What is the time complexity of Find K Closest Elements?

The optimal solution has a time complexity of O(log n).

What data structures are used to solve Find K Closest Elements?

This problem uses Array, Two Pointers, Binary Search, Sliding Window, Sorting, Heap (Priority Queue). Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Find K Closest Elements?

Insight 1: The input array is already sorted, allowing us to use binary search to efficiently find the starting index of the k closest elements. Insight 2: We don't need to compare individual elements with 'x'. Instead, we compare two 'windows' of size k. The comparison effectively identifies the region where the k closest elements reside.

Which companies ask LeetCode 658. Find K Closest Elements in interviews?

This problem is frequently asked at Atlassian, Coupang, DoorDash, LinkedIn, Yandex.

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Find K Closest Elements - LeetCode 658 Solution

Q: How to solve Find K Closest Elements?

The solution uses a binary search algorithm to find the starting index of the subarray of size `k` that contains the `k` closest elements to `x`. The binary search operates on the possible starting indices of the subarray. In each iteration, it compares the distance between `x` and the element at the start of the current window (`arr[mid]`) with the distance between `x` and the element just after the current window (`arr[mid + k]`). This comparison determines whether the window should be shifted to the right or the left. Once the binary search converges to the correct starting index, the subarray of size `k` starting from that index is returned.

Find K Closest Elements - Complete Solution Guide

Find K Closest Elements is LeetCode problem 658, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

Given a sorted integer array arr , two integers k and x , return the k closest integers to x in the array. The result should also be sorted in ascending order. An integer a is closer to x than an integer b if: |a - x| < |b - x| , or |a - x| == |b - x| and a < b Example 1: Input: arr = [1,2,3,4,5], k = 4, x = 3 Output: [1,2,3,4] Example 2: Input: arr = [1,1,2,3,4,5], k = 4, x = -1 Output: [1,1,2,3] Constraints: 1 <= k <= arr.length 1 <= arr.length <= 10 4 arr is sorted in ascending order. -10 4 <

Detailed Explanation

The problem asks us to find the `k` integers in a sorted array `arr` that are closest to a given value `x`. Closeness is defined as having the smallest absolute difference with `x`. If two numbers have the same absolute difference with `x`, the smaller number is considered closer. The returned list of `k` closest elements must also be sorted in ascending order. The input includes a sorted array `arr`, an integer `k` representing the number of closest elements to find, and an integer `x` representing the target value.

Solution Approach

The solution uses a binary search algorithm to find the starting index of the subarray of size `k` that contains the `k` closest elements to `x`. The binary search operates on the possible starting indices of the subarray. In each iteration, it compares the distance between `x` and the element at the start of the current window (`arr[mid]`) with the distance between `x` and the element just after the current window (`arr[mid + k]`). This comparison determines whether the window should be shifted to the right or the left. Once the binary search converges to the correct starting index, the subarray of size `k` starting from that index is returned.

Step-by-Step Algorithm

Step 1: Initialize `left` to 0 and `right` to `len(arr) - k`. `left` and `right` define the search space for the starting index of the k-element window.
Step 2: Perform binary search while `left < right`.
Step 3: Calculate the middle index `mid = (left + right) // 2`.
Step 4: Compare `x - arr[mid]` with `arr[mid + k] - x`. This comparison determines if the window starting at `mid` is too far to the left or too far to the right.
Step 5: If `x - arr[mid] > arr[mid + k] - x`, it means the current window is too far to the left. Shift the left boundary to `mid + 1` (`left = mid + 1`).
Step 6: Otherwise, the current window is too far to the right, or it's the optimal window. Shift the right boundary to `mid` (`right = mid`).
Step 7: After the binary search completes, `left` will be the starting index of the `k` closest elements. Return the subarray `arr[left:left + k]`.

Key Insights

Insight 1: The input array is already sorted, allowing us to use binary search to efficiently find the starting index of the k closest elements.
Insight 2: We don't need to compare individual elements with 'x'. Instead, we compare two 'windows' of size k. The comparison effectively identifies the region where the k closest elements reside.
Insight 3: The comparison `x - arr[mid] > arr[mid + k] - x` is crucial for determining whether the 'optimal' window should be shifted to the right. This cleverly avoids having to compute and sort all differences.

Complexity Analysis

Time Complexity: O(log n)

Space Complexity: O(1)

Topics

This problem involves: Array, Two Pointers, Binary Search, Sliding Window, Sorting, Heap (Priority Queue).

Companies

Asked at: Atlassian, Coupang, DoorDash, LinkedIn, Yandex.

Find K Closest Elements | LeetCode 658

Medium

Array Two Pointers Binary Search Sliding Window Sorting Heap (Priority Queue)

Time: O(log n)Space: O(1)

Solution Code

Available in:

import java.util.ArrayList;
import java.util.List;

class Solution {
    public List<Integer> findClosestElements(int[] arr, int k, int x) {
        int left = 0;
        int right = arr.length - k;

        while (left < right) {
            int mid = (left + right) / 2;
            if (x - arr[mid] > arr[mid + k] - x) {
                left = mid + 1;
            } else {
                right = mid;
            }
        }

        List<Integer> result = new ArrayList<>();
        for (int i = left; i < left + k; i++) {
            result.add(arr[i]);
        }
        return result;
    }
}

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Problem Info

Description

Problem 658: Find K Closest Elements - Detailed explanation and solution approaches.

