> ## Documentation Index
> Fetch the complete documentation index at: https://leetcode-py.wisl.dev/llms.txt
> Use this file to discover all available pages before exploring further.

> ## Agent Instructions
> leetcode-py is a Python LeetCode practice environment generator with one CLI: lcpy. It is not a service or platform.
> Each problem is a directory under leetcode/ with README.md, solution.py, test_solution.py, helpers.py, and playground.ipynb. lcpy gen creates them from JSON templates bundled with the package.
> Examples are backed by tests; copy them verbatim.

# Time Based Key-Value Store Python Solution

> Tested Python solution for LeetCode 981 with 12 pytest cases. Generate a practice environment with lcpy.

LeetCode 981, Medium. Topics: Hash Table, String, Binary Search, Design. [View on LeetCode](https://leetcode.com/problems/time-based-key-value-store/description/).

Generate this problem as a practice environment: tested reference solution, 12 [parametrized pytest cases](/practice/testing), and a playground notebook:

```bash theme={"theme":{"light":"github-light","dark":"github-dark"}}
lcpy gen -n 981   # by problem number
lcpy gen -s time_based_key_value_store   # by problem name
```

## Problem

Design a time-based key-value data structure that can store multiple values for the same key at different time stamps and retrieve the key's value at a certain timestamp.

Implement the `TimeMap` class:

* `TimeMap()` Initializes the object of the data structure.
* `void set(String key, String value, int timestamp)` Stores the key `key` with the value `value` at the given time `timestamp`.
* `String get(String key, int timestamp)` Returns a value such that `set` was called previously, with `timestamp_prev <= timestamp`. If there are multiple such values, it returns the value associated with the largest `timestamp_prev`. If there are no values, it returns `""`.

### Examples

```
Input
["TimeMap", "set", "get", "get", "set", "get", "get"]
[[], ["foo", "bar", 1], ["foo", 1], ["foo", 3], ["foo", "bar2", 4], ["foo", 4], ["foo", 5]]
Output
[null, null, "bar", "bar", null, "bar2", "bar2"]
```

**Explanation:**

```
TimeMap timeMap = new TimeMap();
timeMap.set("foo", "bar", 1);  // store the key "foo" and value "bar" along with timestamp = 1.
timeMap.get("foo", 1);         // return "bar"
timeMap.get("foo", 3);         // return "bar", since there is no value corresponding to foo at timestamp 3 and timestamp 2, then the only value is at timestamp 1 is "bar".
timeMap.set("foo", "bar2", 4); // store the key "foo" and value "bar2" along with timestamp = 4.
timeMap.get("foo", 4);         // return "bar2"
timeMap.get("foo", 5);         // return "bar2"
```

### Constraints

* `1 <= key.length, value.length <= 100`
* `key` and `value` consist of lowercase English letters and digits.
* `1 <= timestamp <= 10^7`
* All the timestamps `timestamp` of `set` are strictly increasing.
* At most `2 * 10^5` calls will be made to `set` and `get`.

## Solution

Reference implementation from [solution.py on GitHub](https://github.com/wislertt/leetcode-py/blob/main/leetcode/time_based_key_value_store/solution.py), full suite in [test\_solution.py](https://github.com/wislertt/leetcode-py/blob/main/leetcode/time_based_key_value_store/test_solution.py):

```python theme={"theme":{"light":"github-light","dark":"github-dark"}}
class TimeMap:
    # Time: O(1)
    # Space: O(n)
    def __init__(self) -> None:
        self.store: dict[str, list[tuple[int, str]]] = {}

    # Time: O(1)
    # Space: O(1)
    def set(self, key: str, value: str, timestamp: int) -> None:
        if key not in self.store:
            self.store[key] = []
        self.store[key].append((timestamp, value))

    # Time: O(log n)
    # Space: O(1)
    def get(self, key: str, timestamp: int) -> str:
        if key not in self.store:
            return ""

        values = self.store[key]
        left, right = 0, len(values) - 1
        result = ""

        while left <= right:
            mid = (left + right) // 2
            if values[mid][0] <= timestamp:
                result = values[mid][1]
                left = mid + 1
            else:
                right = mid - 1

        return result
```

## Complexity

| Time | Space |
| ---- | ----- |
| O(1) | O(n)  |

## Tags

[Grind 75](/catalog/grind-75), [Grind](/catalog/grind), [NeetCode 150](/catalog/neetcode-150), [NeetCode 250](/catalog/neetcode-250), [NeetCode All](/catalog/neetcode).