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Find 4 points with equal Manhattan distance between any pair

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  • Difficulty Level : Easy
  • Last Updated : 01 Apr, 2022

Given an integer N, Find 4 points in a 2D plane having integral coordinates, such that the Manhattan distance between any pair of points is equal to N.

Examples:

Input: N = 6
Output: { {0, -3}, {3, 0}, {-3, 0}, {0, 3} }
Explanation: It can be easily calculated that Manhattan distance between all possible pairs is 6.

Input: N = 11
Output: -1
Explanation: It is not possible to locate 4 points such that Manhattan distance between all pairs is 11

 

Approach: The idea to solve the problem is based on the following observation:

Square having all 4 vertices on all 4 co-ordinate axes (1 vertex on each axis) equidistant from origin have the same distance between any pair of vertices which is the double of the distance between any vertex and origin.
If N is odd, it can not be divided into 2 equal integral parts. So, a square can not be formed such that opposite vertices are at equal distance from the origin (i.e. N/2), with distance between them being N, for such a case 4 points satisfying given condition cannot be found.

Follow the steps mentioned below to solve the problem:

  • Check if N is odd or even.
  • If N is odd then such 4 points cannot be found.
  • Otherwise set the four points as {N/2, 0}, {-N/2, 0}, {0, N/2}, {0, -N/2}

Below is the implementation of the above approach.

C++




// C++ code for above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to find 4 points such that
// manhattan distance between
// any two of them is equal
vector<pair<int, int> > findPoints(int N)
{
    // Initializing vector of pairs to
    // store the 4 pairs
    vector<pair<int, int> > points;
 
    // If N is odd, it is impossible
    // to find 4 such points
    if (N % 2 == 1)
        return points;
 
    // Initializing a variable
    // with value equal to N/2
    int point = N / 2;
 
    // Pushing all the 4 pairs into
    // vector "points" such that distance
    // between any two is equal
    points.push_back({ 0, point });
    points.push_back({ 0, -point });
    points.push_back({ point, 0 });
    points.push_back({ -point, 0 });
 
    // Returning "points" vector
    return points;
}
 
// Function to print
void print(int N)
{
    vector<pair<int, int> > ans
        = findPoints(N);
    if (ans.size() == 0)
        cout << -1;
    for (int i = 0; i < ans.size(); i++) {
        cout << ans[i].first << ", "
             << ans[i].second << "\n";
    }
}
 
// Driver Code
int main()
{
    int N = 6;
 
    // Calling the print function
    print(N);
    return 0;
}

Java




// Java code for above approach
import java.io.*;
 
class GFG {
 
  // Function to find 4 points such that
  // manhattan distance between
  // any two of them is equal
  static int[][] findPoints(int N)
  {
 
    // Initializing vector of pairs to
    // store the 4 pairs
    int[][]points=new int[4][2];
 
    // If N is odd, it is impossible
    // to find 4 such points
    if (N % 2 == 1)
      return points;
 
    // Initializing a variable
    // with value equal to N/2
    int point = N / 2;
 
    // Pushing all the 4 pairs into
    // vector "points" such that distance
    // between any two is equal
    points[0][0] = 0;
    points[0][ 1] = point;
    points[1][0] = 0;
    points[1][1] = -point;
    points[2][0] = point;
    points[2][1] = 0;
    points[3][0] = -point;
    points[3][1] = 0;
 
    // Returning "points" vector
    return points;
  }
 
  // Function to print
  static void print(int N)
  {
    int[][] ans = findPoints(N);
    if (ans.length == 0)
      System.out.print(-1);
    for (int i = 0; i < ans.length; i++) {
      System.out.println(ans[i][0] + ", " + ans[i][1]);
    }
  }
 
  // Driver Code
  public static void main (String[] args) {
    int N = 6;
 
    // Calling the print function
    print(N);
  }
}
 
// This code is contributed by hrithikgarg03188.

Python3




# Python code for the above approach
 
# Function to find 4 points such that
# manhattan distance between
# any two of them is equal
def findPoints(N):
   
    # Initializing vector of pairs to
    # store the 4 pairs
    points = []
 
    # If N is odd, it is impossible
    # to find 4 such points
    if (N % 2 == 1):
        return points
 
    # Initializing a variable
    # with value equal to N/2
    point = (N // 2)
 
    # Pushing all the 4 pairs into
    # vector "points" such that distance
    # between any two is equal
    points.append([0,point ])
    points.append([0,-1 * point ])
    points.append([point,0 ])
    points.append([-1 * point,0 ])
 
    # Returning "points" vector
    return points
 
# Function to print
def Print(N):
    ans = findPoints(N)
    if (len(ans) == 0):
        print(-1)
    for i in range(len(ans)):
        print(str(ans[i][0]) + ", " + str(ans[i][1]))
 
# Driver Code
N = 6
 
# Calling the print function
Print(N)
 
# This code is contributed by shinjanpatra

C#




// C# code for above approach
using System;
class GFG {
 
    // Function to find 4 points such that
    // manhattan distance between
    // any two of them is equal
    static int[, ] findPoints(int N)
    {
       
        // Initializing vector of pairs to
        // store the 4 pairs
        int[, ] points = new int[4, 2];
 
        // If N is odd, it is impossible
        // to find 4 such points
        if (N % 2 == 1)
            return points;
 
        // Initializing a variable
        // with value equal to N/2
        int point = N / 2;
 
        // Pushing all the 4 pairs into
        // vector "points" such that distance
        // between any two is equal
        points[0, 0] = 0;
        points[0, 1] = point;
        points[1, 0] = 0;
        points[1, 1] = -point;
        points[2, 0] = point;
        points[2, 1] = 0;
        points[3, 0] = -point;
        points[3, 1] = 0;
 
        // Returning "points" vector
        return points;
    }
 
    // Function to print
    static void print(int N)
    {
        int[, ] ans = findPoints(N);
        if (ans.GetLength(0) == 0)
            Console.Write(-1);
        for (int i = 0; i < ans.GetLength(0); i++) {
            Console.WriteLine(ans[i, 0] + ", " + ans[i, 1]);
        }
    }
 
    // Driver Code
    public static void Main()
    {
        int N = 6;
 
        // Calling the print function
        print(N);
    }
}
 
// This code is contributed by Samim Hossain Mondal.

Javascript




<script>
    // JavaScript code for the above approach
 
    // Function to find 4 points such that
    // manhattan distance between
    // any two of them is equal
    function findPoints(N) {
        // Initializing vector of pairs to
        // store the 4 pairs
        let points = [];
 
        // If N is odd, it is impossible
        // to find 4 such points
        if (N % 2 == 1)
            return points;
 
        // Initializing a variable
        // with value equal to N/2
        let point = Math.floor(N / 2);
 
        // Pushing all the 4 pairs into
        // vector "points" such that distance
        // between any two is equal
        points.push({ first: 0, second: point });
        points.push({ first: 0, second: -1 * point });
        points.push({ first: point, second: 0 });
        points.push({ first: -1 * point, second: 0 });
 
        // Returning "points" vector
        return points;
    }
 
    // Function to print
    function print(N) {
        let ans = findPoints(N);
        if (ans.length == 0)
            document.write(-1);
        for (let i = 0; i < ans.length; i++) {
            document.write(ans[i].first + ", "
                + ans[i].second + '<br>');
        }
    }
 
    // Driver Code
    let N = 6;
 
    // Calling the print function
    print(N);
 
    // This code is contributed by Potta Lokesh
</script>

Output

0, 3
0, -3
3, 0
-3, 0

Time Complexity: O(1)
Auxiliary Space: O(1)


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