Check if a line at 45 degree can divide the plane into two equal weight parts
Given a set of n points (xi, yi) in 2D coordinate. Each point has some weight wi. The task is to check whether a line at 45 degrees can be drawn so that sum of weights of points on each side is equal.
Examples:
Input : x1 = -1, y1 = 1, w1 = 3
x2 = -2, y2 = 1, w2 = 1
x3 = 1, y3 = -1, w3 = 4
Output : Yes
Input : x1 = 1, y1 = 1, w1 = 2
x2 = -1, y2 = 1, w2 = 1
x3 = 1, y3 = -1, w3 = 2
Output : No
First, let’s try to solve the above problem for a vertical line i.e if a line x = i can divide the plane into two-part such that the sum of weight at each side is equal.
Observe, multiple points with the same x-coordinate can be treated as one point with a weight equal to the sum of weights of all points with the same x-coordinate.
Now, traverse through all x-coordinates from the minimum x-coordinate to maximum x-coordinate. So, make an array prefix_sum[], which will store the sum of weights till the point x = i.
So, there can be two options for which the answer can be ‘Yes’:
- Either prefix_sum[1, 2, …, i-1] = prefix_sum[i+1, …, n]
- or there exist a point i such that a line passes somewhere in between
x = i and x = i+1 and prefix_sum[1, …, i] = prefix_sum[i+1, …, n],
where prefix_sum[i, …, j] is the sum of weight of points from i to j.
int is_possible = false;
for (int i = 1; i < prefix_sum.size(); i++)
if (prefix_sum[i] == total_sum - prefix_sum[i])
is_possible = true
if (prefix_sum[i-1] == total_sum - prefix_sum[i])
is_possible = true
Now, to solve for a line at 45 degrees, we will rotate each point by 45 degrees.
Refer: 2D Transformation or Rotation of objects
So, point at (x, y), after 45 degree rotation will become ((x – y)/sqrt(2), (x + y)/sqrt(2)).
We can ignore the sqrt(2) since it is the scaling factor. Also, we don’t need to care about y-coordinate after rotation because a vertical line cannot distinguish between the point having the same x-coordinate. (x, y1) and (x, y2) will lie to the right, left or on any line of the form x = k.
C++
#include <bits/stdc++.h>
using namespace std;
void is_partition_possible( int n, int x[],
int y[], int w[])
{
map< int , int > weight_at_x;
int max_x = -2e3, min_x = 2e3;
for ( int i = 0; i < n; i++) {
int new_x = x[i] - y[i];
max_x = max(max_x, new_x);
min_x = min(min_x, new_x);
weight_at_x[new_x] += w[i];
}
vector< int > sum_till;
sum_till.push_back(0);
for ( int x = min_x; x <= max_x; x++) {
sum_till.push_back(sum_till.back() +
weight_at_x[x]);
}
int total_sum = sum_till.back();
int partition_possible = false ;
for ( int i = 1; i < sum_till.size(); i++) {
if (sum_till[i] == total_sum - sum_till[i])
partition_possible = true ;
if (sum_till[i - 1] == total_sum - sum_till[i])
partition_possible = true ;
}
printf (partition_possible ? "YES\n" : "NO\n" );
}
int main()
{
int n = 3;
int x[] = { -1, -2, 1 };
int y[] = { 1, 1, -1 };
int w[] = { 3, 1, 4 };
is_partition_possible(n, x, y, w);
return 0;
}
|
Java
import java.util.*;
class GFG
{
static void is_partition_possible( int n, int x[],
int y[], int w[])
{
Map<Integer, Integer> weight_at_x = new HashMap<Integer, Integer>();
int max_x = ( int ) -2e3, min_x = ( int ) 2e3;
for ( int i = 0 ; i < n; i++)
{
int new_x = x[i] - y[i];
max_x = Math.max(max_x, new_x);
min_x = Math.min(min_x, new_x);
if (weight_at_x.containsKey(new_x))
{
weight_at_x.put(new_x, weight_at_x.get(new_x) + w[i]);
}
else
{
weight_at_x.put(new_x,w[i]);
}
}
Vector<Integer> sum_till = new Vector<>();
sum_till.add( 0 );
for ( int s = min_x; s <= max_x; s++)
{
if (weight_at_x.get(s) == null )
sum_till.add(sum_till.lastElement());
else
sum_till.add(sum_till.lastElement() +
weight_at_x.get(s));
}
int total_sum = sum_till.lastElement();
