Given an array arr[] consisting of (4 * N + 1) pairs of coordinates representing the coordinates of the corners of any N squares such that only one coordinate doesn't belong to any square, the task is to find that coordinate that doesn't belong to any square.
Examples:
Input: N = 2, arr[] = { {0, 0}, {0, 1}, {0, 2}, {1, 0}, {1, 1}, {1, 2}, {2, 0}, {2, 1}, {2, 2} }
Output: 1 1
Explanation:
The square has four sides: x = 0, x = 2, y = 0, y = 2, now all the points belong to the square except one point (1, 1).Input: N = 2, arr[] = { {0, 0}, {0, 1}, {0, 2}, {1, 0}, {0, 3}, {1, 2}, {2, 0}, {2, 1}, {2, 2} }
Output: 0 3
Approach: The given problem can be solved based on the following observations:
- The coordinates of the sides of the square will appear at least two times because N ? 2. Therefore, since there is only one point not on the boundary, the maximum between x-coordinates which appear at least twice will give us the x-coordinate of the right side of the square.
- The other three sides can be obtained similarly with different combinations of maximum/minimum and x-/ycoordinates.
- After knowing the sides of the square, it is easy to identify the point not on the boundary.
Follow the steps below to solve the problem :
- Iterate over a range [0, N] using the variable i and perform the following steps:
- Initialize the variables x1, y1 with 2e9 and x2, y2 with -2e9 to store the points of the upper and lower boundary of the square.
- Iterate over a range [0, N] using the variable j and perform the following steps:
- If i is not equal to j, then, perform the following steps:
- Set x1 to the minimum of x1 or p[j].first.
- Set x2 to the maximum of x2 or p[j].first.
- Set y1 to the minimum of y1 or p[j].second.
- Set y2 to the minimum of y2 or p[j].second.
- If i is not equal to j, then, perform the following steps:
- Initialize a variable, say ok to true and variables cnt1, cnt2, cnt3, cnt4 as 0 to store the count of points with maximum and minimum as x1, x2, y1, y2.
- Iterate over a range [0, N] using the variable j and perform the following steps:
- If i is not equal to j, then, perform the following steps:
- If p[j].first is equal to x1, then, increase the value of cnt1 by 1.
- If p[j].first is equal to x2, then, increase the value of cnt2 by 1.
- If p[j].second is equal to y1, then, increase the value of cnt3 by 1.
- If p[j].second is equal to y2, then, increase the value of cnt4 by 1.
- Else, set the value of ok to false.
- If i is not equal to j, then, perform the following steps:
- If the value of ok is true and the values of cnt1, cnt2, cnt3, cnt4 are greater than equal to N, and x2 - x2 is equal to y2 - y1, then, p[i] is the required point. Print the answer.
Below is the implementation of the above approach:
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define fi first
#define se second
// Function to find the point that is
// not a part of the side of a square
void findPoint(int n,
vector<pair<int, int> > p)
{
// Traverse each pair of coordinates
for (int i = 0; i < n * 4 + 1; ++i) {
int x1 = 2e9, x2 = -2e9;
int y1 = 2e9, y2 = -2e9;
for (int j = 0; j < n * 4 + 1; ++j)
if (i != j) {
// Minimize x-coordinate
// in all the points
// except current point
x1 = min(x1, p[j].fi);
// Maximize x-coordinate in
// all the points except
// the current point
x2 = max(x2, p[j].fi);
// Minimize y-coordinate
// in all the points
// except current point
y1 = min(y1, p[j].se);
// Maximize y-coordinate
// in all the points
// except current point
y2 = max(y2, p[j].se);
}
bool ok = 1;
int c1 = 0, c2 = 0;
int c3 = 0, c4 = 0;
for (int j = 1; j < n * 4 + 1; ++j)
if (i != j) {
// If x-coordinate matches
// with other same line
if ((p[j].fi == x1
|| p[j].fi == x2)
|| ((p[j].se == y1
|| p[j].se == y2))) {
if (p[j].fi == x1)
++c1;
if (p[j].fi == x2)
++c2;
// If y coordinate matches
// with other same line
if (p[j].se == y1)
++c3;
