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Home Code Solutions Hackerrank Algorithms

Points in a Plane – HackerRank Solution

Points in a Plane - HackerRank Solution Java , Python 3, Python 2 , C , C++, Best and Optimal Solutions , All you need.

bhautik bhalala by bhautik bhalala
May 31, 2022
Reading Time: 1 min read
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Table of Contents

  • Points in a Plane – HackerRank Solution Java , Python 3, Python 2 , C , C++, Best and Optimal Solutions , All you need.
  • Solutions of Algorithms Data Structures Hard HackerRank:
    • Here are all the Solutions of Hard , Advanced , Expert Algorithms of Data Structure of Hacker Rank , Leave a comment for similar posts
  • C++ Points in a Plane HackerRank Solution
  • Java Points in a Plane HackerRank Solution
  • Python 3 Points in a Plane HackerRank Solution
  • Python 2 Points in a Plane HackerRank Solution
  • C Points in a Plane HackerRank Solution
    • Warmup Implementation Strings Sorting Search Graph Theory Greedy Dynamic Programming Constructive Algorithms Bit Manipulation Recursion Game Theory NP Complete Debugging
    • Leave a comment below
      • Related posts:

Points in a Plane – HackerRank Solution Java , Python 3, Python 2 , C , C++, Best and Optimal Solutions , All you need.

Solutions of Algorithms Data Structures Hard HackerRank:

Here are all the Solutions of Hard , Advanced , Expert Algorithms of Data Structure of Hacker Rank , Leave a comment for similar posts

C++ Points in a Plane HackerRank Solution


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/* Enter your code here. Read input from STDIN. Print output to STDOUT */#include<iostream>
#include<set>
#include<map>
#include<string>
#include<stdio.h>
#include<sstream>
#include<algorithm>
#include<queue>
#include<cmath>
#include<string.h>
using namespace std ;
#define MOD 1000000007
#define INF (int)1e9
#define MAXN 20
typedef pair<int,int> P ;

int n,pre[MAXN],fac[MAXN],x[MAXN],y[MAXN] ;
int col[MAXN][MAXN] ;

char bit[1 << MAXN] ;
char good[1 << MAXN] ;
char best[1 << MAXN] ;
char valid[1 << MAXN] ;

char vid,id[1 << MAXN] ;
int memo[1 << MAXN] ;
int solve(int mask)
{
 if(bit[mask] <= 2) return 1 ;
 if(id[mask] == vid) return memo[mask] ;
 if(good[mask]) return pre[bit[mask]] ;
 id[mask] = vid ;

 int j,nmask = mask ;
 for(j = 0;j < n;j++) if(mask & 1 << j)
 {
  nmask ^= 1 << j ;
  break ;
 }

 int ways = 0,can = best[mask] ;
 if(best[nmask] == can - 1) ways = solve(nmask) ;
 for(int i = nmask;i > 0;i = ((i - 1) & nmask))
 {
  int k = i | 1 << j ;
  if(valid[k] && best[mask ^ k] == can - 1)
  {
   ways += solve(mask ^ k) ;
   if(ways >= MOD) ways -= MOD ;
  }
 }
 return memo[mask] = ways ;
}

void generate()
{
 for(int tt = 0;tt < 10;tt++)
 {
  char in[] = "in .txt" ;
  in[2] = tt + '0' ;
  FILE * fout = fopen(in,"w") ;
  
  int runs = 50 ;
  fprintf(fout,"%d\n",runs) ;
  for(int j = 0;j < runs;j++)
  {
   n = rand() % 18 ;
   if(tt == 8) n = 18 ;
   
   char vis[102][102] ;
   memset(vis,0,sizeof vis) ;
   for(int i = 0;i < n;i++)
   {
    if(tt < 3 && j < 15)
    {
     x[i] = rand() % 10 ;
     y[i] = rand() % 10 ;     
    }
    else if(tt < 6 && j < 15)
    {
     x[i] = rand() % 5 ;
     y[i] = rand() % 5 ;
    }
    else if(tt < 10 && j < 15)
    {
     x[i] = i ;
     y[i] = i + (rand() % 5 - 2) ;
     if(y[i] < 0) y[i] = i ;
    }
    else
    {
     x[i] = rand() % 100 + 1 ;
     y[i] = rand() % 100 + 1 ;
    }
    
