list nodes in-place by manipulating pointers without allocating
extra space. This technique uses three pointers to track the previous, current,
and next nodes while systematically reversing the direction of links.
when_to_use: |
- Reversing an entire linked list
- Reversing a portion of a linked list
- Reversing in groups of K nodes
- Palindrome linked list verification
- Reordering list problems
metaphor: |
Imagine a conga line where everyone faces forward. To reverse it, you don't
rearrange people—you have each person turn around and grab the shoulders of
whoever was behind them. You process one person at a time: they turn around,
the next person steps forward, and so on until everyone faces the opposite direction.
Another analogy: reversing a chain of paper clips. You unclip each one from its
forward neighbor and clip it to its backward neighbor, working through the chain.
core_concept: |
Linked list reversal uses **three pointers** moving through the list:
- **prev**: Points to the already-reversed portion (starts as null)
- **curr**: The node currently being processed
- **next**: Temporarily stores the next node before we break the link
At each step:
1. Save `curr.next` in `next` (before we lose it)
2. Reverse the link: `curr.next = prev`
3. Advance: `prev = curr`, `curr = next`
The key insight is that we're not moving nodes—we're redirecting pointers.
This achieves O(n) time with O(1) space.
For **partial reversal** (reversing between positions m and n), we:
1. Navigate to position m-1 (the node before reversal starts)
2. Reverse nodes from m to n
3. Reconnect the reversed portion to the rest of the list
visualization: |
**Full list reversal:**
```
Initial: 1 → 2 → 3 → 4 → null
prev=null, curr=1
Step 1: Save next=2, reverse 1's link
null ← 1 2 → 3 → 4
prev curr
Step 2: Save next=3, reverse 2's link
null ← 1 ← 2 3 → 4
prev curr
Step 3: Save next=4, reverse 3's link
null ← 1 ← 2 ← 3 4
prev curr
Step 4: Save next=null, reverse 4's link
null ← 1 ← 2 ← 3 ← 4
prev curr=null
Result: 4 → 3 → 2 → 1 → null
```
**Partial reversal (positions 2 to 4):**
```
Initial: 1 → 2 → 3 → 4 → 5
positions: 1 2 3 4 5
Goal: 1 → 4 → 3 → 2 → 5
Step 1: Find node before position 2
before = node 1
Step 2: Reverse nodes 2, 3, 4
1 null ← 2 ← 3 ← 4 5
↑ ↑ ↑
before prev curr
Step 3: Reconnect
before.next.next = curr (2 → 5)
before.next = prev (1 → 4)
Result: 1 → 4 → 3 → 2 → 5
```
code_template: |
class ListNode:
def __init__(self, val=0, next=None):
self.val = val
self.next = next
def reverse_list(head: ListNode) -> ListNode:
"""Reverse entire linked list."""
prev = None
curr = head
while curr:
next_node = curr.next # Save next
curr.next = prev # Reverse link
prev = curr # Advance prev
curr = next_node # Advance curr
return prev # New head
def reverse_between(head: ListNode, m: int, n: int) -> ListNode:
"""Reverse nodes from position m to n (1-indexed)."""
if not head or m == n:
return head
dummy = ListNode(0)
dummy.next = head
before = dummy
# Move to node before reversal starts
for _ in range(m - 1):
before = before.next
# Reverse n - m + 1 nodes
prev = None
curr = before.next
for _ in range(n - m + 1):
next_node = curr.next
curr.next = prev
prev = curr
curr = next_node
# Reconnect
before.next.next = curr # tail of reversed → rest of list
before.next = prev # before → new head of reversed
return dummy.next
def reverse_k_group(head: ListNode, k: int) -> ListNode:
"""Reverse nodes in groups of k."""
# Count total nodes
count = 0
node = head
while node:
count += 1
node = node.next
dummy = ListNode(0)
dummy.next = head
before = dummy
while count >= k:
# Reverse k nodes
prev = None
curr = before.next
for _ in range(k):
next_node = curr.next
curr.next = prev
prev = curr
curr = next_node
# Reconnect
tail = before.next
tail.next = curr
before.next = prev
# Move to next group
before = tail
count -= k
return dummy.next
def reverse_list_recursive(head: ListNode) -> ListNode:
"""Reverse list using recursion."""
if not head or not head.next:
return head
new_head = reverse_list_recursive(head.next)
head.next.next = head # Reverse link
head.next = None # Prevent cycle
return new_head
recognition_signals:
- "reverse linked list"
- "reverse between"
- "reverse in groups"
- "reverse k-group"
- "palindrome linked list"
- "reorder list"
- "swap nodes"
- "rotate list"
common_mistakes:
- title: Losing reference to next node
description: |
Reversing `curr.next` before saving it means you can't advance to the
next node.
