2857. Count Pairs of Points With Distance k
Problem Statement
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
Intuition
Approach:
Since K constraint is less,
We can iterate throught the k
x1^x2 = i , y1^y2 = k-i
a^b = c -> b = c^a
Go through all k's
Links
https://leetcode.com/problems/count-pairs-of-points-with-distance-k/description/
Video Links
https://www.youtube.com/watch?v=1Hp6Hsghtno&ab_channel=AryanMittal
Approach 1:
class Solution {
public:
int countPairs(vector<vector<int>>& arr, int k) {
map<pair<int,int> , int> mp;
int ans = 0;
for(auto &it: arr){
int x1 = it[0], y1 = it[1];
for(int i=0; i<=k; i++){
int x2 = x1^i;
int y2 = y1^(k-i);
ans += mp[{x2,y2}];
}
mp[{x1,y1}]++;
}
return ans;
}
};
Approach 2:
Approach 3:
Approach 4:
Similar Problems
Last updated