Topics

Array Two Pointers Binary Search Sliding Window Sorting Heap (Priority Queue)

Companies

AtlassianCoupangDoorDashLinkedInYandex

Stats

Difficulty:Medium

Languages:4

Companies:5

Problem #:658

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

7 steps

1
Step 1: Initialize `left` to 0 and `right` to `len(arr) - k`. `left` and `right` define the search space for the starting index of the k-element window.
2
Step 2: Perform binary search while `left < right`.
3
Step 3: Calculate the middle index `mid = (left + right) // 2`.
4
Step 4: Compare `x - arr[mid]` with `arr[mid + k] - x`. This comparison determines if the window starting at `mid` is too far to the left or too far to the right.
5
Step 5: If `x - arr[mid] > arr[mid + k] - x`, it means the current window is too far to the left. Shift the left boundary to `mid + 1` (`left = mid + 1`).
6
Step 6: Otherwise, the current window is too far to the right, or it's the optimal window. Shift the right boundary to `mid` (`right = mid`).
7
Step 7: After the binary search completes, `left` will be the starting index of the `k` closest elements. Return the subarray `arr[left:left + k]`.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(log n)

🚀 Logarithmic - Very efficient! Halves the problem each step.

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The dominant operation is the binary search, which has a time complexity of O(log N) where N is the range of possible starting indices for the window of size k, which is arr.length - k + 1. This simplifies to O(log n) as k is at most the length of arr. The space complexity is O(1) because we are modifying the array in place using window based approaches (except Java/C which allocate the final list).

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Sorting with Lambda

Sort the entire array based on the absolute difference from `x` and then select the first `k` elements. Sort the selected elements.

Time: O(n log n) for sorting the entire array.Space: O(n) in some implementations due to sorting algorithms.

Trade-offs: Simpler to implement, but less efficient than the binary search approach, especially for large arrays. Use this if the array size is small, and readability is prioritized over performance.

Heap (Priority Queue)

Use a max-heap to store the `k` closest elements seen so far. Iterate through the array and for each element, compare its distance to `x` with the distance of the farthest element in the heap. If the current element is closer, replace the farthest element in the heap. Finally, extract the elements from the heap and sort them.

Time: O(n log k) since inserting into a heap of size k takes log k time, and we do this for each of the n elements. The sorting step is O(k log k), so the overall time complexity is dominated by O(n log k).Space: O(k) to store the heap.

Trade-offs: Useful if the input is a stream of data, as the algorithm can maintain the `k` closest elements as new elements are added. More complex to implement than the sorting approach.

✨

Key Insights

1Insight 1: The input array is already sorted, allowing us to use binary search to efficiently find the starting index of the k closest elements.
2Insight 2: We don't need to compare individual elements with 'x'. Instead, we compare two 'windows' of size k. The comparison effectively identifies the region where the k closest elements reside.
3Insight 3: The comparison `x - arr[mid] > arr[mid + k] - x` is crucial for determining whether the 'optimal' window should be shifted to the right. This cleverly avoids having to compute and sort all differences.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Two Pointers

Using two indices to traverse data from different positions simultaneously.

Binary Search

Halving the search space each step. Like finding a word in a dictionary!

Sliding Window

A moving "window" over data to process subsets efficiently.

Sorting

Arranging elements in a specific order (ascending/descending).

Edge Cases

Edge case 1: `k` is equal to the length of the array. In this case, the entire array is the answer, and the binary search will still correctly converge to the starting index 0.
Edge case 2: `x` is smaller than all elements in the array. The binary search will converge to the beginning of the array, correctly selecting the first `k` elements.
Edge case 3: `x` is larger than all elements in the array. The binary search will converge to the end of the array, again resulting in the k largest elements being returned (which are the closest).

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not handling the case where two numbers have the same absolute difference with `x`. The tie-breaking rule (smaller number is closer) must be correctly implemented, which this binary search does by implicitly favoring the left window when the distances are equal.

✗Common mistake 2: Not using binary search and instead trying to iterate through the entire array to find the `k` closest elements. This leads to a time complexity of O(n), which is less efficient than O(log n) when using binary search.

✗Common mistake 3: Not sorting the final result. The problem statement requires the output to be sorted, so after obtaining the k closest elements, it must be sorted to meet the output requirement if the method used does not maintain order.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Recommendation systems. Finding the `k` most relevant items to a user based on a similarity score (e.g., collaborative filtering). In cases where similarity scores are partially sorted or can be searched efficiently, this algorithm can be adapted.

🏦Real-world scenario 2: Data analysis and anomaly detection. Identifying the `k` data points closest to a particular value or distribution. For example, in financial analysis, finding the `k` most similar stock price patterns to a target pattern.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

If stuck, try working through a small example by hand. The pattern often reveals itself.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: September 27, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 13

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 6

❓

Frequently Asked Questions

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time complexity.

When is the two-pointer technique most effective?

Two pointers work best on sorted arrays or when searching for pairs/triplets that satisfy a condition. Common patterns include: moving pointers inward from both ends (like in two-sum variants), fast/slow pointers for cycle detection, or left/right pointers for partitioning.

What techniques are commonly tested in medium-difficulty problems?

Medium problems often require combining multiple techniques: two-pointer with hash maps, BFS/DFS with state tracking, dynamic programming with optimization, or divide-and-conquer approaches. They test your ability to recognize patterns and choose the right combination of techniques.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Find K Closest Elements"?

For this medium-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether two pointers techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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