grid47 Exploring patterns and algorithms
Sep 2, 2024
5 min read
Solution to LeetCode 658: Find K Closest Elements Problem
Given a sorted integer array and two integers k and x, return the k closest integers to x in the array, sorted in ascending order. If two integers are equally close to x, the smaller integer should be preferred.
Problem
Approach
Steps
Complexity
Input: The input consists of a sorted array of integers arr, and two integers k and x.
Example: arr = [10, 20, 30, 40, 50], k = 3, x = 35
Constraints:
• 1 <= k <= arr.length
• 1 <= arr.length <= 10^4
• arr is sorted in ascending order.
• -10^4 <= arr[i], x <= 10^4
Output: The output is a sorted array of the k integers closest to x from the input array arr.
Example: [30, 40, 50]
Constraints:
• The output array will contain exactly k integers.
Goal: The goal is to find the k closest integers to x by comparing the absolute differences of each integer in the array to x, and selecting the k smallest differences.
Steps:
• 1. Use a binary search to efficiently locate the closest element to x.
• 2. Expand outward from the closest element to gather k elements (both left and right).
• 3. Sort the selected k elements and return them.
Goal: The input array is sorted, and the number of elements is between 1 and 10^4. The integer values and x are between -10^4 and 10^4.
Steps:
• 1 <= arr.length <= 10^4
• 1 <= k <= arr.length
• -10^4 <= arr[i], x <= 10^4
Assumptions:
• The input array is always sorted in ascending order.
• The value of k will always be valid (1 <= k <= arr.length).
• Input: arr = [5, 10, 15, 20, 25, 30], k = 2, x = 12
• Explanation: The closest integers to 12 are 10 and 15. Sorted, the result is [10, 15].
Approach: This problem can be solved using binary search to quickly find the nearest element to x, followed by expanding to collect the k closest integers.
Observations:
• Binary search can help quickly locate the closest integer to x in the sorted array.
• Once the closest element is found, we can expand outward to select the k closest integers.
• Efficiently finding the closest element using binary search can save time in large arrays. Afterward, we just need to compare adjacent elements to gather the k closest integers.
Steps:
• 1. Use binary search to find the position of the element closest to x.
• 2. Initialize two pointers, one to the left and one to the right of the closest element, to select the k closest integers.
• 3. Sort the selected integers and return them as the result.
Empty Inputs:
• The input array will never be empty as per the problem constraints.
Large Inputs:
• For large arrays (up to 10^4 elements), binary search ensures the solution remains efficient.
Special Values:
• The array can contain both negative and positive integers, and edge cases where all elements are either greater or less than x should be considered.
Constraints:
• Ensure that the selected k elements are always the closest ones to x, and handle edge cases where the array contains values that are equally close to x.
vector<int> findClosestElements(vector<int>& arr, int k, int x) {
int l =0, r = arr.size() -k;
while(l < r) {
int m = (l + r) /2;
if(x - arr[m] > arr[m + k] - x) {
l = m +1;
} else r = m;
}
return vector<int>(arr.begin() + l, arr.begin() + l + k);
}
1 : Function Definition
vector<int> findClosestElements(vector<int>& arr, int k, int x) {
This is the function definition for `findClosestElements`, which takes an array `arr`, an integer `k`, and a target value `x`. It returns a vector containing the `k` closest elements to `x` from the sorted array.
2 : Variable Initialization
int l =0, r = arr.size() - k;
Initialize two pointers: `l` starts at 0 (left boundary), and `r` is set to the index of the last valid element that can form a window of size 'k'.
3 : Binary Search Loop
while(l < r) {
A binary search loop is started to find the best position for the `k` closest elements. The loop continues until the left pointer is no longer less than the right pointer.
4 : Calculate Midpoint
int m = (l + r) /2;
Calculate the middle index `m` as the average of the left (`l`) and right (`r`) pointers.
5 : Compare Differences
if(x - arr[m] > arr[m + k] - x) {
Compare the distance from `x` to the current element `arr[m]` with the distance from `x` to the element `arr[m + k]`. This is used to decide whether to move the left or right pointer.
6 : Update Left Pointer
l = m +1;
If the distance to `arr[m]` is greater, it means the closest `k` elements are to the right, so we move the left pointer `l` to `m + 1`.
7 : Update Right Pointer
} else r = m;
Otherwise, the closest `k` elements are either at or before the midpoint, so we move the right pointer `r` to `m`.
8 : Return Result
return vector<int>(arr.begin() + l, arr.begin() + l + k);
The function returns a vector containing the `k` closest elements, starting from index `l` to `l + k` in the original array.
Best Case: O(log(n) + k)
Average Case: O(log(n) + k)
Worst Case: O(log(n) + k)
Description: The time complexity is O(log(n) + k) due to the binary search and the subsequent expansion to gather k elements.
Best Case: O(k)
Worst Case: O(k)
Description: The space complexity is O(k) due to the storage required for the k closest integers.
