LeetCode 658: Find K Closest Elements Solution in Python – A Step-by-Step Guide

What Is LeetCode 658: Find K Closest Elements?

In LeetCode 658: Find K Closest Elements, you’re given a sorted integer array arr (ascending order), an integer k, and a target value x. Your task is to find the k elements in arr closest to x, returning them in a list sorted in ascending order. Closeness is measured by absolute difference (|a - x|), and if two elements are equally distant, the smaller one is preferred (due to ascending order requirement). For example, with arr = [1, 2, 3, 4, 5], k = 4, and x = 3, the closest elements are [1, 2, 3, 4]. Since the array is sorted, we can leverage this property for efficiency. This problem tests your ability to optimize searches in a sorted structure.

Problem Statement

• Input:
• arr: Sorted list of integers (ascending).
• k: Integer (number of elements to find).
• x: Integer (target value).
• Output: A list of k integers—closest elements to x, sorted ascending.
• Rules:
• Closeness = |element - x|.
• If equal distance, prefer smaller element.
• Return k elements in ascending order.

Constraints

• 1 ≤ k ≤ arr.length ≤ 10⁴.
• -10⁴ ≤ arr[i], x ≤ 10⁴.
• arr is sorted in ascending order.

Examples

• Input: arr = [1, 2, 3, 4, 5], k = 4, x = 3
• Output: [1, 2, 3, 4] (Closest to 3)
• Input: arr = [1, 2, 3, 4, 5], k = 4, x = -1
• Output: [1, 2, 3, 4] (Leftmost 4 elements)
• Input: arr = [1, 1, 1, 10, 10, 10], k = 1, x = 9
• Output: [10] (Closest to 9, smallest if tied)

These examples set the array—let’s solve it!

Understanding the Problem: Finding Closest Elements

To solve LeetCode 658: Find K Closest Elements in Python, we need to find k elements in a sorted array arr closest to a target x, returning them in ascending order. A brute-force approach—computing distances for all elements and sorting—would be O(n log n), inefficient for n ≤ 10⁴. Since the array is sorted, we can optimize by finding the optimal starting point of a k-length window. We’ll use:

• Best Solution (Binary Search with Sliding Window): O(log n + k) time, O(1) space—fast and efficient.
• Alternative Solution (Sorting with Heap): O(n log k) time, O(k) space—simpler but slower.

Let’s start with the binary search solution, breaking it down for beginners.

Best Solution: Binary Search with Sliding Window

Why This Is the Best Solution

The binary search with sliding window approach is the top choice for LeetCode 658 because it’s highly efficient—O(log n + k) time with O(1) space—and leverages the sorted nature of the array to quickly find the leftmost index of the optimal k-length window, then extracts the result in order. It fits constraints (n ≤ 10⁴) perfectly and is elegant once you grasp the binary search logic. It’s like pinpointing the perfect window and sliding it into place!

How It Works

Think of this as a window finder:

• Step 1: Define Window:
• We need a contiguous k-length subarray.
• Goal: Find left index left where arr[left:left+k] is closest to x.
• Step 2: Binary Search:
• Search range [0, n-k] for left.
• Compare |arr[left] - x| vs |arr[left+k] - x|.
• If arr[left] is farther, move left up (left too small).
• If arr[left+k] is farther, move right down (left too big).
• Converge to optimal left.
• Step 3: Extract Result:
• Return arr[left:left+k] (already sorted).
• Step 4: Return Result:
• List of k elements.

It’s like a window slider—search and grab!

Step-by-Step Example

Example: arr = [1, 2, 3, 4, 5], k = 4, x = 3

• Step 1: Window size = 4, n = 5, range = [0, 1].
• Step 2: Binary search:
• Left = 0, Right = 1, Mid = 0.
• Check: |arr[0] - 3| = 2 vs |arr[4] - 3| = 2.
• Equal, but need leftmost, try left = 0.
• Mid = 1: |arr[1] - 3| = 1 vs |arr[5] - 3| = inf (out of bounds), left = 0.
• Converge: left = 0 (distances: 2, 1, 0, 1 vs 1, 0, 1, 2).
• Step 3: Extract: arr[0:4] = [1, 2, 3, 4].
• Output: [1, 2, 3, 4].

