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2857. Count Pairs of Points With Distance k

Description

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.

 

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 <= xi, yi <= 106
  • 0 <= k <= 100

Solutions

Solution 1: Hash Table + Enumeration

We can use a hash table $cnt$ to count the occurrence of each point in the array $coordinates$.

Next, we enumerate each point $(x_2, y_2)$ in the array $coordinates$. Since the range of $k$ is $[0, 100]$, and the result of $x_1 \oplus x_2$ or $y_1 \oplus y_2$ is always greater than or equal to $0$, we can enumerate the result $a$ of $x_1 \oplus x_2$ in the range $[0,..k]$. Then, the result of $y_1 \oplus y_2$ is $b = k - a$. In this way, we can calculate the values of $x_1$ and $y_1$, and add the occurrence of $(x_1, y_1)$ to the answer.

The time complexity is $O(n \times k)$, and the space complexity is $O(n)$. Here, $n$ is the length of the array $coordinates$.

  • 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;
    }
}

  • class Solution {
public:
    int countPairs(vector<vector<int>>& coordinates, int k) {
        unordered_map<long long, int> cnt;
        auto f = [](int x, int y) {
            return x * 1000000L + y;
        };
        int ans = 0;
        for (auto& c : coordinates) {
            int x2 = c[0], y2 = c[1];
            for (int a = 0; a <= k; ++a) {
                int b = k - a;
                int x1 = a ^ x2, y1 = b ^ y2;
                ans += cnt[f(x1, y1)];
            }
            ++cnt[f(x2, y2)];
        }
        return ans;
    }
};

  • 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


  • 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;
}

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