K Closest Points to Origin

🎯

K Closest Points to Origin

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Imagine you are building a location-based app that needs to quickly find the nearest stores to a user’s current location.

💡 This problem is about selecting the closest points from a list based on their distance to the origin. Beginners often struggle because they try to sort all points fully or don’t know how to efficiently keep track of only the closest points without sorting everything.

📋

Problem Statement

Given an array of points where points[i] = [x_i, y_i] represents a point on the X-Y plane and an integer k, return the k closest points to the origin (0, 0). The distance between two points on the X-Y plane is the Euclidean distance: sqrt(x^2 + y^2). You may return the answer in any order. The answer is guaranteed to be unique (except for the order that it is in).

1 ≤ points.length ≤ 10^5-10^4 ≤ x_i, y_i ≤ 10^41 ≤ k ≤ points.length

💡

Example

Input"points = [[1,3],[-2,2]], k = 1"

Output[[-2,2]]

The point (-2, 2) is closer to the origin than (1, 3).

Input"points = [[3,3],[5,-1],[-2,4]], k = 2"

Output[[3,3],[-2,4]]

The two closest points to the origin are (3,3) and (-2,4).

• k equals the number of points → return all points
• All points are at the same distance → any k points are valid
• Points with negative coordinates → distance calculation still valid
• k is 1 → return the single closest point

⚠️

Common Mistakes

Using sqrt in distance comparison

Unnecessary computation slows down the solution

✅ Compare squared distances instead of actual distances

Using min heap instead of max heap for k closest points

Heap grows too large or logic fails to keep closest points

✅ Use max heap to keep track of the farthest point among the k closest

Not handling k equal to number of points

Code may fail or return incorrect results

✅ Add condition to return all points if k equals points length

Incorrect partitioning in quickselect

Algorithm may not converge or return wrong points

✅ Ensure partition places points with smaller distance before pivot and swap correctly

Sorting the entire array inside the heap approach

Negates the efficiency gain of the heap approach

✅ Maintain heap property without sorting entire array repeatedly

🧠

Brute Force (Sort All Points by Distance)

💡 This approach is the most straightforward and helps beginners understand the problem by sorting all points by their distance to the origin, even though it is not the most efficient.

Intuition

Calculate the distance of every point from the origin, then sort all points based on these distances, and finally pick the first k points.

Algorithm

1. Calculate the Euclidean distance for each point.
2. Sort the points array based on these distances.
3. Select the first k points from the sorted array.
4. Return these k points as the result.

💡 Sorting all points is easy to understand but inefficient for large inputs because it does more work than needed.

</>

Code

from typing import List
import math

def kClosest(points: List[List[int]], k: int) -> List[List[int]]:
    # Calculate distance and sort
    points.sort(key=lambda point: point[0]**2 + point[1]**2)
    return points[:k]

# Driver code
if __name__ == '__main__':
    pts = [[1,3],[-2,2]]
    k = 1
    print(kClosest(pts, k))  # Expected output: [[-2,2]]

Line Notes

points.sort(key=lambda point: point[0]**2 + point[1]**2)Sort points by squared distance to avoid costly sqrt and maintain relative order

return points[:k]Return the first k points after sorting as the closest points

def kClosest(points: List[List[int]], k: int) -> List[List[int]]:Function signature with type hints for clarity

if __name__ == '__main__':Allows running the example test when script is executed

import java.util.*;

public class KClosestPoints {
    public static int[][] kClosest(int[][] points, int k) {
        Arrays.sort(points, (a, b) -> Integer.compare(a[0]*a[0] + a[1]*a[1], b[0]*b[0] + b[1]*b[1]));
        return Arrays.copyOfRange(points, 0, k);
    }

    public static void main(String[] args) {
        int[][] points = {{1,3},{-2,2}};
        int k = 1;
        int[][] res = kClosest(points, k);
        System.out.println(Arrays.deepToString(res)); // Expected: [[-2,2]]
    }
}

