Given a number N, the task is to find the integer points (x, y) such that 0 <= x, y <= N and Manhattan distance between any two points will be atleast N.
Examples:
Input: N = 3 Output: (0, 0) (0, 3) (3, 0) (3, 3) Input: N = 4 Output: (0, 0) (0, 4) (4, 0) (4, 4) (2, 2)
Approach:
- Manhattan Distance between two points (x1, y1) and (x2, y2) is:
|x1 – x2| + |y1 – y2| - Here for all pair of points this distance will be atleast N.
- As 0 <= x <= N and 0 <= y <= N so we can imagine a square of side length N whose bottom left corner is (0, 0) and top right corner is (N, N).
- So if we place 4 points in this corner then Manhattan distance will be atleast N.
- Now as we have to maximize the number of the point we have to check is there any available point inside the square.
- If N is even then middle point of the square which is (N/2, N/2) is integer point, otherwise, it will be float value as N/2 is not a integer when N is odd.
- So the only available position is the middle point and we can put a point there only if N is even.
- So number of points will be 4 if N is odd and if N is even then the number of points will be 5.
Below is the implementation of the above approach:
C++
// C++ code to Find the integer points (x, y)// with Manhattan distance atleast N#include <bits/stdc++.h>using namespace std;// C++ function to find all possible pointvector<pair<int, int> > FindPoints(int n){ vector<pair<int, int> > v; // Find all 4 corners of the square // whose side length is n v.push_back({ 0, 0 }); v.push_back({ 0, n }); v.push_back({ n, 0 }); v.push_back({ n, n }); // If n is even then the middle point // of the square will be an integer, // so we will take that point if (n % 2 == 0) v.push_back({ n / 2, n / 2 }); return v;}// Driver Codeint main(){ int N = 8; vector<pair<int, int> > v = FindPoints(N); // Printing all possible points for (auto i : v) { cout << "(" << i.first << ", " << i.second << ") "; } return 0;} |
Java
// Java code to Find the integer points (x, y)// with Manhattan distance atleast Nimport java.util.*;class GFG{static class pair{ int first, second; public pair(int first, int second) { this.first = first; this.second = second; } }// Java function to find all possible pointstatic Vector<pair> FindPoints(int n){ Vector<pair> v = new Vector<pair>(); // Find all 4 corners of the square // whose side length is n v.add(new pair( 0, 0 )); v.add(new pair( 0, n )); v.add(new pair( n, 0 )); v.add(new pair( n, n )); // If n is even then the middle point // of the square will be an integer, // so we will take that point if (n % 2 == 0) v.add(new pair( n / 2, n / 2 )); return v;}// Driver Codepublic static void main(String[] args){ int N = 8; Vector<pair > v = FindPoints(N); // Printing all possible points for (pair i : v) { System.out.print("(" + i.first + ", " + i.second + ") "); }}}// This code is contributed by PrinciRaj1992 |
Python3
# Python3 code to Find the integer points (x, y) # with Manhattan distance atleast N # function to find all possible point def FindPoints(n) : v = []; # Find all 4 corners of the square # whose side length is n v.append([ 0, 0 ]); v.append([ 0, n ]); v.append([ n, 0 ]); v.append([ n, n ]); # If n is even then the middle point # of the square will be an integer, # so we will take that point if (n % 2 == 0) : v.append([ n // 2, n // 2 ]); return v; # Driver Code if __name__ == "__main__" : N = 8; v = FindPoints(N); # Printing all possible points for element in v : print("(", element[0], ",", element[1], ")", end = " "); # This code is contributed by AnkitRai01 |
C#
// C# code to Find the integer points (x, y)// with Manhattan distance atleast Nusing System;using System.Collections.Generic;class GFG{class pair{ public int first, second; public pair(int first, int second) { this.first = first; this.second = second; } }// Function to find all possible pointstatic List<pair> FindPoints(int n){ List<pair> v = new List<pair>(); // Find all 4 corners of the square // whose side length is n v.Add(new pair( 0, 0 )); v.Add(new pair( 0, n )); v.Add(new pair( n, 0 )); v.Add(new pair( n, n )); // If n is even then the middle point // of the square will be an integer, // so we will take that point if (n % 2 == 0) v.Add(new pair( n / 2, n / 2 )); return v;}// Driver Codepublic static void Main(String[] args){ int N = 8; List<pair > v = FindPoints(N); // Printing all possible points foreach (pair i in v) { Console.Write("(" + i.first + ", " + i.second + ") "); }}}// This code is contributed by Rajput-Ji |
Javascript
<script>// Javascript code to Find the integer points (x, y)// with Manhattan distance atleast N// C++ function to find all possible pointfunction FindPoints(n){ var v = []; // Find all 4 corners of the square // whose side length is n v.push([ 0, 0 ]); v.push([ 0, n ]); v.push([ n, 0 ]); v.push([ n, n ]); // If n is even then the middle point // of the square will be an integer, // so we will take that point if (n % 2 == 0) v.push([ n / 2, n / 2 ]); return v;}// Driver Codevar N = 8;var v = FindPoints(N);// Printing all possible pointsv.forEach(i => { document.write( "(" + i[0] + ", " + i[1] + ") ");});// This code is contributed by rrrtnx.</script> |
Output:
(0, 0) (0, 8) (8, 0) (8, 8) (4, 4)
Time Complexity: O(1)
Auxiliary Space: O(1)
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