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Java-based LeetCode algorithm problem solutions, regularly updated
1622\. Fancy Sequence
Hard
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`.