Fancy Sequence Solution in Java

Ah, the Fancy Sequence problem! It’s like trying to impress your friends with your cooking skills, but instead of a delicious meal, you’re serving them a sequence of numbers that can be manipulated in various ways. Imagine you’re at a party, and you want to show off your fancy new recipe. You can add ingredients (or numbers) to your dish, multiply them, and even adjust the flavors (or values) on the fly. But wait! What if you want to retrieve the original taste of your dish after all that mixing? That’s where the Fancy Sequence comes in, allowing you to perform operations on a sequence of numbers while keeping track of the original values.

Problem Description

In this problem, you are tasked with implementing a class called Fancy that supports the following operations:

Append a value to the sequence.
Add a constant to all values in the sequence.
Multiply all values in the sequence by a constant.
Get the value at a specific index in the sequence.

The challenge lies in efficiently managing these operations without losing track of the original values, especially after multiple additions and multiplications.

Code Solution


class Fancy {
    public void append(int val) {
        final long x = (val - b + kMod) % kMod;
        vals.add(x * modPow(a, kMod - 2) % kMod);
    }

    public void addAll(int inc) {
        b = (b + inc) % kMod;
    }

    public void multAll(int m) {
        a = (a * m) % kMod;
        b = (b * m) % kMod;
    }

    public int getIndex(int idx) {
        return idx >= vals.size() ? -1 : (int) ((a * vals.get(idx) + b) % kMod);
    }

    private static final int kMod = 1_000_000_007;
    private List vals = new ArrayList<>();
    private long a = 1;
    private long b = 0;

    private int modPow(long x, long n) {
        if (n == 0)
            return 1;
        if (n % 2 == 1)
            return (int) (x * modPow(x % kMod, (n - 1)) % kMod);
        return modPow(x * x % kMod, (n / 2)) % kMod;
    }
}

Approach Explanation

The approach taken in this solution revolves around maintaining two variables, a and b, which represent the current state of the sequence. When you append a value, it’s transformed based on the current a and b values. The addAll and multAll methods adjust these variables accordingly, allowing for efficient updates without needing to modify the entire sequence. The getIndex method then retrieves the value at a specific index by applying the transformations back to the stored values.

Time and Space Complexity

Operation	Time Complexity	Space Complexity
append	O(1)	O(n)
addAll	O(1)	O(1)
multAll	O(1)	O(1)
getIndex	O(1)	O(1)

Real-World Example

Imagine you’re a barista at a coffee shop. You have a special drink that can be customized with different flavors (append), sweeteners (addAll), and sizes (multAll). Each time a customer orders, you adjust the recipe based on their preferences. However, if they want to know what the original drink tasted like before all those modifications, you need a way to backtrack. The Fancy Sequence is like your secret recipe book that allows you to keep track of all these changes while still being able to serve the original drink if requested.