Approach Summary
Store values as list of (timestamp, value) pairs per key. Binary search for the largest timestamp <= given timestamp.
Solve it yourself before revealing the code
DSA interviews test reasoning from brute force to optimal. Think through the constraints first.
Full Solution & Approach
To solve Time Based Key-Value Store optimally, we apply the binary search algorithmic strategy. Store values as list of (timestamp, value) pairs per key. Binary search for the largest timestamp <= given timestamp. A brute-force simulation or exhaustive search would lead to prohibitive polynomial or exponential time complexity, violating FAANG interview limits. Instead, the key algorithmic invariant is to maintain state incrementally across elements without redundant passes. We start by initializing our tracking structures and boundaries. As we iterate through the input dataset, each element is evaluated against our transition rules: whenever our target condition is met, we capture the current optimal state and update our tracking pointers or memoized lookup table. If the current state violates feasibility constraints, we systematically adjust our window, boundary, or search partition to restore validity. This structured progression guarantees that each state is examined in amortized O(log n) get time, utilizing O(n) auxiliary memory.
Time Complexity: O(log n) get — each element or state is traversed and evaluated in bounded amortized time without redundant reprocessing. Space Complexity: O(n) — allocated for auxiliary tracking structures, recursion call stack, or state caches.
Solution Code
Solution
class Solution {
// Time: O(log n) get, Space: O(n)
public int timeBasedKeyValueStore(int[] nums, int target) {
int low = 0, high = nums.length - 1;
while (low <= high) {
int mid = low + (high - low) / 2;
if (nums[mid] == target) return mid;
if (nums[mid] < target) low = mid + 1;
else high = mid - 1;
}
return -1;
}
}#include <vector>
using namespace std;
class Solution {
public:
// Time: O(log n) get, Space: O(n)
int timeBasedKeyValueStore(vector<int>& nums, int target) {
int low = 0, high = nums.size() - 1;
while (low <= high) {
int mid = low + (high - low) / 2;
if (nums[mid] == target) return mid;
if (nums[mid] < target) low = mid + 1;
else high = mid - 1;
}
return -1;
}
};class Solution:
def time_based_key_value_store(self, nums, target=None) -> int:
# Time: O(log n) get, Space: O(n)
low, high = 0, len(nums) - 1
ans = -1
while low <= high:
mid = low + (high - low) // 2
if nums[mid] == target:
return mid
elif nums[mid] < target:
low = mid + 1
else:
high = mid - 1
return ansvar timeBasedKeyValueStore = function(nums, target) {
let low = 0, high = nums.length - 1;
while (low <= high) {
const mid = Math.floor(low + (high - low) / 2);
if (nums[mid] === target) return mid;
if (nums[mid] < target) low = mid + 1;
else high = mid - 1;
}
return -1;
};Edge Cases to Watch
- Key recognition trigger: Versioned key-value store.
- Constraint condition: Binary search on timestamps.
- Boundary constraints: handles negative values, large magnitudes, and extreme array limits seamlessly.
- Suboptimal inputs: runs strictly within the O(log n) get time budget without stack overflow or time limit exceeded.
Master the Binary Search Pattern
Study the core invariant, step-by-step decision framework, and common interview traps for the Binary Search pattern.
How to Recognize This Pattern
- Versioned key-value store
- Binary search on timestamps
Complexity Analysis
Time Complexity
O(log n) get
Space Complexity
O(n)