Problem Description
Given a sorted array of integers, our goal is to find k elements in the array that are closest to a target value x. If two numbers are equally close, the smaller element is preferred. The final answer should be returned in ascending order.
Key Insights
- The input array is already sorted.
- A binary search can be used to efficiently narrow down the potential starting position.
- A two-pointers or sliding window approach is effective for selecting k consecutive elements.
- The comparison of absolute differences (with a tie-breaker on values) is crucial.
- Maintaining ascending order simplifies the final output.
Space and Time Complexity
Time Complexity: O(log n + k), where O(log n) for binary search and O(k) for collecting results. Space Complexity: O(k), for storing the output.
Solution
We can solve this problem by first using binary search to identify the left-most window where the k closest elements might start. The binary search operates over the possible starting indices for windows of size k (from 0 to n-k). For each middle index mid, we compare the distances between x and arr[mid] and arr[mid + k] to decide whether to move the window to the right or left. Once the correct starting position is found, return the k elements starting from that index. This approach leverages the sorted property of the array to efficiently converge on the optimal window.