Merge pull request #1133 from vivekvardhan2810/main

Fancy Sequence (Leetcode) Added problem number 1622

Author: ajay-dhangar

dsa-solutions/lc-solutions/1600-1699/1622- Fancy Sequence.md

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+---
+id: fancy-sequence
+title: Fancy Sequence
+sidebar_label: 1622 Fancy Sequence
+tags:
+- Math
+- Design
+- Segment Tree
+description: "This document provides a solution where we need to Write an API that generates fancy sequences using the append, addAll, and multAll operations. "
+---
+
+## 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.
+
+- 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** $10^9$ + $7$. If the index is greater or equal than the length of the sequence, return $-1$.
+
+### Examples
+
+**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 <= 10^5`
+- At most $10^5$ calls total will be made to $append$, $addAll$, $multAll$, and $getIndex$.
+
+---
+
+## Approach
+
+To solve the problem, we need to understand the nature of the allowed moves:
+
+1. **Using Lazy Operations**:
+
+   - Instead of updating the entire sequence for $addAll$ and $multAll$, we maintain cumulative addition and multiplication factors. This way, we can defer the actual updates to when we need to access a particular element.
+
+2. **Tracking Inverse Operations**:
+
+   - When appending a new value, we need to account for the current cumulative operations, so we store values in a form that can be easily adjusted when retrieving them.
+
+3. **Efficient Index Retrieval**:
+
+   - By maintaining cumulative operations and only applying them when accessing an index, we ensure that our operations are efficient and avoid unnecessary updates to the sequence.
+  
+## Solution for Fancy Sequence
+
+The problem involves three types of operations on a sequence of numbers: appending a value, adding a value to all elements, and multiplying all elements by a value. A naive approach would directly modify the sequence for each operation, but this would be inefficient given the constraints. Instead, we can use lazy propagation-like techniques to efficiently handle the operations.
+
+#### Code in Java
+
+```java
+import java.util.ArrayList;
+import java.util.List;
+
+class Fancy {
+    private static final int MOD = 1000000007;
+    private List<Long> sequence;
+    private long add;
+    private long mult;
+
+    public Fancy() {
+        sequence = new ArrayList<>();
+        add = 0;
+        mult = 1;
+    }
+
+    public void append(int val) {
+        // Apply the inverse of the current add and mult to keep the value as its original form
+        long adjustedVal = ((val - add) * modInverse(mult, MOD)) % MOD;
+        if (adjustedVal < 0) adjustedVal += MOD;
+        sequence.add(adjustedVal);
+    }
+
+    public void addAll(int inc) {
+        add = (add + inc) % MOD;
+    }
+
+    public void multAll(int m) {
+        add = (add * m) % MOD;
+        mult = (mult * m) % MOD;
+    }
+
+    public int getIndex(int idx) {
+        if (idx >= sequence.size()) return -1;
+        long result = (sequence.get(idx) * mult + add) % MOD;
+        return (int) result;
+    }
+
+    // Function to compute the modular inverse using Fermat's Little Theorem
+    private long modInverse(long a, int mod) {
+        return power(a, mod - 2, mod);
+    }
+
+    // Function to compute (x^y) % mod
+    private long power(long x, int y, int mod) {
+        if (y == 0) return 1;
+        long p = power(x, y / 2, mod) % mod;
+        p = (p * p) % mod;
+        return (y % 2 == 0) ? p : (x * p) % mod;
+    }
+
+    public static void main(String[] args) {
+        Fancy fancy = new Fancy();
+        fancy.append(2);   
+        fancy.addAll(3);   
+        fancy.append(7);   
+        fancy.multAll(2);  
+        System.out.println(fancy.getIndex(0));
+        fancy.addAll(3);   
+        fancy.append(10);  
+        fancy.multAll(2);  
+        System.out.println(fancy.getIndex(0)); 
+        System.out.println(fancy.getIndex(1)); 
+        System.out.println(fancy.getIndex(2)); 
+    }
+}
+    
+```
+
+### Complexity Analysis
+
+#### Time Complexity: $O(1)$
+
+> **Reason**: Time Complexity is $O(1)$, because append involves adding an element to the list, Whereas addAll only updates a cumulative variable, multAll updates cumulative variables and lastly getIndex, it basically involves getIndex calculating the result using the cumulative variables.
+
+#### Space Complexity: $O(n)$
+
+> **Reason**: $O(n)$, where n is the number of elements in the sequence, as we store each element.
+
+# References
+
+- **LeetCode Problem:** [Fancy Sequence](https://leetcode.com/problems/fancy-sequence/description/)
+- **Solution Link:** [Fancy Sequence Solution on LeetCode](https://leetcode.com/problems/fancy-sequence/solutions/)
+- **Authors LeetCode Profile:** [Vivek Vardhan](https://leetcode.com/u/vivekvardhan43862/)