Leetcode 2857: Count Pairs of Points With Distance k

Input: The input consists of a list of 2D points and an integer k.

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

Constraints:

• 2 <= coordinates.length <= 50000

• 0 <= xi, yi <= 10^6

• 0 <= k <= 100

Output: Return the number of pairs (i, j) where i < j and the distance between points i and j equals k.

Example: Output: 2

Constraints:

• The output is an integer representing the number of valid pairs.

Goal: Calculate the number of valid pairs of points whose XOR-based distance equals k.

Steps:

• Initialize a hash map to store the count of points encountered so far.

• Iterate through the points and for each point, calculate the potential pairings that result in the required distance k.

• For each valid pair, increment the result count.

• Return the final result.

Goal: Constraints for the input array and k.

Steps:

• The array contains between 2 and 50,000 points.

• Each point's coordinates are between 0 and 1,000,000.

• The distance value k is between 0 and 100.

Assumptions:

• The input coordinates are non-negative integers.

• The value of k is not negative.

• Input: coordinates = [[1, 2], [4, 2], [1, 3], [5, 2]], k = 5

• Explanation: Valid pairs (0, 1) and (2, 3) both have a distance of 5.

• Input: coordinates = [[1, 3], [1, 3], [1, 3], [1, 3], [1, 3]], k = 0

• Explanation: Any two points have a distance of 0, so there are 10 ways to select two points from the list.

Approach: Use a hash map to efficiently count valid pairs of points whose XOR distance is equal to k.

Observations:

• The problem requires efficiently finding pairs of points that satisfy the XOR distance condition.

• Using a hash map allows us to track previously encountered points and quickly check if a valid pair can be formed.

• The key insight is to iterate through the points and use XOR operations to find potential valid pairs that give the desired distance.

Steps:

• Initialize a hash map to store the frequency of encountered points.

• For each point, calculate the distance from all previous points using XOR and check if it matches k.

• Count the valid pairs and return the result.

Empty Inputs:

• If the input array contains fewer than 2 points, no pairs can be formed.

Large Inputs:

• For large input sizes, ensure that the solution runs efficiently within the time limit.

Special Values:

• If all points have the same coordinates, only pairs with distance 0 can be formed.

Constraints:

• Make sure the solution can handle the maximum possible value for k (100) and large input sizes.

int countPairs(vector<vector<int>>& coordinates, int k) {
    unordered_map<int, unordered_map<int, int>> count;
    int res = 0;
    for (auto& c : coordinates) {
        for (int x = 0; x <= k; x++)
            if (count[c[0] ^ x].count(c[1] ^ (k - x)))
                res += count[c[0] ^ x][c[1] ^ (k - x)];
        count[c[0]][c[1]]++;
    }
    return res;
}

1 : Initialization

int countPairs(vector<vector<int>>& coordinates, int k) {

Defines the function `countPairs` that takes in a 2D vector `coordinates` and an integer `k`. It will calculate the number of pairs of coordinates with specific properties.

2 : Data Structure Initialization

    unordered_map<int, unordered_map<int, int>> count;

Initializes a nested unordered map `count` to keep track of coordinate pairs and their counts using XOR as the key.

3 : Variable Initialization

    int res = 0;

Declares and initializes an integer variable `res` to store the count of valid pairs found during the loop.

4 : Outer Loop

    for (auto& c : coordinates) {

Iterates over each coordinate `c` in the given 2D vector `coordinates`.

5 : Inner Loop

        for (int x = 0; x <= k; x++)

Nested loop that iterates over possible values of `x` from `0` to `k`.

6 : Condition Check

            if (count[c[0] ^ x].count(c[1] ^ (k - x)))

Checks if there is a previously seen coordinate pair (after XORing with `x` and `k-x`) in the `count` map.

7 : Result Update

                res += count[c[0] ^ x][c[1] ^ (k - x)];

If the condition is true, it adds the count of matching coordinate pairs to the result `res`.

8 : Map Update

        count[c[0]][c[1]]++;

Updates the count of the current coordinate pair `c[0]` and `c[1]` in the map `count`.

9 : Return Statement

    return res;

Returns the final count of valid pairs stored in `res`.

Best Case: O(n)

Average Case: O(n)

Worst Case: O(n)

Description: The time complexity is O(n) due to the use of a hash map and iterating over the points once.

Best Case: O(n)

Worst Case: O(n)

Description: The space complexity is O(n) because we store the points in a hash map.

Leetcode 2857: Count Pairs of Points With Distance k

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