fix: |
Always save the next node first:
```python
next_node = curr.next # Save FIRST
curr.next = prev # Then reverse
```
- title: Forgetting to update connections in partial reversal
description: |
Reversing the middle portion without reconnecting it to the beginning
and end of the list breaks the list.
fix: |
After reversing, reconnect both ends:
```python
before.next.next = curr # tail → rest
before.next = prev # before → new head
```
- title: Not using a dummy node
description: |
When the reversal might include the head, handling the head separately
adds complexity and edge cases.
fix: |
Use a dummy node pointing to head. This simplifies edge cases:
```python
dummy = ListNode(0)
dummy.next = head
# ... reversal logic ...
return dummy.next
```
- title: Off-by-one with positions
description: |
Confusing 0-indexed vs 1-indexed positions causes reversal of wrong nodes.
fix: |
Clarify indexing convention. For 1-indexed positions, loop `m-1` times
to reach the node *before* position m.
variations:
- name: Full reversal
description: |
Reverse the entire linked list. Simplest form—just walk through and
reverse each link.
example: "Reverse Linked List"
- name: Partial reversal
description: |
Reverse only nodes between positions m and n. Need to track connection
points before and after the reversed section.
example: "Reverse Linked List II"
- name: K-group reversal
description: |
Reverse every k consecutive nodes. Often requires counting total nodes
first to know when to stop.
example: "Reverse Nodes in k-Group"
- name: Alternating reversal
description: |
Reverse every other group of k nodes. Combines k-group logic with
skip logic.
example: "Reverse Alternate K Nodes"
- name: Recursive reversal
description: |
Elegant recursive solution that reverses by relying on the recursive
call to reverse the rest, then fixing up the current node.
example: "Reverse Linked List (recursive approach)"
related_patterns:
- fast-slow-pointers
- two-pointers
prerequisite_patterns: []

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name: Prefix Sum
slug: prefix-sum
difficulty_level: 2
description: >
Precompute cumulative sums to answer range sum queries in O(1) time. This
transforms repeated O(n) range calculations into O(n) preprocessing plus
O(1) per query, making it essential for problems involving subarray sums.
when_to_use: |
- Range sum queries
- Subarray sum equals target
- Count subarrays with given sum
- Product of array except self
- 2D matrix region sums
metaphor: |
Imagine tracking your total running distance over a year. Instead of adding up
daily distances each time someone asks "how far did you run from day 50 to day
75?", you keep a running total. The cumulative distance on day 75 minus day 49
instantly gives you the answer.
Another analogy: a bank account balance. To find how much you spent between
two dates, you subtract the earlier balance from the later balance—no need to
sum individual transactions.
core_concept: |
A **prefix sum array** stores cumulative sums where `prefix[i]` = sum of all
elements from index 0 to i-1. This enables O(1) range sum queries:
**sum(i, j) = prefix[j+1] - prefix[i]**
The key insight is that any range sum can be computed from two prefix sums.
Instead of iterating through the range (O(n) per query), we do O(n) preprocessing
once and answer unlimited queries in O(1) each.
This pattern extends to:
- **Prefix products**: For multiplication-based problems
- **Prefix counts**: For counting occurrences
- **2D prefix sums**: For matrix region queries
- **Prefix sum + hash map**: For "subarray sum equals K"
visualization: |
**Building prefix sum array:**
```
Array: [3, 1, 4, 1, 5, 9]
Index: 0 1 2 3 4 5
prefix[0] = 0 (empty prefix)
prefix[1] = 0 + 3 = 3 (sum of first 1 element)
prefix[2] = 3 + 1 = 4 (sum of first 2 elements)
prefix[3] = 4 + 4 = 8
prefix[4] = 8 + 1 = 9
prefix[5] = 9 + 5 = 14
prefix[6] = 14 + 9 = 23
Prefix: [0, 3, 4, 8, 9, 14, 23]
Index: 0 1 2 3 4 5 6
```
**Range sum query: sum(2, 4) = elements at indices 2, 3, 4**
```
sum(2, 4) = prefix[5] - prefix[2]
= 14 - 4
= 10
Verification: arr[2] + arr[3] + arr[4] = 4 + 1 + 5 = 10 ✓
```
**Subarray sum equals K using hash map:**
```
Array: [1, 2, 3, -2, 5] K = 4
As we iterate, track prefix sums and look for prefix_sum - K:
i=0: prefix=1, need 1-4=-3, not found
i=1: prefix=3, need 3-4=-1, not found
i=2: prefix=6, need 6-4=2, not found
i=3: prefix=4, need 4-4=0, found! (empty prefix)
→ subarray [0:4] sums to 4
i=4: prefix=9, need 9-4=5, not found
Wait, also: prefix at i=1 is 3, at i=4 is 9-4=5... let me recalculate
Actually arr[1:3] = 2+3-2=3, arr[2:4]=3-2+5=6... checking for K=4:
subarray [0:4] → 1+2+3-2=4 ✓
```
code_template: |
def build_prefix_sum(arr: list[int]) -> list[int]:
"""Build prefix sum array. prefix[i] = sum of arr[0:i]."""
