Fancy Sequence — LeetCode 1622 Python Solution

HardDesignSegment TreeMath
Problem
#1622
Reading time
19 min

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

Python
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

MeasureComplexity
TimeO(n) or O(1)
SpaceO(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.

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