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To solve this follow-up, we can modify our space-efficient solution to support range queries efficiently. Here's how we can approach this:

Keep the existing structure (fixed-size array) for space efficiency.
Use binary search to find the start and end points of the range.
Implement the range query method.

Here's the modified implementation:
javaCopyclass TimeMap {
    private static final int N = 5; // Number of recent values to keep per key
    private Map<String, TimestampedValue[]> map;

    private static class TimestampedValue {
        int timestamp;
        String value;

        TimestampedValue(int timestamp, String value) {
            this.timestamp = timestamp;
            this.value = value;
        }
    }

    public TimeMap() {
        map = new HashMap<>();
    }

    // Existing set and get methods remain the same

    public List<String> getRange(String key, int startTime, int endTime) {
        List<String> result = new ArrayList<>();
        TimestampedValue[] values = map.get(key);
        if (values == null) return result;

        int startIndex = findLowerBound(values, startTime);
        int endIndex = findUpperBound(values, endTime);

        for (int i = startIndex; i <= endIndex; i++) {
            if (values[i] != null && values[i].timestamp >= startTime && values[i].timestamp <= endTime) {
                result.add(values[i].value);
            }
        }

        return result;
    }

    private int findLowerBound(TimestampedValue[] values, int timestamp) {
        int left = 0, right = N - 1;
        while (left <= right) {
            int mid = left + (right - left) / 2;
            if (values[mid] == null || values[mid].timestamp < timestamp) {
                left = mid + 1;
            } else {
                right = mid - 1;
            }
        }
        return left;
    }

    private int findUpperBound(TimestampedValue[] values, int timestamp) {
        int left = 0, right = N - 1;
        while (left <= right) {
            int mid = left + (right - left) / 2;
            if (values[mid] == null || values[mid].timestamp <= timestamp) {
                left = mid + 1;
            } else {
                right = mid - 1;
            }
        }
        return right;
    }

    // Existing helper methods (findInsertionIndex, findFloorIndex) remain the same
}
Key points of this solution:

Space Efficiency:

Maintains the fixed-size array (N = 5) for each key, preserving space efficiency.


Time Complexity:

getRange(): O(log N + K), where N is the size of the array (5 in this case) and K is the number of entries within the range.
Binary search is used to find the start and end points: O(log N)
Iterating through the range to collect results: O(K)


Range Query Implementation:

findLowerBound: Finds the index of the first element greater than or equal to startTime.
findUpperBound: Finds the index of the last element less than or equal to endTime.
Iterate through the range to collect values within the specified time range.


Maintaining Existing Operations:

The set() and get() methods remain unchanged, preserving their original time and space complexities.



Trade-offs and Considerations:

The solution maintains the space efficiency of the previous version.
Range queries are efficient for small N, but if N were to increase significantly, we might need to consider more advanced data structures like a segment tree.
This implementation assumes that timestamps are unique for each key. If multiple values could have the same timestamp, we might need to modify the binary search slightly.

In an interview, you could discuss these points and potentially explore how the solution would change if the requirements were modified (e.g., if N were much larger, or if we needed to support even more complex queries).