K Closest Points to Origin

▶

Steps

setup

Start Quickselect on full array

We begin with the full points array [[1,3],[-2,2]] and k=1. The algorithm will partition to find the closest point to the origin.

💡 Starting with the entire array is necessary to narrow down to the k closest points.

Line:

quickselect(0, len(points) - 1, k - 1)

💡 The algorithm will recursively partition the array to isolate the closest k points.

setup

Choose pivot index randomly

Randomly select a pivot index between left=0 and right=1. Suppose pivot_index=1 is chosen, pointing to (-2,2).

💡 Choosing a pivot is key to partitioning the array around a reference point.

Line:

pivot_index = random.randint(left, right)

💡 The pivot point's distance will be the threshold for partitioning.

partition

Swap pivot with rightmost element

Swap the pivot point (-2,2) at index 1 with the rightmost element at index 1 (itself). No change in array.

💡 Moving pivot to the end simplifies partitioning logic.

Line:

points[pivot_index], points[right] = points[right], points[pivot_index]

💡 Pivot is now at the end, ready for partitioning.

partition

Initialize store_index for partitioning

Set store_index to left=0. This index tracks where to place points closer than pivot.

💡 store_index helps segregate points closer than pivot to the left.

Line:

store_index = left

💡 Points closer than pivot will be swapped to store_index.

compare

Compare point at index 0 to pivot distance

Calculate distance of point (1,3) at index 0 and compare to pivot distance (distance of (-2,2)).

💡 Comparing distances determines if point should be moved left of pivot.

Line:

if dist(i) < pivot_dist:

💡 This comparison decides if a swap is needed to move closer points left.

compare

Decision: no swap since point is farther

Distance(1,3)=10 is greater than or equal to pivot distance(8), so no swap occurs and store_index remains 0.

💡 No swap means the point stays on the right side of the pivot.

Line:

if dist(i) < pivot_dist: (false)

💡 Points farther than pivot remain to the right, preserving partitioning.

partition

Finish scanning, swap pivot to final position

Swap pivot (-2,2) from right index 1 with point at store_index 0 (1,3).

💡 Placing pivot at store_index partitions array into closer and farther points.

Line:

points[right], points[store_index] = points[store_index], points[right]

💡 Pivot is now at index 0, points to left are closer (none here), right are farther.

compare

Check pivot index against k-1

Pivot index is 0, which equals k-1=0, so partitioning found the k closest points.

💡 If pivot index matches k-1, first k points are closest.

Line:

if k_smallest == pivot_index: return

💡 No further partitioning needed; closest points isolated.

reconstruct

Return first k points as closest

Return the first k=1 points from the partitioned array, which is [[-2,2]].

💡 After partitioning, the first k points are guaranteed closest.

Line:

return points[:k]

💡 The algorithm efficiently found the closest point without full sort.

final

Final confirmation of closest points

The closest point to the origin is confirmed as (-2,2), matching the expected output.

💡 Final step confirms the correctness of the algorithm's output.

Line:

return points[:k]

💡 The algorithm's partitioning and selection steps successfully isolated the closest point.

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)  # STEP 2
        points[pivot_index], points[right] = points[right], points[pivot_index]  # STEP 3
        store_index = left  # STEP 4
        for i in range(left, right):  # STEP 5
            if dist(i) < pivot_dist:  # STEP 5 decision
                points[store_index], points[i] = points[i], points[store_index]  # STEP 5 swap
                store_index += 1
        points[right], points[store_index] = points[store_index], points[right]  # STEP 6
        return store_index

    def quickselect(left, right, k_smallest):
        if left == right:  # STEP 1
            return
        pivot_index = random.randint(left, right)  # STEP 2
        pivot_index = partition(left, right, pivot_index)  # STEP 3-6
        if k_smallest == pivot_index:  # STEP 7
            return
        elif k_smallest < pivot_index:  # STEP 7 decision
            quickselect(left, pivot_index - 1, k_smallest)
        else:
            quickselect(pivot_index + 1, right, k_smallest)

    quickselect(0, len(points) - 1, k - 1)  # STEP 1
    return points[:k]  # STEP 8

📊

K Closest Points to Origin - Watch the Algorithm Execute, Step by Step

Watching each partition and pivot decision helps you understand how Quickselect efficiently finds the k closest points without sorting the entire array.

peekmin-heap

(1,3)

0

(-2,2)

1

peekmin-heap

(1,3)

0

(-2,2)

1

insertmin-heapswap [1↔1]

(1,3)

0

(-2,2)

1

peekmin-heap

(1,3)

0

(-2,2)

1

comparemin-heap

(1,3)

0

(-2,2)

1

peekmin-heap

(1,3)

0

(-2,2)

1

insertmin-heapswap [0↔1]

(-2,2)

0

(1,3)

1

peekmin-heap

(-2,2)

0

(1,3)

1

peekmin-heap

(-2,2)

0

(1,3)

1

Added: [["-2,2"]]

peekmin-heap

(-2,2)

0

(1,3)

1

Added: [["-2,2"]]

Key Takeaways

✓ Quickselect efficiently finds the k closest points without sorting the entire array.

Reading code alone hides how partitioning narrows the search space step-by-step.

✓ Partitioning rearranges points so those closer than the pivot come before it, isolating the closest points.