Line Notes

Arrays.sort(points, (a, b) -> Integer.compare(a[0]*a[0] + a[1]*a[1], b[0]*b[0] + b[1]*b[1]))Sort points by squared distance to origin using a comparator with safe integer comparison

return Arrays.copyOfRange(points, 0, k)Return the first k points after sorting

public static int[][] kClosest(int[][] points, int k)Method signature for kClosest

System.out.println(Arrays.deepToString(res))Print the result array for verification

#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;

vector<vector<int>> kClosest(vector<vector<int>>& points, int k) {
    sort(points.begin(), points.end(), [](const vector<int>& a, const vector<int>& b) {
        return (a[0]*a[0] + a[1]*a[1]) < (b[0]*b[0] + b[1]*b[1]);
    });
    return vector<vector<int>>(points.begin(), points.begin() + k);
}

int main() {
    vector<vector<int>> points = {{1,3},{-2,2}};
    int k = 1;
    vector<vector<int>> res = kClosest(points, k);
    for (auto& p : res) {
        cout << '[' << p[0] << ',' << p[1] << ']';
    }
    cout << endl; // Expected output: [-2,2]
    return 0;
}

Line Notes

sort(points.begin(), points.end(), [](const vector<int>& a, const vector<int>& b)Sort points by squared distance using a lambda comparator

return vector<vector<int>>(points.begin(), points.begin() + k)Return first k points after sorting

int main()Driver code to test the function

cout << '[' << p[0] << ',' << p[1] << ']'Print each point in the result for verification

function kClosest(points, k) {
    points.sort((a, b) => (a[0]*a[0] + a[1]*a[1]) - (b[0]*b[0] + b[1]*b[1]));
    return points.slice(0, k);
}

// Test
const points = [[1,3],[-2,2]];
const k = 1;
console.log(kClosest(points, k)); // Expected output: [[-2,2]]

Line Notes

points.sort((a, b) => (a[0]*a[0] + a[1]*a[1]) - (b[0]*b[0] + b[1]*b[1]))Sort points by squared distance to origin

return points.slice(0, k)Return first k points after sorting

function kClosest(points, k)Function declaration for kClosest

console.log(kClosest(points, k))Print output for quick verification

⏱

Complexity

TimeO(n log n)

SpaceO(log n) due to sorting stack space

Sorting all n points takes O(n log n) time; space is O(log n) due to recursion stack in sorting

💡 For n=100,000 points, sorting means about 1.5 million operations, which can be slow in interviews.

Interview Verdict: Accepted but inefficient for large n

This approach works but is not optimal; it’s a good starting point to understand the problem.

🧠

Max Heap of Size k

💡 This approach improves efficiency by maintaining only the k closest points seen so far using a max heap, avoiding sorting all points.

Intuition

Use a max heap to keep track of the k closest points. For each new point, if it is closer than the farthest point in the heap, replace the farthest with the new point.

Algorithm

1. Initialize a max heap to store points by their distance.
2. Iterate over each point:
3. - Calculate its distance.
4. - If heap size < k, add the point.
5. - Else if current point is closer than the max distance in heap, pop max and add current point.
6. Return all points in the heap.

💡 The heap ensures we only keep k points, so we avoid sorting the entire list.

</>

Code

from typing import List
import heapq

def kClosest(points: List[List[int]], k: int) -> List[List[int]]:
    max_heap = []
    for (x, y) in points:
        dist = -(x*x + y*y)  # negative for max heap
        if len(max_heap) < k:
            heapq.heappush(max_heap, (dist, x, y))
        else:
            if dist > max_heap[0][0]:
                heapq.heappop(max_heap)
                heapq.heappush(max_heap, (dist, x, y))
    return [[x, y] for (_, x, y) in max_heap]

# Driver code
if __name__ == '__main__':
    pts = [[1,3],[-2,2],[2,-2]]
    k = 2
    print(kClosest(pts, k))  # Expected output: [[-2,2],[2,-2]] or similar