    if(vis[x[i]][y[i]]) { i-- ; continue ; }
    vis[x[i]][y[i]] = 1 ;
   }


   fprintf(fout,"%d\n",n) ;
   for(int i = 0;i < n;i++) fprintf(fout,"%d %d\n",x[i],y[i]) ;
  }
  fclose(fout) ;
 }
}

int main()
{
 fac[0] = 1 ;
 for(int i = 1;i < MAXN;i++) fac[i] = 1LL * i * fac[i - 1] % MOD ;
 for(int i = 1;i < 1 << MAXN;i++) bit[i] = bit[i >> 1] + (i & 1) ;
 pre[0] = pre[1] = 1 ;
 for(int i = 2;i < MAXN;i++)
 {
  pre[i] = 1LL * pre[i - 2] * (i - 1) % MOD ;
  if(i % 2 == 1) pre[i] += pre[i - 1] ;
  pre[i] %= MOD ;
 }

// generate() ; return 0 ;

 int runs ; 
 scanf("%d",&runs) ;
 while(runs--)
 {
  scanf("%d",&n) ;
  for(int i = 0;i < n;i++) scanf("%d%d",&x[i],&y[i]) ;
  
  memset(col,0,sizeof col) ;
  for(int k1 = 0;k1 < n;k1++)
   for(int k2 = 0;k2 < n;k2++)
   {
    for(int j = 0;j < n;j++)
    {
     int area = x[j] * (y[k1] - y[k2]) + x[k1] * (y[k2] - y[j]) + x[k2] * (y[j] - y[k1]) ;
     if(area == 0) col[k1][k2] |= 1 << j ;
    }
   }
  
  for(int i = 0;i < 1 << n;i++)
  {
   if(bit[i] <= 2) { valid[i] = true ; continue ; }
   for(int j = 0;j < n;j++) if(i & 1 << j)
   {
    int k1 = -1 ;
    for(int k = j + 1;k < n;k++) if(i & 1 << k) { k1 = k ; break ; }
    if((col[j][k1] | i) == col[j][k1]) valid[i] = true ;
    else valid[i] = false ;
    break ;
   }
  }
  
  best[0] = 0 ;
  for(int i = 1;i < 1 << n;i++)
  {
   if(bit[i] == 1) { best[i] = 1 ; continue ; }
   int j;
   for(j = 0;j < n;j++) if(i & 1 << j) break ;
   
   int cret = n ;
   for(int k = j + 1;k < n;k++)
    if(i & 1 << k)
     cret = min(cret,1 + best[i & ~col[j][k]]) ;
   best[i] = cret ;
  }
  
  for(int i = 0;i < 1 << n;i++)
  {
   good[i] = 1 ;
   if(bit[i] <= 2) continue ;
   int j;
   for(j = 0;j < n;j++) if(i & 1 << j) break ;
   if(!good[i ^ 1 << j]) { good[i] = 0 ; continue ; }
   
   for(int k = j + 1;k < n;k++)
    if(i & 1 << k)
     if(bit[i & col[j][k]] > 2)
      good[i] = 0 ;
  }
  
  int tot = best[(1 << n) - 1] ;
  vid++ ;
  int ret = solve((1 << n) - 1) ;
  ret = 1LL * ret * fac[tot] % MOD ;
  printf("%d %d\n",tot,ret) ;
 }
 
 return 0 ;
}

Java Points in a Plane HackerRank Solution


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import java.util.*;
import java.io.*;

class Solution
{
	BufferedReader input;
	BufferedWriter out;
	StringTokenizer token;

	int N;
	int[] x,y;
	int[] dp,dp3;
	boolean[] ok;
	int[] member;
	int mod = 1000000007;

	int BitCount(int x)
	{
		int ret = 0;
		while(x > 0)
		{
			if( (x&1) != 0 ) ret++;
			x >>= 1;
		}
		return ret;
	}

	boolean collinear(int set)
	{
		int ctr = 0;
		for(int i = 0; set > 0; i++)
		{
			if( (set&1) != 0 )
				member[ctr++] = i;
			set >>= 1;
		}
		if(ctr <= 2)return true;
		int a = x[member[0]]-x[member[1]];
		int b = y[member[0]]-y[member[1]];
		for(int i = 2; i < ctr; i++)
		{
			int aa = x[member[0]]-x[member[i]];
			int bb = y[member[0]]-y[member[i]];
			if(aa*b != a*bb)return false;
		}
		return true;
	}