Visualizing swaps and comparisons clarifies how the array is segmented.

✓ The pivot index check determines whether to recurse left or right, or stop when the k closest points are found.

Seeing the pivot index compared to k-1 helps understand the stopping condition concretely.

Practice

1. You are given a 2D matrix and need to find the kth largest value among all XOR coordinate values, where each coordinate value is the XOR of all elements from the top-left corner to that coordinate. Which approach guarantees an optimal solution in terms of time complexity?

easy

A. Use nested loops to compute XOR for each coordinate independently, then sort all values.

B. Apply dynamic programming with memoization to store XOR values and then sort them.

C. Use a greedy approach to pick the largest elements first and XOR them until k values are found.

D. Compute a prefix XOR matrix and maintain a min-heap of size k to track the kth largest value.

Solution

1. Step 1: Understand the problem constraints

   Computing XOR values for each coordinate independently with nested loops leads to O(m^2 * n^2) time, which is inefficient.

2. Step 2: Recognize prefix XOR and min-heap usage

   Using a prefix XOR matrix allows O(m * n) computation of all XOR coordinate values. Maintaining a min-heap of size k ensures we track the kth largest value efficiently, leading to O(m * n * log k) time.

3. Final Answer:

   Option D -> Option D

4. Quick Check:

   Prefix XOR + min-heap is optimal for kth largest XOR coordinate value [OK]

Hint: Prefix XOR + min-heap is optimal for kth largest XOR values [OK]

Common Mistakes:

• Thinking nested loops with sorting is efficient enough
• Assuming greedy selection works for XOR
• Confusing DP with prefix XOR usage

2. What is the time complexity of the pop() operation in the optimal frequency stack implementation that uses a max-heap keyed by frequency and timestamp, assuming n total operations?

medium

A. O(1) amortized, since the heap is balanced automatically

B. O(log n), due to heap pop operation on at most n elements

C. O(n), because frequencies must be recomputed on each pop

D. O(log k), where k is the number of distinct elements

Solution

1. Step 1: Identify heap size and operations

   The heap stores one entry per push operation, so its size can be up to n.

2. Step 2: Analyze pop complexity

   Each pop calls heapq.heappop which takes O(log n) time in worst case. Frequency map updates are O(1).

3. Final Answer:

   Option B -> Option B

4. Quick Check:

   Heap pop dominates time complexity, so O(log n) per pop [OK]

Hint: Heap pop is O(log n), not O(1) or O(n) [OK]

Common Mistakes:

• Assuming heap pop is O(1)
• Confusing frequency recomputation cost
• Assuming complexity depends on distinct elements only

3. The following code attempts to implement the optimal frequency stack using a max-heap with timestamp. Identify the bug that causes incorrect behavior when popping elements.

import heapq

class FreqStack:
    def __init__(self):
        self.freq = {}
        self.heap = []
        self.time = 0

    def push(self, val: int) -> None:
        self.freq[val] = self.freq.get(val, 0) + 1
        self.time += 1
        heapq.heappush(self.heap, (-self.freq[val], -self.time, val))

    def pop(self) -> int:
        freq, time, val = heapq.heappop(self.heap)
        # BUG: Missing frequency decrement
        if self.freq[val] == 0:
            del self.freq[val]
        return val

medium

A. Line with heapq.heappop(self.heap) does not pop the correct element.

B. Frequency map deletion condition is incorrect; should delete when freq[val] == 1.

C. Timestamp is not updated on push, so recency is lost.

D. Frequency map is not decremented after popping, causing stale frequencies.

Solution

1. Step 1: Identify frequency update after pop

   The code does not decrement frequency after popping an element, so freq[val] remains stale.

2. Step 2: Consequences of missing decrement

   Without decrement, future pushes and pops will have incorrect frequency counts, breaking the max frequency logic.

3. Final Answer:

   Option D -> Option D

4. Quick Check:

   Frequency must be decremented after pop to maintain correctness [OK]

Hint: Always decrement frequency after popping an element [OK]

Common Mistakes:

• Forgetting to update frequency map after pop
• Mismanaging timestamp updates
• Incorrect deletion condition for frequency map

4. 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

5. Suppose the problem changes: the matrix is no longer sorted by columns, only rows are sorted, and you still need to find the kth smallest element. Which approach is the most suitable to solve this variant efficiently?

hard

A. Use binary search on the value range with the same counting method as before.

B. Use a greedy approach picking the smallest element from each row repeatedly.

C. Flatten the matrix and sort all elements to find the kth smallest.

D. Use a min heap to merge the k sorted rows, extracting the kth smallest element.

Solution

1. Step 1: Understand changed constraints

   Only rows are sorted, columns are not sorted, so binary search counting on columns is invalid.

2. Step 2: Identify suitable approach

   Min heap merging k sorted lists (rows) efficiently finds kth smallest in O(k log n) time.

3. Step 3: Reject other approaches

   Binary search counting fails without column sorting; flattening is O(n² log n²); greedy is incorrect.

4. Final Answer:

   Option D -> Option D

5. Quick Check:

   Min heap merge exploits row sorting without column sorting [OK]

Hint: Without column sorting, binary search counting fails [OK]

Common Mistakes:

• Trying binary search counting without column sorting
• Assuming greedy works