Line Notes

dist = -(x*x + y*y)Use negative distance to simulate max heap with Python's min heap

if len(max_heap) < k:Add points until heap size reaches k

if dist > max_heap[0][0]:Check if current point is closer than the farthest in heap

return [[x, y] for (_, x, y) in max_heap]Extract points from heap tuples for final output

import java.util.*;

public class KClosestPoints {
    public static int[][] kClosest(int[][] points, int k) {
        // PriorityQueue stores int[] of {distance, x, y} to maintain max heap by distance
        PriorityQueue<int[]> maxHeap = new PriorityQueue<>((a, b) -> Integer.compare(b[0], a[0]));
        for (int[] point : points) {
            int dist = point[0]*point[0] + point[1]*point[1];
            if (maxHeap.size() < k) {
                maxHeap.offer(new int[]{dist, point[0], point[1]});
            } else if (dist < maxHeap.peek()[0]) {
                maxHeap.poll();
                maxHeap.offer(new int[]{dist, point[0], point[1]});
            }
        }
        int[][] res = new int[k][2];
        int i = 0;
        for (int[] entry : maxHeap) {
            res[i][0] = entry[1];
            res[i][1] = entry[2];
            i++;
        }
        return res;
    }

    public static void main(String[] args) {
        int[][] points = {{1,3},{-2,2},{2,-2}};
        int k = 2;
        int[][] res = kClosest(points, k);
        System.out.println(Arrays.deepToString(res)); // Expected: two closest points, e.g. [[-2,2],[2,-2]]
    }
}

Line Notes

PriorityQueue<int[]> maxHeap = new PriorityQueue<>((a, b) -> Integer.compare(b[0], a[0]))Max heap by distance using comparator on distance (first element)

maxHeap.offer(new int[]{dist, point[0], point[1]})Add current point with its distance to heap

if (maxHeap.size() > k) { maxHeap.poll(); }Remove farthest point if heap exceeds size k (replaced by conditional insert logic)

for (int[] entry : maxHeap)Extract points from heap entries for result

#include <iostream>
#include <vector>
#include <queue>
using namespace std;

vector<vector<int>> kClosest(vector<vector<int>>& points, int k) {
    auto comp = [](const vector<int>& a, const vector<int>& b) {
        return (a[0]*a[0] + a[1]*a[1]) < (b[0]*b[0] + b[1]*b[1]);
    };
    priority_queue<vector<int>, vector<vector<int>>, decltype(comp)> maxHeap(comp);
    for (auto& point : points) {
        maxHeap.push(point);
        if ((int)maxHeap.size() > k) {
            maxHeap.pop();
        }
    }
    vector<vector<int>> res;
    while (!maxHeap.empty()) {
        res.push_back(maxHeap.top());
        maxHeap.pop();
    }
    return res;
}

int main() {
    vector<vector<int>> points = {{1,3},{-2,2},{2,-2}};
    int k = 2;
    vector<vector<int>> res = kClosest(points, k);
    for (auto& p : res) {
        cout << '[' << p[0] << ',' << p[1] << ']';
    }
    cout << endl; // Expected output: two closest points
    return 0;
}

Line Notes

auto comp = [](const vector<int>& a, const vector<int>& b)Comparator for max heap by distance

priority_queue<vector<int>, vector<vector<int>>, decltype(comp)> maxHeap(comp)Declare max heap with custom comparator

maxHeap.push(point)Add point to heap

if ((int)maxHeap.size() > k) { maxHeap.pop(); }Remove farthest point if heap size exceeds k

function kClosest(points, k) {
    const maxHeap = [];
    function dist(point) {
        return point[0]*point[0] + point[1]*point[1];
    }
    function heapPush(point) {
        maxHeap.push(point);
        maxHeap.sort((a,b) => dist(b) - dist(a));
        if (maxHeap.length > k) maxHeap.pop();
    }
    for (const point of points) {
        if (maxHeap.length < k) {
            maxHeap.push(point);
            maxHeap.sort((a,b) => dist(b) - dist(a));
        } else if (dist(point) < dist(maxHeap[0])) {
            maxHeap[0] = point;
            maxHeap.sort((a,b) => dist(b) - dist(a));
        }
    }
    return maxHeap;
}

// Test
const points = [[1,3],[-2,2],[2,-2]];
const k = 2;
console.log(kClosest(points, k)); // Expected output: two closest points

Line Notes

function dist(point)Helper to calculate squared distance

maxHeap.sort((a,b) => dist(b) - dist(a))Sort heap to maintain max at front (inefficient but simulates max heap)

if (maxHeap.length > k) maxHeap.pop()Remove farthest point if heap size exceeds k

if (dist(point) < dist(maxHeap[0]))Replace farthest point if current is closer

⏱

Complexity

TimeO(n log k)

SpaceO(k)

Each insertion/removal in heap is O(log k), done n times, so O(n log k) total; space is O(k) for heap

💡 For n=100,000 and k=10, this means about 230,000 operations, much faster than sorting all points.