if (p[j].se == y2)
++c4;
}
else
ok = 0;
}
// Check if the condition
// for square exists or not
if (ok && c1 >= n && c2 >= n
&& c3 >= n && c4 >= n
&& x2 - x1 == y2 - y1) {
// Print the output
cout << p[i].fi << " "
<< p[i].se << "\n";
}
}
}
// Driver Code
int main()
{
int N = 2;
vector<pair<int, int> > arr
= { { 0, 0 }, { 0, 1 }, { 0, 2 },
{ 1, 0 }, { 1, 1 }, { 1, 2 },
{ 2, 0 }, { 2, 1 }, { 2, 2 } };
findPoint(N, arr);
return 0;
}
import java.util.ArrayList;
//Java program for above approach
class GFG{
static class pair<T, V>{
T fi;
V se;
pair(T a, V b){
this.fi = a;
this.se = b;
}
}
// Function to find the point that is
// not a part of the side of a square
static void findPoint(int n,
ArrayList<pair<Integer, Integer> > p)
{
// Traverse each pair of coordinates
for (int i = 0; i < n * 4 + 1; ++i) {
int x1 = (int) 2e9, x2 = (int) -2e9;
int y1 = (int) 2e9, y2 = (int) -2e9;
for (int j = 0; j < n * 4 + 1; ++j)
if (i != j) {
// Minimize x-coordinate
// in all the points
// except current point
x1 = Math.min(x1, p.get(j).fi);
// Maximize x-coordinate in
// all the points except
// the current point
x2 = Math.max(x2, p.get(j).fi);
// Minimize y-coordinate
// in all the points
// except current point
y1 = Math.min(y1, p.get(j).se);
// Maximize y-coordinate
// in all the points
// except current point
y2 = Math.max(y2, p.get(j).se);
}
boolean ok = true;
int c1 = 0, c2 = 0;
int c3 = 0, c4 = 0;
for (int j = 1; j < n * 4 + 1; ++j)
if (i != j) {
// If x-coordinate matches
// with other same line
if ((p.get(j).fi == x1
|| p.get(j).fi == x2)
|| ((p.get(j).se == y1
|| p.get(j).se == y2))) {
if (p.get(j).fi == x1)
++c1;
if (p.get(j).fi == x2)
++c2;
// If y coordinate matches
// with other same line
if (p.get(j).se == y1)
++c3;
if (p.get(j).se == y2)
++c4;
}
else
ok = false;
}
// Check if the condition
// for square exists or not
if (ok && c1 >= n && c2 >= n
&& c3 >= n && c4 >= n
&& x2 - x1 == y2 - y1) {
// Print the output
System.out.println(p.get(i).fi + " " + p.get(i).se);
}
}
}
//Driver Code
public static void main(String[] args) {
int N = 2;
ArrayList<pair<Integer, Integer> > arr = new ArrayList<>();
arr.add(new pair<Integer, Integer>(0,0));
arr.add(new pair<Integer, Integer>(0,1));
arr.add(new pair<Integer, Integer>(0,2));
arr.add(new pair<Integer, Integer>(1,0));
arr.add(new pair<Integer, Integer>(1,1));
arr.add(new pair<Integer, Integer>(1,2));
arr.add(new pair<Integer, Integer>(2,0));
arr.add(new pair<Integer, Integer>(2,1));
arr.add(new pair<Integer, Integer>(2,2));
findPoint(N, arr);
}
}
// This code is contributed by hritikrommie.
# Python 3 program for the above approach
# Function to find the point that is
# not a part of the side of a square
def findPoint(n, p):
# Traverse each pair of coordinates
for i in range(n * 4 + 1):
x1 = 2e9
x2 = -2e9
y1 = 2e9
y2 = -2e9
for j in range(n * 4 + 1):
if (i != j):
# Minimize x-coordinate
# in all the points
# except current point
x1 = min(x1, p[j][0])
# Maximize x-coordinate in
# all the points except
# the current point
x2 = max(x2, p[j][0])
# Minimize y-coordinate
# in all the points
# except current point
y1 = min(y1, p[j][1])
# Maximize y-coordinate
# in all the points
# except current point
y2 = max(y2, p[j][1])
ok = 1
c1 = 0
c2 = 0
c3 = 0
c4 = 0
for j in range(1,n * 4 + 1,1):
if (i != j):
# If x-coordinate matches
# with other same line
if ((p[j][0] == x1 or p[j][0] == x2) or (p[j][1] == y1 or p[j][1] == y2)):
if (p[j][0] == x1):
c1 += 1
if (p[j][0] == x2):
c2 += 1
# If y coordinate matches
# with other same line
if (p[j][1] == y1):
c3 += 1
if (p[j][1] == y2):
c4 += 1
else:
ok = 0
# Check if the condition
# for square exists or not
if (ok and c1 >= n and c2 >= n and c3 >= n and c4 >= n and x2 - x1 == y2 - y1):
# Print the output
print(p[i][0],p[i][1])
# Driver Code
if __name__ == '__main__':
N = 2
arr = [[0, 0],[0, 1],[0, 2],[1, 0],[1, 1],[1, 2],[2, 0],[2, 1],[2, 2]]
findPoint(N, arr)
