How do you solve Fancy Sequence on LeetCode?

Fancy Sequence (LeetCode #1622) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Using cumulative operations reduces the need for repeated iterations over the sequence.

What is the brute force solution for Fancy Sequence?

The brute force approach for Fancy Sequence is Brute Force, with O(n²) time complexity and O(n) space complexity. The brute force approach involves directly applying each operation to the sequence, modifying the entire list for each operation. This is straightforward but inefficient, especially for larger sequences. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Fancy Sequence?

Fancy Sequence uses the following concepts: Math, Design, Segment Tree, Number Theory. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Fancy Sequence?

Explain your thought process clearly while coding. Discuss the trade-offs of different approaches before diving into code. Practice dry runs of your code with sample inputs to catch errors early.

What are common mistakes when solving Fancy Sequence?

Failing to account for the cumulative effects of addAll and multAll. Not using modular arithmetic correctly, leading to overflow.

#1622

Fancy Sequence

Hard

MathDesignSegment TreeNumber TheoryHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(n)	O(n)

💡

Intuition

Time O(n)Space O(n)

The optimal solution uses two variables to track cumulative additions and multiplications, allowing operations to be applied in constant time, rather than modifying the entire sequence each time.

⚙️

Algorithm

5 steps

1Step 1: Initialize an empty list to store the sequence and variables for cumulative add and multiply.
2Step 2: For 'append', store the adjusted value considering the current add and multiply.
3Step 3: For 'addAll', simply update the cumulative add variable.
4Step 4: For 'multAll', update both the cumulative add and multiply variables.
5Step 5: For 'getIndex', compute the current value using the stored sequence and the cumulative adjustments.

solution.py20 lines

1class Fancy:
2    def __init__(self):
3        self.sequence = []
4        self.add = 0
5        self.mul = 1
6
7    def append(self, val: int) -> None:
8        self.sequence.append((val - self.add) // self.mul)
9
10    def addAll(self, inc: int) -> None:
11        self.add += inc
12
13    def multAll(self, m: int) -> None:
14        self.add *= m
15        self.mul *= m
16
17    def getIndex(self, idx: int) -> int:
18        if idx >= len(self.sequence):
19            return -1
20        return (self.sequence[idx] * self.mul + self.add) % (10**9 + 7)

ℹ

Complexity note: The time complexity is O(n) for the append operation, but all other operations (addAll, multAll, getIndex) are O(1). This allows us to efficiently manage the sequence without modifying it directly for each operation.

1Using cumulative operations reduces the need for repeated iterations over the sequence.
2Understanding modular arithmetic is crucial for handling large numbers.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.