Interview Verdict: Accepted and efficient for large n when k is small

This approach balances efficiency and simplicity, making it a strong candidate in interviews.

🧠

Quickselect Algorithm (Average O(n))

💡 Quickselect is an advanced approach that finds the k closest points without fully sorting, using partitioning similar to quicksort, which is optimal on average.

Intuition

Use a partitioning method to place points closer than a pivot to the left and farther to the right, then recurse only on the side containing the kth closest point until the k closest points are found.

Algorithm

1. Choose a pivot point and compute its distance.
2. Partition points so those closer than pivot come before it.
3. Check the pivot’s final position:
4. - If pivot index == k, first k points are closest.
5. - Else recurse on the side containing the kth point.
6. Return first k points after partitioning.

💡 Quickselect reduces work by focusing only on the part of the array that contains the kth closest point.

</>

Code

from typing import List
import random

def kClosest(points: List[List[int]], k: int) -> List[List[int]]:
    def dist(i):
        return points[i][0]*points[i][0] + points[i][1]*points[i][1]

    def partition(left, right, pivot_index):
        pivot_dist = dist(pivot_index)
        points[pivot_index], points[right] = points[right], points[pivot_index]
        store_index = left
        for i in range(left, right):
            if dist(i) < pivot_dist:
                points[store_index], points[i] = points[i], points[store_index]
                store_index += 1
        points[right], points[store_index] = points[store_index], points[right]
        return store_index

    def quickselect(left, right, k_smallest):
        if left == right:
            return
        pivot_index = random.randint(left, right)
        pivot_index = partition(left, right, pivot_index)
        if k_smallest == pivot_index:
            return
        elif k_smallest < pivot_index:
            quickselect(left, pivot_index - 1, k_smallest)
        else:
            quickselect(pivot_index + 1, right, k_smallest)

    quickselect(0, len(points) - 1, k - 1)
    return points[:k]

# Driver code
if __name__ == '__main__':
    pts = [[3,3],[5,-1],[-2,4]]
    k = 2
    print(kClosest(pts, k))  # Expected output: [[3,3],[-2,4]] or similar

Line Notes

pivot_index = random.randint(left, right)Random pivot to avoid worst-case performance

points[pivot_index], points[right] = points[right], points[pivot_index]Move pivot to end for partitioning

if dist(i) < pivot_dist:Compare distances to pivot to partition points

quickselect(0, len(points) - 1, k - 1)Start quickselect to find kth closest point

import java.util.*;

public class KClosestPoints {
    private static int dist(int[] point) {
        return point[0]*point[0] + point[1]*point[1];
    }

    private static int partition(int[][] points, int left, int right, int pivotIndex) {
        int pivotDist = dist(points[pivotIndex]);
        swap(points, pivotIndex, right);
        int storeIndex = left;
        for (int i = left; i < right; i++) {
            if (dist(points[i]) < pivotDist) {
                swap(points, storeIndex, i);
                storeIndex++;
            }
        }
        swap(points, storeIndex, right);
        return storeIndex;
    }

    private static void swap(int[][] points, int i, int j) {
        int[] temp = points[i];
        points[i] = points[j];
        points[j] = temp;
    }

    private static void quickselect(int[][] points, int left, int right, int k) {
        if (left >= right) return;
        Random rand = new Random();
        int pivotIndex = left + rand.nextInt(right - left + 1);
        pivotIndex = partition(points, left, right, pivotIndex);
        if (k == pivotIndex) return;
        else if (k < pivotIndex) quickselect(points, left, pivotIndex - 1, k);
        else quickselect(points, pivotIndex + 1, right, k);
    }

    public static int[][] kClosest(int[][] points, int k) {
        quickselect(points, 0, points.length - 1, k - 1);
        return Arrays.copyOfRange(points, 0, k);
    }

    public static void main(String[] args) {
        int[][] points = {{3,3},{5,-1},{-2,4}};
        int k = 2;
        int[][] res = kClosest(points, k);
        System.out.println(Arrays.deepToString(res)); // Expected: two closest points
    }
}