# This code is contributed by ipg2016107.
// C# code to implement the approach
using System;
using System.Collections.Generic;
class GFG {
// pair class definition
public class pair<T, V> {
public T fi;
public V se;
// constructor
public pair(T a, V b)
{
this.fi = a;
this.se = b;
}
}
// Function to find the point that is
// not a part of the side of a square
static void findPoint(int n, List<pair<int, int> > p)
{
// Traverse each pair of coordinates
for (int i = 0; i < n * 4 + 1; ++i) {
int x1 = int.MaxValue, x2 = int.MinValue;
int y1 = int.MaxValue, y2 = int.MinValue;
for (int j = 0; j < n * 4 + 1; ++j)
if (i != j) {
// Minimize x-coordinate
// in all the points
// except current point
x1 = Math.Min(x1, p[j].fi);
// Maximize x-coordinate in
// all the points except
// the current point
x2 = Math.Max(x2, p[j].fi);
// Minimize y-coordinate
// in all the points
// except current point
y1 = Math.Min(y1, p[j].se);
// Maximize y-coordinate
// in all the points
// except current point
y2 = Math.Max(y2, p[j].se);
}
bool ok = true;
int c1 = 0, c2 = 0;
int c3 = 0, c4 = 0;
for (int j = 1; j < n * 4 + 1; ++j)
if (i != j) {
// If x-coordinate matches
// with other same line
if ((p[j].fi == x1 || p[j].fi == x2)
|| ((p[j].se == y1
|| p[j].se == y2))) {
if (p[j].fi == x1)
++c1;
if (p[j].fi == x2)
++c2;
// If y coordinate matches
// with other same line
if (p[j].se == y1)
++c3;
if (p[j].se == y2)
++c4;
}
else
ok = false;
}
// Check if the condition
// for square exists or not
if (ok && c1 >= n && c2 >= n && c3 >= n
&& c4 >= n && x2 - x1 == y2 - y1) {
// Print the output
Console.WriteLine(p[i].fi + " " + p[i].se);
}
}
}
// Driver Code
static void Main(string[] args)
{
int N = 2;
List<pair<int, int> > arr
= new List<pair<int, int> >();
arr.Add(new pair<int, int>(0, 0));
arr.Add(new pair<int, int>(0, 1));
arr.Add(new pair<int, int>(0, 2));
arr.Add(new pair<int, int>(1, 0));
arr.Add(new pair<int, int>(1, 1));
arr.Add(new pair<int, int>(1, 2));
arr.Add(new pair<int, int>(2, 0));
arr.Add(new pair<int, int>(2, 1));
arr.Add(new pair<int, int>(2, 2));
// Function call
findPoint(N, arr);
}
}
// This code is contributed by phasing17
// JavaScript program for the above approach
// Function to find the point that is
// not a part of the side of a square
function findPoint(n, p)
{
// Traverse each pair of coordinates
for (let i = 0; i < n * 4 + 1; ++i) {
let x1 = 2e9, x2 = -2e9;
let y1 = 2e9, y2 = -2e9;
for (let j = 0; j < n * 4 + 1; ++j){
if (i != j) {
// Minimize x-coordinate
// in all the points
// except current point
x1 = Math.min(x1, p[j][0]);
// Maximize x-coordinate in
// all the points except
// the current point
x2 = Math.max(x2, p[j][0]);
// Minimize y-coordinate
// in all the points
// except current point
y1 = Math.min(y1, p[j][1]);
// Maximize y-coordinate
// in all the points
// except current point
y2 = Math.max(y2, p[j][1]);
}
}
let ok = 1;
let c1 = 0, c2 = 0;
let c3 = 0, c4 = 0;
for (let j = 1; j < n * 4 + 1; ++j){
if (i != j) {
// If x-coordinate matches
// with other same line
if ((p[j][0] == x1 || p[j][0] == x2) || (p[j][1] == y1 || p[j][1] == y2)) {
if (p[j][0] == x1)
++c1;
if (p[j][0] == x2)
++c2;
// If y coordinate matches
// with other same line
if (p[j][1] == y1)
++c3;
if (p[j][1] == y2)
++c4;
}
else{
ok = 0;
}
}
}
// Check if the condition
// for square exists or not
if (ok && c1 >= n && c2 >= n && c3 >= n && c4 >= n && x2 - x1 == y2 - y1) {
// Print the output
console.log(p[i][0] + " " + p[i][1]);
}
}
}
// Driver Code
let N = 2;
let arr = [[0, 0], [0, 1], [0, 2],
[1, 0], [1, 1], [1, 2 ],
[2, 0], [2, 1], [2, 2]];
findPoint(N, arr);
// The code is contributed by Gautam goel(gautamgoel962)
Output:
1 1
Time Complexity: O(N2)
Auxiliary Space: O(1)