	String binary(int x)
	{
		String ret = "";
		for(int i = 0; i < N; i++)
		{
			if( ((x>>i)&1) == 0) ret = "0"+ret;
				else ret = "1"+ret;
		}
		return ret;
	}

	void solve() throws IOException
	{
		long qq = System.currentTimeMillis();
		input = new BufferedReader(new InputStreamReader(System.in));
		out = new BufferedWriter(new OutputStreamWriter(System.out));
		int T = nextInt();
		int twoMax = (1<<16);
		dp = new int[twoMax];
		x = new int[16];
		y = new int[16];
		ok = new boolean[twoMax];
		dp3 = new int[twoMax];
		member = new int[16];
		ArrayList<Integer> o;
		for(int t = 0; t < T; t++)
		{
			N = nextInt();
			int twoN = (1<<N);
			for(int i = 0; i < N; i++)
			{
				x[i] = nextInt();
				y[i] = nextInt();
			}
			o = new ArrayList<Integer>();
			for(int i = twoN-1; i > 0; i--)
			{
				ok[i] = false;
				if(collinear(i))
				{
					ok[i] = true;
					o.add(i);
				}
			}
			Arrays.fill(dp,-1);
			dp[0] = 0;
			dp3[0] = 1;
			int m = 0;
			for(int i = 0; i < o.size(); i++)
			{
				int ii = o.get(i);
				for(int j = m; j >= 0; j--)
				{
					if((ii&j) == 0 && dp[j] != -1)
					{
						m = Math.max(m,j|ii);
						if(dp[j|ii] == -1 || dp[j|ii] > 1+dp[j])
						{
							dp[j|ii] = 1+dp[j];
							dp3[j|ii] = (int)(((long)(dp[j]+1)*dp3[j])%mod);
						}
						else if(dp[j|ii] == 1+dp[j])
						{
							dp3[j|ii] += ((long)(dp[j]+1)*dp3[j])%mod;
							dp3[j|ii] %= mod;
						}
					}
				}
			}
			out.write(""+ dp[(twoN)-1] + " " + dp3[(twoN)-1]);
			out.newLine();
		}
		out.flush();
		out.close();
		input.close();

	}

	int nextInt() throws IOException
	{
		if(token == null || !token.hasMoreTokens())
			token = new StringTokenizer(input.readLine());
		return Integer.parseInt(token.nextToken());
	}

	Long nextLong() throws IOException
	{
		if(token == null || !token.hasMoreTokens())
			token = new StringTokenizer(input.readLine());
		return Long.parseLong(token.nextToken());
	}

	String next() throws IOException
	{
		if(token == null || !token.hasMoreTokens())
			token = new StringTokenizer(input.readLine());
		return token.nextToken();
	}

	public static void main(String[] args) throws Exception
	{
		new Solution().solve();
	}
}



Python 3 Points in a Plane HackerRank Solution


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Python 2 Points in a Plane HackerRank Solution


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# Enter your code here. Read input from STDIN. Print output to STDOUT


def constdr1(Points) :
	Droites = []	
	i = 0
	j = i + 1
	while j < len(Points) :
		S = True
		for D in Droites :
			if (Points[D[1]][0]-Points[D[0]][0])*(Points[j][1]-Points[i][1])-(Points[D[1]][1]-Points[D[0]][1])*(Points[j][0]-Points[i][0])== 0 :
				S=False
				if not (j in D) :
					D.append(j)
					break
		if S :
			Droites.append([i,j])
		j+=1
	return Droites

def sousparties(D) :
	N = len(D)
	D2 = D +[]
	if N==0 :
		return [[]]
	else :
		i =D2[0]
		D2.pop(0)
		Sol = sousparties(D2)
		Sol2 = []		
		for P in Sol :
			Sol2.append([i]+P)
		return Sol2 + Sol