Line Notes

import java.util.Random;Import Random class for pivot selection

int pivotIndex = left + rand.nextInt(right - left + 1)Random pivot selection to avoid worst-case

swap(points, pivotIndex, right)Move pivot to end for partitioning

if (dist(points[i]) < pivotDist)Partition points by comparing distances

quickselect(points, 0, points.length - 1, k - 1)Start quickselect to find kth closest

#include <iostream>
#include <vector>
#include <cstdlib>
#include <ctime>
using namespace std;

int dist(const vector<int>& point) {
    return point[0]*point[0] + point[1]*point[1];
}

int partition(vector<vector<int>>& points, int left, int right, int pivotIndex) {
    int pivotDist = dist(points[pivotIndex]);
    swap(points[pivotIndex], points[right]);
    int storeIndex = left;
    for (int i = left; i < right; i++) {
        if (dist(points[i]) < pivotDist) {
            swap(points[storeIndex], points[i]);
            storeIndex++;
        }
    }
    swap(points[storeIndex], points[right]);
    return storeIndex;
}

void quickselect(vector<vector<int>>& points, int left, int right, int k) {
    if (left >= right) return;
    int pivotIndex = left + rand() % (right - left + 1);
    pivotIndex = partition(points, left, right, pivotIndex);
    if (k == pivotIndex) return;
    else if (k < pivotIndex) quickselect(points, left, pivotIndex - 1, k);
    else quickselect(points, pivotIndex + 1, right, k);
}

int main() {
    srand(time(0));
    vector<vector<int>> points = {{3,3},{5,-1},{-2,4}};
    int k = 2;
    quickselect(points, 0, points.size() - 1, k - 1);
    for (int i = 0; i < k; i++) {
        cout << '[' << points[i][0] << ',' << points[i][1] << ']';
    }
    cout << endl; // Expected output: two closest points
    return 0;
}

Line Notes

int pivotIndex = left + rand() % (right - left + 1)Random pivot to avoid worst-case

swap(points[pivotIndex], points[right])Move pivot to end for partitioning

if (dist(points[i]) < pivotDist)Partition points by distance comparison

quickselect(points, 0, points.size() - 1, k - 1)Start quickselect to find kth closest

function kClosest(points, k) {
    function dist(i) {
        return points[i][0]*points[i][0] + points[i][1]*points[i][1];
    }
    function swap(i, j) {
        [points[i], points[j]] = [points[j], points[i]];
    }
    function partition(left, right, pivotIndex) {
        let pivotDist = dist(pivotIndex);
        swap(pivotIndex, right);
        let storeIndex = left;
        for (let i = left; i < right; i++) {
            if (dist(i) < pivotDist) {
                swap(storeIndex, i);
                storeIndex++;
            }
        }
        swap(storeIndex, right);
        return storeIndex;
    }
    function quickselect(left, right, kSmallest) {
        if (left === right) return;
        let pivotIndex = left + Math.floor(Math.random() * (right - left + 1));
        pivotIndex = partition(left, right, pivotIndex);
        if (kSmallest === pivotIndex) return;
        else if (kSmallest < pivotIndex) quickselect(left, pivotIndex - 1, kSmallest);
        else quickselect(pivotIndex + 1, right, kSmallest);
    }
    quickselect(0, points.length - 1, k - 1);
    return points.slice(0, k);
}

// Test
const points = [[3,3],[5,-1],[-2,4]];
const k = 2;
console.log(kClosest(points, k)); // Expected output: two closest points

Line Notes

let pivotIndex = left + Math.floor(Math.random() * (right - left + 1))Random pivot selection to avoid worst-case

swap(pivotIndex, right)Move pivot to end for partitioning

if (dist(i) < pivotDist)Partition points by comparing distances

quickselect(0, points.length - 1, k - 1)Start quickselect to find kth closest

⏱

Complexity

TimeAverage O(n), worst O(n^2)

SpaceO(1) additional space

Quickselect partitions the array and recurses only on one side, reducing average time to linear; worst case is quadratic but rare with random pivots.

