Leetcode #1622: Fancy Sequence
In this guide, we solve Leetcode #1622 Fancy Sequence in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
Write an API that generates fancy sequences using the append, addAll, and multAll operations. Implement the Fancy class: Fancy() Initializes the object with an empty sequence.
Quick Facts
- Difficulty: Hard
- Premium: No
- Tags: Design, Segment Tree, Math
Intuition
There is a mathematical invariant or formula that directly leads to the result.
Using math avoids unnecessary loops and reduces complexity.
Approach
Derive the formula or update rule, then compute the answer directly.
Handle edge cases like overflow or zero carefully.
Steps:
- Identify the math relationship.
- Compute the result with a loop or formula.
- Handle edge cases.
Example
Input
["Fancy", "append", "addAll", "append", "multAll", "getIndex", "addAll", "append", "multAll", "getIndex", "getIndex", "getIndex"]
[[], [2], [3], [7], [2], [0], [3], [10], [2], [0], [1], [2]]
Output
[null, null, null, null, null, 10, null, null, null, 26, 34, 20]
Explanation
Fancy fancy = new Fancy();
fancy.append(2); // fancy sequence: [2]
fancy.addAll(3); // fancy sequence: [2+3] -> [5]
fancy.append(7); // fancy sequence: [5, 7]
fancy.multAll(2); // fancy sequence: [5*2, 7*2] -> [10, 14]
fancy.getIndex(0); // return 10
fancy.addAll(3); // fancy sequence: [10+3, 14+3] -> [13, 17]
fancy.append(10); // fancy sequence: [13, 17, 10]
fancy.multAll(2); // fancy sequence: [13*2, 17*2, 10*2] -> [26, 34, 20]
fancy.getIndex(0); // return 26
fancy.getIndex(1); // return 34
fancy.getIndex(2); // return 20
Python Solution
MOD = int(1e9 + 7)
class Node:
def __init__(self, l, r):
self.left = None
self.right = None
self.l = l
self.r = r
self.mid = (l + r) >> 1
self.v = 0
self.add = 0
self.mul = 1
class SegmentTree:
def __init__(self):
self.root = Node(1, int(1e5 + 1))
def modifyAdd(self, l, r, inc, node=None):
if l > r:
return
if node is None:
node = self.root
if node.l >= l and node.r <= r:
node.v = (node.v + (node.r - node.l + 1) * inc) % MOD
node.add += inc
return
self.pushdown(node)
if l <= node.mid:
self.modifyAdd(l, r, inc, node.left)
if r > node.mid:
self.modifyAdd(l, r, inc, node.right)
self.pushup(node)
def modifyMul(self, l, r, m, node=None):
if l > r:
return
if node is None:
node = self.root
if node.l >= l and node.r <= r:
node.v = (node.v * m) % MOD
node.add = (node.add * m) % MOD
node.mul = (node.mul * m) % MOD
return
self.pushdown(node)
if l <= node.mid:
self.modifyMul(l, r, m, node.left)
if r > node.mid:
self.modifyMul(l, r, m, node.right)
self.pushup(node)
def query(self, l, r, node=None):
if l > r:
return 0
if node is None:
node = self.root
if node.l >= l and node.r <= r:
return node.v
self.pushdown(node)
v = 0
if l <= node.mid:
v = (v + self.query(l, r, node.left)) % MOD
if r > node.mid:
v = (v + self.query(l, r, node.right)) % MOD
return v
def pushup(self, node):
node.v = (node.left.v + node.right.v) % MOD
def pushdown(self, node):
if node.left is None:
node.left = Node(node.l, node.mid)
if node.right is None:
node.right = Node(node.mid + 1, node.r)
left, right = node.left, node.right
if node.add != 0 or node.mul != 1:
left.v = (left.v * node.mul + (left.r - left.l + 1) * node.add) % MOD
right.v = (right.v * node.mul + (right.r - right.l + 1) * node.add) % MOD
left.add = (left.add * node.mul + node.add) % MOD
right.add = (right.add * node.mul + node.add) % MOD
left.mul = (left.mul * node.mul) % MOD
right.mul = (right.mul * node.mul) % MOD
node.add = 0
node.mul = 1
class Fancy:
def __init__(self):
self.n = 0
self.tree = SegmentTree()
def append(self, val: int) -> None:
self.n += 1
self.tree.modifyAdd(self.n, self.n, val)
def addAll(self, inc: int) -> None:
self.tree.modifyAdd(1, self.n, inc)
def multAll(self, m: int) -> None:
self.tree.modifyMul(1, self.n, m)
def getIndex(self, idx: int) -> int:
return -1 if idx >= self.n else self.tree.query(idx + 1, idx + 1)
# Your Fancy object will be instantiated and called as such:
# obj = Fancy()
# obj.append(val)
# obj.addAll(inc)
# obj.multAll(m)
# param_4 = obj.getIndex(idx)
Complexity
The time complexity is O(n) or O(1). The space complexity is O(1).
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.