Fancy Sequence — LeetCode 1622 Python Solution
- Problem
- #1622
- Pattern
- Math and Number Theory
- Reading time
- 19 min
- Source
- leetcode.com
The problem
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.
Example
- Input
- ["Fancy", "append", "addAll", "append", "multAll", "getIndex", "addAll", "append", "multAll", "getIndex", "getIndex", "getIndex"]
- Output
- [null, null, null, null, null, 10, null, null, null, 26, 34, 20]
- Explanation
- Fancy fancy = new Fancy();
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
| Measure | Complexity |
|---|---|
| Time | O(n) or O(1) |
| Space | O(1) auxiliary |
Pattern: Math and Number Theory
Find the closed form, the invariant, or the modular identity — and skip the loop entirely. LeetCode 1622. Fancy Sequence is filed here on both counts: the reference solution below belongs to the algorithm family this hub collects, and LeetCode tags it Math.
The math and number theory guide has the Python template for the pattern and the 485 LeetCode problems that use it.
Related problems
Frequently asked questions
- How hard is LeetCode 1622. Fancy Sequence?
- LeetCode 1622. Fancy Sequence is rated Hard on LeetCode.
- What topics does LeetCode 1622. Fancy Sequence cover?
- LeetCode 1622. Fancy Sequence is tagged Design, Segment Tree and Math on LeetCode.