💡 For n=100,000, this approach can find k closest points in about 100,000 operations on average, much faster than sorting.

Interview Verdict: Accepted and optimal average time

This is the best approach to mention and implement if comfortable, showing mastery of selection algorithms.

📊

All Approaches - One-Glance Tradeoffs

💡 In most interviews, the max heap approach is a good balance of efficiency and coding simplicity, while quickselect is optimal but more complex.

Approach	Time	Space	Stack Risk	Reconstruct	Use In Interview
1. Brute Force	O(n log n)	O(log n)	No	N/A	Mention only - never code
2. Max Heap of Size k	O(n log k)	O(k)	No	N/A	Good to code if k is small or medium
3. Quickselect Algorithm	Average O(n)	O(1)	Yes (due to recursion depth)	N/A	Best to mention and code if comfortable

💼

Interview Strategy

💡 Use this guide to understand the problem deeply before your interview. Start by clarifying the problem, then explain the brute force approach, and progressively discuss more efficient solutions.

How to Present

Step 1: Clarify the problem and constraints with the interviewer.Step 2: Describe the brute force approach using sorting all points.Step 3: Explain the max heap approach to improve efficiency.Step 4: Discuss the quickselect algorithm as the optimal solution.Step 5: Code the chosen approach carefully and test with examples.

Time Allocation

Clarify: 3min → Approach: 5min → Code: 10min → Test: 5min. Total ~23min

What the Interviewer Tests

The interviewer tests your ability to optimize from brute force to efficient solutions, your understanding of heaps and selection algorithms, and your coding accuracy.

Common Follow-ups

• What if points are streaming and you need to maintain k closest? → Use a max heap to maintain k closest dynamically.
• Can you solve this in O(n) time? → Use quickselect algorithm.

💡 These follow-ups test your knowledge of data streams and advanced selection algorithms, common in real-world scenarios.

🔍

Pattern Recognition

When to Use

1) Need to find k elements based on a ranking or metric, 2) Input size is large, 3) Sorting entire input is expensive, 4) Problem involves distances or comparable keys.

Signature Phrases

k closest pointsdistance from originreturn k points closest

NOT This Pattern When

Sorting entire array without selection, or problems requiring all elements sorted fully.

Similar Problems

Top K Frequent Elements - uses heaps to find top k by frequencyKth Largest Element in an Array - selection problem using quickselectFind Median from Data Stream - dynamic k-selection with heaps

Practice

1. What is the time complexity of the add(val) operation in the optimal min-heap approach for maintaining the kth largest element in a stream, assuming the heap size is k and the stream length is n?

medium

A. O(n ∙ log n) per add call, because the stream grows with n

B. O(1) per add call, as heap root access is constant time

C. O(k) per add call, because we may need to scan the heap to find the position

D. O(log k) per add call, since heap operations depend only on heap size k

Solution

1. Step 1: Identify heap size and operations

   The heap size is fixed at k, independent of stream length n. Each add involves either a push or a replace operation on the heap.

2. Step 2: Analyze complexity of heap operations

   Heap push or replace operations take O(log k) time because the heap has size k. Accessing the root is O(1), but add involves heapify steps.

3. Final Answer:

   Option D -> Option D

4. Quick Check:

   Heap operations depend on k, not n, so O(log k) per add [OK]

Hint: Heap operations scale with heap size k, not stream length n [OK]

Common Mistakes:

• Confusing n with k in complexity
• Assuming linear scan of heap
• Ignoring heapify cost

2. What is the time complexity of the optimal sweep line algorithm for finding the maximum CPU load given n jobs, and why might a candidate incorrectly think it is higher?

medium

A. O(n^2) because each job is compared with every other job to find overlaps.

B. O(n log n) because of heap operations for each job's start and end.

C. O(n log n) due to sorting all start and end events, then a linear sweep.

D. O(n) because we only iterate through the jobs once.

Solution

1. Step 1: Identify the main cost is sorting 2n events.

   Sorting events takes O(n log n) time; sweeping through events is O(n).