Code with Detailed Line-by-Line Explanation

Here’s the Python code, explained clearly:

class Solution:
    def findClosestElements(self, arr: List[int], k: int, x: int) -> List[int]:
        # Step 1: Binary search for leftmost index of k-window
        left = 0
        right = len(arr) - k

        while left < right:
            mid = (left + right) // 2
            # Compare distances at window edges
            left_dist = x - arr[mid]
            right_dist = arr[mid + k] - x

            if left_dist > right_dist:
                left = mid + 1  # Move left up if left is farther
            else:
                right = mid     # Move right down if right is farther or equal

        # Step 2: Extract k elements from left index
        return arr[left:left + k]

• Lines 4-6: Init: Search range for left index of k-window.
• Lines 8-15: Binary search:
• Compute mid, compare distances at mid and mid+k.
• Adjust left/right based on which is farther.
• Line 18: Extract: Return k elements from optimal left.
• Time Complexity: O(log n + k)—binary search O(log n), extraction O(k).
• Space Complexity: O(1)—no extra space beyond output.

This is like a window spotter—search and slide!

Alternative Solution: Sorting with Heap

Why an Alternative Approach?

The sorting with heap approach uses a min-heap—O(n log k) time, O(k) space. It’s simpler conceptually, sorting all elements by distance and picking the top k, but less efficient due to heap operations. It’s like gathering all treasures and selecting the closest ones!

How It Works

Picture this as a treasure sorter:

• Step 1: Build heap of (distance, value) pairs.
• Step 2: Extract k smallest distances.
• Step 3: Sort result ascending.
• Step 4: Return list.

It’s like a heap picker—sort and grab!

Step-by-Step Example

Example: Same as above

• Step 1: Heap = [(2, 1), (1, 2), (0, 3), (1, 4), (2, 5)].
• Step 2: Extract k=4: [(0, 3), (1, 2), (1, 4), (2, 1)].
• Step 3: Sort by value: [1, 2, 3, 4].
• Output: [1, 2, 3, 4].

Code for Heap Approach

import heapq

class Solution:
    def findClosestElements(self, arr: List[int], k: int, x: int) -> List[int]:
        # Step 1: Build heap of (distance, value)
        heap = [(abs(num - x), num) for num in arr]
        heapq.heapify(heap)

        # Step 2: Extract k smallest distances
        result = [heapq.heappop(heap)[1] for _ in range(k)]

        # Step 3: Sort result ascending
        return sorted(result)

• Line 1: Import heapq for min-heap.
• Lines 6-7: Build heap with (distance, value).
• Line 10: Extract k elements.
• Line 13: Sort for ascending order.
• Time Complexity: O(n log k)—heapify O(n), k extractions O(k log n), sort O(k log k).
• Space Complexity: O(k)—heap size.

It’s a heap sorter—gather and pick!

Comparing the Two Solutions

• Binary Search (Best):
• Pros: O(log n + k) time, O(1) space, fast and minimal.
• Cons: Binary search logic less obvious.
• Heap (Alternative):
• Pros: O(n log k) time, O(k) space, conceptually simple.
• Cons: Slower, more space.

Binary search wins for efficiency.

Additional Examples and Edge Cases

• Input: arr = [1], k = 1, x = 0
• Output: [1].
• Input: arr = [1, 2, 3], k = 2, x = 2
• Output: [1, 2].

Both handle these well.

Complexity Breakdown

• Binary Search: Time O(log n + k), Space O(1).
• Heap: Time O(n log k), Space O(k).

Binary search is optimal.

Key Takeaways

• Binary Search: Window precision—smart!
• Heap: Heap simplicity—clear!
• Closeness: Searching is fun.
• Python Tip: Binary search rocks—see Python Basics.

Final Thoughts: Find Those Elements

LeetCode 658: Find K Closest Elements in Python is a fun search challenge. Binary search with sliding window offers speed and elegance, while heap sorting provides a clear alternative. Want more? Try LeetCode 1: Two Sum or LeetCode 215: Kth Largest Element. Ready to search? Head to Solve LeetCode 658 on LeetCode and find those closest elements today!