LeetCode #2857 — MEDIUM

Count Pairs of Points With Distance k

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

You are given a 2D integer array coordinates and an integer k, where coordinates[i] = [x_i, y_i] are the coordinates of the i^th point in a 2D plane.

We define the distance between two points (x₁, y₁) and (x₂, y₂) as (x1 XOR x2) + (y1 XOR y2) where XOR is the bitwise XOR operation.

Return the number of pairs (i, j) such that i < j and the distance between points i and j is equal to k.

Example 1:

Input: coordinates = [[1,2],[4,2],[1,3],[5,2]], k = 5
Output: 2
Explanation: We can choose the following pairs:
- (0,1): Because we have (1 XOR 4) + (2 XOR 2) = 5.
- (2,3): Because we have (1 XOR 5) + (3 XOR 2) = 5.

Example 2:

Input: coordinates = [[1,3],[1,3],[1,3],[1,3],[1,3]], k = 0
Output: 10
Explanation: Any two chosen pairs will have a distance of 0. There are 10 ways to choose two pairs.

Constraints:

2 <= coordinates.length <= 50000
0 <= x_i, y_i <= 10⁶
0 <= k <= 100

Step 01

Brute Force Baseline

Problem summary: You are given a 2D integer array coordinates and an integer k, where coordinates[i] = [xi, yi] are the coordinates of the ith point in a 2D plane. We define the distance between two points (x1, y1) and (x2, y2) as (x1 XOR x2) + (y1 XOR y2) where XOR is the bitwise XOR operation. Return the number of pairs (i, j) such that i < j and the distance between points i and j is equal to k.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Hash Map · Bit Manipulation

Example 1

[[1,2],[4,2],[1,3],[5,2]]
5

Example 2

[[1,3],[1,3],[1,3],[1,3],[1,3]]
0

Step 02

Core Insight

What unlocks the optimal approach

<div class="_1l1MA">Suppose that <code>x = x1 XOR x2</code> and y = y1 XOR y2 then we can get <code>x2 = x XOR x1</code> and <code>y2 = y XOR y1</code>.</div>
<div class="_1l1MA">We are supposed to have k = x + y so we can get <code>x2 = x XOR x1</code> and <code>y2 = (k - x) XOR y1</code>.</div>
<div class="_1l1MA">We can iterate over all possible values of <code>x</code> and count the number of points <code>(x1, x2)</code> and <code>(x2, y2)</code>.</div>

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2857: Count Pairs of Points With Distance k
class Solution {
    public int countPairs(List<List<Integer>> coordinates, int k) {
        Map<List<Integer>, Integer> cnt = new HashMap<>();
        int ans = 0;
        for (var c : coordinates) {
            int x2 = c.get(0), y2 = c.get(1);
            for (int a = 0; a <= k; ++a) {
                int b = k - a;
                int x1 = a ^ x2, y1 = b ^ y2;
                ans += cnt.getOrDefault(List.of(x1, y1), 0);
            }
            cnt.merge(c, 1, Integer::sum);
        }
        return ans;
    }
}

// Accepted solution for LeetCode #2857: Count Pairs of Points With Distance k
func countPairs(coordinates [][]int, k int) (ans int) {
	cnt := map[[2]int]int{}
	for _, c := range coordinates {
		x2, y2 := c[0], c[1]
		for a := 0; a <= k; a++ {
			b := k - a
			x1, y1 := a^x2, b^y2
			ans += cnt[[2]int{x1, y1}]
		}
		cnt[[2]int{x2, y2}]++
	}
	return
}

# Accepted solution for LeetCode #2857: Count Pairs of Points With Distance k
class Solution:
    def countPairs(self, coordinates: List[List[int]], k: int) -> int:
        cnt = Counter()
        ans = 0
        for x2, y2 in coordinates:
            for a in range(k + 1):
                b = k - a
                x1, y1 = a ^ x2, b ^ y2
                ans += cnt[(x1, y1)]
            cnt[(x2, y2)] += 1
        return ans

// Accepted solution for LeetCode #2857: Count Pairs of Points With Distance k
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #2857: Count Pairs of Points With Distance k
// class Solution {
//     public int countPairs(List<List<Integer>> coordinates, int k) {
//         Map<List<Integer>, Integer> cnt = new HashMap<>();
//         int ans = 0;
//         for (var c : coordinates) {
//             int x2 = c.get(0), y2 = c.get(1);
//             for (int a = 0; a <= k; ++a) {
//                 int b = k - a;
//                 int x1 = a ^ x2, y1 = b ^ y2;
//                 ans += cnt.getOrDefault(List.of(x1, y1), 0);
//             }
//             cnt.merge(c, 1, Integer::sum);
//         }
//         return ans;
//     }
// }

// Accepted solution for LeetCode #2857: Count Pairs of Points With Distance k
function countPairs(coordinates: number[][], k: number): number {
    const cnt: Map<number, number> = new Map();
    const f = (x: number, y: number): number => x * 1000000 + y;
    let ans = 0;
    for (const [x2, y2] of coordinates) {
        for (let a = 0; a <= k; ++a) {
            const b = k - a;
            const [x1, y1] = [a ^ x2, b ^ y2];
            ans += cnt.get(f(x1, y1)) ?? 0;
        }
        cnt.set(f(x2, y2), (cnt.get(f(x2, y2)) ?? 0) + 1);
    }
    return ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n × k)

Space

O(n)

Approach Breakdown

SORT + SCAN

O(n log n) time

O(n) space

Sort the array in O(n log n), then scan for the missing or unique element by comparing adjacent pairs. Sorting requires O(n) auxiliary space (or O(1) with in-place sort but O(n log n) time remains). The sort step dominates.

BIT MANIPULATION

O(n) time

O(1) space

Bitwise operations (AND, OR, XOR, shifts) are O(1) per operation on fixed-width integers. A single pass through the input with bit operations gives O(n) time. The key insight: XOR of a number with itself is 0, which eliminates duplicates without extra space.

Shortcut: Bit operations are O(1). XOR cancels duplicates. Single pass → O(n) time, O(1) space.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.