2. Step 2: Explain why O(n^2) is incorrect.

   Brute force checking all pairs is O(n^2), but sweep line avoids pairwise comparisons by event sorting.

3. Final Answer:

   Option C -> Option C

4. Quick Check:

   Sorting dominates, linear sweep after sorting [OK]

Hint: Sorting events dominates complexity [OK]

Common Mistakes:

• Confusing brute force O(n^2) with sweep line
• Ignoring sorting cost

3. The following code attempts to implement the optimal bucket sort approach for top k frequent elements. Identify the line that contains a subtle bug that can cause incorrect results or runtime errors.

medium

A. Line updating frequency dictionary with freq.get(num, 0) + 1

B. Line iterating over freq.items() to append elements to buckets

C. Line creating buckets with size max_freq instead of max_freq + 1

D. Line iterating buckets from max_freq - 1 down to 0

Solution

1. Step 1: Check bucket list size

   Buckets are created with size max_freq, but frequencies range from 1 to max_freq inclusive, so index max_freq is needed.

2. Step 2: Consequence of incorrect bucket size

   Accessing buckets[count] where count == max_freq causes IndexError or missing elements if bucket list is too small.

3. Final Answer:

   Option C -> Option C

4. Quick Check:

   Bucket list must have max_freq + 1 slots to index frequencies correctly [OK]

Hint: Bucket size must be max_freq + 1 to include max frequency [OK]

Common Mistakes:

• Off-by-one bucket size
• Misindexing buckets causing runtime errors

4. Suppose you want to find the kth largest element in an array where elements can be repeated and you want to find the kth largest unique element instead. Which modification to the Quickselect approach is correct?

hard

A. Use a min-heap of size k to store unique elements only, ignoring duplicates during insertion.

B. Run Quickselect as usual and then scan left and right to skip duplicates until kth unique is found.

C. Sort the array fully and iterate from largest to smallest counting unique elements until kth unique is reached.

D. Modify partition to skip duplicates and recurse only on unique elements.

Solution

1. Step 1: Understand uniqueness constraint

   Quickselect partitions by value but does not handle uniqueness inherently.

2. Step 2: Evaluate modifications

   Partial scanning or modifying partition is complex and error-prone; min-heap requires extra checks.

3. Step 3: Choose reliable approach

   Sorting fully and counting unique elements is straightforward and guarantees correctness.

4. Final Answer:

   Option C -> Option C

5. Quick Check:

   Full sort enables unique counting easily [OK]

Hint: Uniqueness requires full sort or extra data structures [OK]

Common Mistakes:

• Assuming Quickselect handles uniqueness
• Trying to modify partition without correctness
• Ignoring duplicates in heap approach

5. Suppose the problem changes: you want to find the kth largest XOR coordinate value, but now the matrix can contain negative integers. Which modification to the optimal prefix XOR + min-heap approach is necessary to handle this correctly?

hard

A. Use a 64-bit integer type to store prefix XOR values to avoid sign issues.

B. Convert all negative integers to positive before computing prefix XOR.

C. No modification needed; XOR works the same with negative integers.

D. Use bitwise AND instead of XOR to handle negative values correctly.

Solution

1. Step 1: Understand XOR with negative integers

   In Python, integers are unbounded but negative numbers use two's complement representation internally, which can cause sign extension issues if stored in fixed-width types.

2. Step 2: Use 64-bit integer type for prefix XOR

   To correctly handle negative values and avoid sign extension or overflow, use a 64-bit integer type (e.g., numpy.int64) for prefix XOR computations and heap storage.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Handling negative integers requires careful data type choice, not changing XOR logic [OK]

Hint: XOR logic unchanged; data type must handle sign correctly [OK]

Common Mistakes:

• Assuming XOR fails with negatives
• Replacing XOR with AND
• Ignoring integer overflow/sign issues