LeetCode #1622 — HARD

Fancy Sequence

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

Solve on LeetCode
The Problem

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.
  • void append(val) Appends an integer val to the end of the sequence.
  • void addAll(inc) Increments all existing values in the sequence by an integer inc.
  • void multAll(m) Multiplies all existing values in the sequence by an integer m.
  • int getIndex(idx) Gets the current value at index idx (0-indexed) of the sequence modulo 109 + 7. If the index is greater or equal than the length of the sequence, return -1.

Example 1:

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

Constraints:

  • 1 <= val, inc, m <= 100
  • 0 <= idx <= 105
  • At most 105 calls total will be made to append, addAll, multAll, and getIndex.
Patterns Used
🏗Design Patterns→📐Segment Tree→

Roadmap

  1. Brute Force Baseline
  2. Core Insight
  3. Algorithm Walkthrough
  4. Edge Cases
  5. Full Annotated Code
  6. Interactive Study Demo
  7. Complexity Analysis
Step 01

Brute Force Baseline

Problem summary: 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. void append(val) Appends an integer val to the end of the sequence. void addAll(inc) Increments all existing values in the sequence by an integer inc. void multAll(m) Multiplies all existing values in the sequence by an integer m. int getIndex(idx) Gets the current value at index idx (0-indexed) of the sequence modulo 109 + 7. If the index is greater or equal than the length of the sequence, return -1.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Math · Design · Segment Tree

Example 1

["Fancy","append","addAll","append","multAll","getIndex","addAll","append","multAll","getIndex","getIndex","getIndex"]
[[],[2],[3],[7],[2],[0],[3],[10],[2],[0],[1],[2]]
Step 02

Core Insight

What unlocks the optimal approach

  • Use two arrays to save the cumulative multipliers at each time point and cumulative sums adjusted by the current multiplier.
  • The function getIndex(idx) ask to the current value modulo 10^9+7. Use modular inverse and both arrays to calculate this value.
Interview move: turn each hint into an invariant you can check after every iteration/recursion step.
Step 03

Algorithm Walkthrough

Iteration Checklist

  1. Define state (indices, window, stack, map, DP cell, or recursion frame).
  2. Apply one transition step and update the invariant.
  3. Record answer candidate when condition is met.
  4. Continue until all input is consumed.
Use the first example testcase as your mental trace to verify each transition.
Step 04

Edge Cases

Minimum Input
Single element / shortest valid input
Validate boundary behavior before entering the main loop or recursion.
Duplicates & Repeats
Repeated values / repeated states
Decide whether duplicates should be merged, skipped, or counted explicitly.
Extreme Constraints
Largest constraint values
Re-check complexity target against constraints to avoid time-limit issues.
Invalid / Corner Shape
Empty collections, zeros, or disconnected structures
Handle special-case structure before the core algorithm path.
Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1622: Fancy Sequence
class Node {
    Node left;
    Node right;
    int l;
    int r;
    int mid;
    long v;
    long add;
    long mul = 1;

    public Node(int l, int r) {
        this.l = l;
        this.r = r;
        this.mid = (l + r) >> 1;
    }
}

class SegmentTree {
    private Node root = new Node(1, (int) 1e5 + 1);
    private static final int MOD = (int) 1e9 + 7;

    public SegmentTree() {
    }

    public void modifyAdd(int l, int r, int inc) {
        modifyAdd(l, r, inc, root);
    }

    public void modifyAdd(int l, int r, int inc, Node node) {
        if (l > r) {
            return;
        }
        if (node.l >= l && node.r <= r) {
            node.v = (node.v + (node.r - node.l + 1) * inc) % MOD;
            node.add = (node.add + inc) % MOD;
            return;
        }
        pushdown(node);
        if (l <= node.mid) {
            modifyAdd(l, r, inc, node.left);
        }
        if (r > node.mid) {
            modifyAdd(l, r, inc, node.right);
        }
        pushup(node);
    }

    public void modifyMul(int l, int r, int m) {
        modifyMul(l, r, m, root);
    }

    public void modifyMul(int l, int r, int m, Node node) {
        if (l > r) {
            return;
        }
        if (node.l >= l && node.r <= r) {
            node.v = (node.v * m) % MOD;
            node.add = (node.add * m) % MOD;
            node.mul = (node.mul * m) % MOD;
            return;
        }
        pushdown(node);
        if (l <= node.mid) {
            modifyMul(l, r, m, node.left);
        }
        if (r > node.mid) {
            modifyMul(l, r, m, node.right);
        }
        pushup(node);
    }

    public int query(int l, int r) {
        return query(l, r, root);
    }

    public int query(int l, int r, Node node) {
        if (l > r) {
            return 0;
        }
        if (node.l >= l && node.r <= r) {
            return (int) node.v;
        }
        pushdown(node);
        int v = 0;
        if (l <= node.mid) {
            v = (v + query(l, r, node.left)) % MOD;
        }
        if (r > node.mid) {
            v = (v + query(l, r, node.right)) % MOD;
        }
        return v;
    }

    public void pushup(Node node) {
        node.v = (node.left.v + node.right.v) % MOD;
    }

    public void pushdown(Node node) {
        if (node.left == null) {
            node.left = new Node(node.l, node.mid);
        }
        if (node.right == null) {
            node.right = new Node(node.mid + 1, node.r);
        }
        if (node.add != 0 || node.mul != 1) {
            Node left = node.left, right = node.right;
            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 {
    private int n;
    private SegmentTree tree = new SegmentTree();

    public Fancy() {
    }

    public void append(int val) {
        ++n;
        tree.modifyAdd(n, n, val);
    }

    public void addAll(int inc) {
        tree.modifyAdd(1, n, inc);
    }

    public void multAll(int m) {
        tree.modifyMul(1, n, m);
    }

    public int getIndex(int idx) {
        return idx >= n ? -1 : tree.query(idx + 1, idx + 1);
    }
}

/**
 * Your Fancy object will be instantiated and called as such:
 * Fancy obj = new Fancy();
 * obj.append(val);
 * obj.addAll(inc);
 * obj.multAll(m);
 * int param_4 = obj.getIndex(idx);
 */
Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Press Step or Run All to begin.
Step 07

Complexity Analysis

Time
O(1) per op
Space
O(n)

Approach Breakdown

NAIVE
O(n) per op time
O(n) space

Use a simple list or array for storage. Each operation (get, put, remove) requires a linear scan to find the target element — O(n) per operation. Space is O(n) to store the data. The linear search makes this impractical for frequent operations.

OPTIMIZED DESIGN
O(1) per op time
O(n) space

Design problems target O(1) amortized per operation by combining data structures (hash map + doubly-linked list for LRU, stack + min-tracking for MinStack). Space is always at least O(n) to store the data. The challenge is achieving constant-time operations through clever structure composition.

Shortcut: Combine two data structures to get O(1) for each operation type. Space is always O(n).
Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.

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