prefix = [0] * (len(arr) + 1)
for i in range(len(arr)):
prefix[i + 1] = prefix[i] + arr[i]
return prefix
def range_sum(prefix: list[int], i: int, j: int) -> int:
"""Sum of elements from index i to j (inclusive)."""
return prefix[j + 1] - prefix[i]
def subarray_sum_equals_k(nums: list[int], k: int) -> int:
"""Count subarrays with sum equal to k."""
count = 0
prefix_sum = 0
# Map: prefix_sum -> count of occurrences
sum_count = {0: 1} # Empty prefix has sum 0
for num in nums:
prefix_sum += num
# If (prefix_sum - k) exists, those prefixes form valid subarrays
if prefix_sum - k in sum_count:
count += sum_count[prefix_sum - k]
# Record current prefix sum
sum_count[prefix_sum] = sum_count.get(prefix_sum, 0) + 1
return count
def product_except_self(nums: list[int]) -> list[int]:
"""Product of array except self without division."""
n = len(nums)
result = [1] * n
# Prefix products (left to right)
prefix = 1
for i in range(n):
result[i] = prefix
prefix *= nums[i]
# Suffix products (right to left)
suffix = 1
for i in range(n - 1, -1, -1):
result[i] *= suffix
suffix *= nums[i]
return result
def matrix_region_sum(matrix: list[list[int]],
row1: int, col1: int,
row2: int, col2: int) -> int:
"""2D prefix sum for matrix region queries."""
# Build 2D prefix sum
m, n = len(matrix), len(matrix[0])
prefix = [[0] * (n + 1) for _ in range(m + 1)]
for i in range(1, m + 1):
for j in range(1, n + 1):
prefix[i][j] = (matrix[i-1][j-1]
+ prefix[i-1][j]
+ prefix[i][j-1]
- prefix[i-1][j-1])
# Query region sum using inclusion-exclusion
return (prefix[row2+1][col2+1]
- prefix[row1][col2+1]
- prefix[row2+1][col1]
+ prefix[row1][col1])
recognition_signals:
- "range sum"
- "subarray sum"
- "cumulative"
- "sum equals k"
- "count subarrays"
- "product except self"
- "matrix region sum"
- "running total"
- "continuous subarray"
common_mistakes:
- title: Off-by-one in prefix array indexing
description: |
Confusion about whether prefix[i] includes arr[i] or not leads to
incorrect range sums.
fix: |
Convention: `prefix[i]` = sum of first i elements = `arr[0:i]`.
So `prefix[0] = 0` (empty), and range sum is `prefix[j+1] - prefix[i]`.
- title: Forgetting the empty prefix for "sum equals K"
description: |
Not initializing the hash map with `{0: 1}` misses subarrays starting
from index 0.
fix: |
Always initialize: `sum_count = {0: 1}`. This handles the case where the
subarray from the beginning has sum K.
- title: Integer overflow with large sums
description: |
Prefix sums can grow very large when array elements are big, causing
overflow in some languages.
fix: |
In Python this isn't an issue. In Java/C++, use `long` for prefix sums
or check constraints carefully.
- title: Not handling negative numbers
description: |
Prefix sum works with negative numbers, but some may expect only positive
sums and use wrong optimization (like sliding window).
fix: |
Prefix sum handles negatives correctly. But problems like "minimum sum
subarray of size k" need different approaches when negatives are present.
variations:
- name: Basic prefix sum
description: |
Precompute cumulative sums for O(1) range queries. Foundation for other
variations.
example: "Range Sum Query - Immutable, Running Sum of 1d Array"
- name: Prefix sum with hash map
description: |
Track prefix sums in a hash map to find subarrays summing to a target.
Key insight: `prefix[j] - prefix[i] = target` means subarray [i,j] works.
example: "Subarray Sum Equals K, Contiguous Array"
- name: Prefix product
description: |
Same idea but with multiplication. Watch for zeros and consider using
left/right products separately.
example: "Product of Array Except Self"
- name: 2D prefix sum
description: |
Extend to matrices for O(1) region sum queries. Uses inclusion-exclusion
principle.
example: "Range Sum Query 2D, Matrix Block Sum"
- name: Difference array
description: |
Inverse of prefix sum. Store differences to support O(1) range updates.
Prefix sum of difference array gives original.
example: "Range Addition, Corporate Flight Bookings"
related_patterns:
- sliding-window
- two-pointers
prerequisite_patterns: []