int partition_possible = 0 ;
for ( int i = 1 ; i < sum_till.size(); i++)
{
if (sum_till.get(i) == total_sum - sum_till.get(i))
partition_possible = 1 ;
if (sum_till.get(i- 1 ) == total_sum - sum_till.get(i))
partition_possible = 1 ;
}
System.out.printf(partition_possible == 1 ? "YES\n" : "NO\n" );
}
public static void main(String[] args)
{
int n = 3 ;
int x[] = { - 1 , - 2 , 1 };
int y[] = { 1 , 1 , - 1 };
int w[] = { 3 , 1 , 4 };
is_partition_possible(n, x, y, w);
}
}
|
Python3
from collections import defaultdict
def is_partition_possible(n, x, y, w):
weight_at_x = defaultdict( int )
max_x = - 2e3
min_x = 2e3
for i in range (n):
new_x = x[i] - y[i]
max_x = max (max_x, new_x)
min_x = min (min_x, new_x)
weight_at_x[new_x] + = w[i]
sum_till = []
sum_till.append( 0 )
for x in range (min_x, max_x + 1 ):
sum_till.append(sum_till[ - 1 ] +
weight_at_x[x])
total_sum = sum_till[ - 1 ]
partition_possible = False
for i in range ( 1 , len (sum_till)):
if (sum_till[i] = = total_sum - sum_till[i]):
partition_possible = True
if (sum_till[i - 1 ] = = total_sum - sum_till[i]):
partition_possible = True
if partition_possible:
print ( "YES" )
else :
print ( "NO" )
if __name__ = = "__main__" :
n = 3
x = [ - 1 , - 2 , 1 ]
y = [ 1 , 1 , - 1 ]
w = [ 3 , 1 , 4 ]
is_partition_possible(n, x, y, w)
|
C#
using System;
using System.Collections.Generic;
public class GFG{
static void is_partition_possible( int n, int [] x, int [] y, int [] w)
{
Dictionary< int , int > weight_at_x = new Dictionary< int , int >();
int max_x = ( int ) -2e3, min_x = ( int ) 2e3;
for ( int i = 0; i < n; i++)
{
int new_x = x[i] - y[i];
max_x = Math.Max(max_x, new_x);
min_x = Math.Min(min_x, new_x);
if (weight_at_x.ContainsKey(new_x))
{
weight_at_x[new_x]+=w[i];
}
else
{
weight_at_x.Add(new_x,w[i]);
}
}
List< int > sum_till = new List< int >();
sum_till.Add(0);
for ( int s = min_x; s <= max_x; s++)
{
if (!weight_at_x.ContainsKey(s))
{
sum_till.Add(sum_till[sum_till.Count - 1]);
}
else
{
sum_till.Add(sum_till[sum_till.Count-1] + weight_at_x[s]);
}
}
int total_sum = sum_till[sum_till.Count-1];
int partition_possible = 0;
for ( int i = 1; i < sum_till.Count; i++)
{
if (sum_till[i] == total_sum - sum_till[i])
partition_possible = 1;
if (sum_till[i-1] == total_sum - sum_till[i])
partition_possible = 1;
}
Console.WriteLine(partition_possible == 1 ? "YES" : "NO" );
}
static public void Main (){
int n = 3;
int [] x = { -1, -2, 1 };
int [] y = { 1, 1, -1 };
int [] w = { 3, 1, 4 };
is_partition_possible(n, x, y, w);
}
}
|
Javascript
<script>
function is_partition_possible(n,x,y,w)
{
let weight_at_x = new Map();
let max_x = -2e3, min_x = 2e3;
for (let i = 0; i < n; i++)
{
let new_x = x[i] - y[i];
max_x = Math.max(max_x, new_x);
min_x = Math.min(min_x, new_x);
if (weight_at_x.has(new_x))
{
weight_at_x.set(new_x, weight_at_x.get(new_x)
+ w[i]);
}
else
{
weight_at_x.set(new_x,w[i]);
}
}
let sum_till = [];
sum_till.push(0);
for (let s = min_x; s <= max_x; s++)
{
if (weight_at_x.get(s) == null )
sum_till.push(sum_till[sum_till.length-1]);
else
sum_till.push(sum_till[sum_till.length-1] +
weight_at_x.get(s));
}
let total_sum = sum_till[sum_till.length-1];
let partition_possible = 0;
for (let i = 1; i < sum_till.length; i++)
{
if (sum_till[i] == total_sum - sum_till[i])
partition_possible = 1;
if (sum_till[i-1] == total_sum - sum_till[i])
partition_possible = 1;
}
document.write(partition_possible == 1 ? "YES\n" : "NO\n" );
}
let n = 3;
let x=[ -1, -2, 1 ];
let y=[1, 1, -1 ];
let w=[ 3, 1, 4 ];
is_partition_possible(n, x, y, w);
</script>
|
Output:
Yes
Time Complexity: O(nlogn + max) where max = 4*103
Auxiliary Space: O(n + max)
Last Updated :
23 Jun, 2022
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