def factorielle(n) :
	if n == 0 :
		return 1
	else :
		return n*factorielle(n-1)
								
def main(Points, d,nbp) :
	if len(Points)< 3 :
		nb= nbp + len(Points)
		k = (nb+1)/2
		if k > d :
			return [d,0]
		else :
			return [k,factorielle(2*k)/2**k]
	elif d == 0 :
		#print "plop2"
		return [0,0]
	else :
		Droites = constdr1(Points)
		i = 0
		sol = 0
		while i< len(Droites) :
			Droites[i].pop(0)
			for P in sousparties(Droites[i]) :
				if len(P) < 2 :
					continue
				j=1
				Points2 = []
				while j < len(Points) :
					if not j in P :
						Points2.append(Points[j])
					j+=1			
				A = main(Points2,d-1,nbp)
				if A[0]<d-1:
					d = A[0] +1
					sol = d*A[1]
				else :
					sol +=d*A[1]
			i+=1
		A = main(Points[1:],d, nbp+1)
		if A[0]<d:
			d = A[0]
			sol = A[1]
		else:
			sol +=A[1]
		#print sol
		return [d,int(sol %1000000007)]

T = map(eval, raw_input().split())					
i = 0
while i < T[0] :
	M = map(eval, raw_input().split())
	N= M[0]	
	j = 0
	points = []
	while j < N:
		X, Y = map(eval, raw_input().split())
		points += [[X,Y]]
		j+=1
	Sol = main(points, (N+1)/2,0)
	print Sol[0], Sol[1]
	i+=1



C Points in a Plane HackerRank Solution


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/* Enter your code here. Read input from STDIN. Print output to STDOUT */

#include <stdio.h>

#define P 1000000007

long long g=1,p[20][2],t,tt,v,kon,a[20][70000][2],b[70000],i,j,k,l,m,n,maz[70000],mmaz[70000];


void uloz(long long mam, long long ind, long long vv)
{
if(ind ==n) {mmaz[mam]=1;return;}

uloz(mam, ind+1,vv);

if(vv&(1<<ind)) uloz(mam+(1<<ind),ind+1,vv);

return;
}

void priamka(long long xx, long long yy)
{
long long vv=0,ii;

for(ii=0;ii<n;ii++)
  {
  vv*=2;

  if((p[ii][0]-p[xx][0])*(p[yy][1]-p[xx][1]) == (p[ii][1]-p[xx][1])*(p[yy][0]-p[xx][0]))
     vv++;
  }


if(maz[vv]==0) uloz(0,0,vv);

maz[vv]=1;

return;
}

/*
void makaj1(long long ind, long long ii, long long vv, long long mam, long long dl)
{

if(dl==n)
   {
      if(mam==0) return;

      if(mam > vv) {printf("chuj\n"); return;}

           if(b[mam]!=g)
           {
//           printf("..%lld: %lld %lld  %lld=vv\n",ind,ii,mam,vv);

             b[mam]=g;
             a[ind][ii^mam][1] = 1;
             a[ind][ii^mam][0] = (a[ind][ii^mam][0] + a[ind-1][ii][0])%P;
           }
    return;
   }

makaj1(ind,ii,vv,mam,dl+1);
 if(vv&(1<<dl)) makaj1(ind,ii,vv,mam+(1<<dl),dl+1);

return;
}
*/
void pocitaj(long long ind)
{
long long ii,jj,kk,min;

for(ii=0;ii<(1<<n);ii++) {a[ind][ii][0]=0;a[ind][ii][1]=0;}

for(ii=0;ii<(1<<n);ii++)
  if(a[ind-1][ii][1] && mmaz[ii])
    {
    a[ind][0][1] = 1;
    a[ind][0][0] = (a[ind][0][0] + a[ind-1][ii][0])%P;
    kon=1;
   }

//printf("%lld=kon\n",kon);

if(kon) return;

a[ind][0][0]=0;
a[ind][0][1]=0;



for(ii=0;ii<(1<<n);ii++)
  if(a[ind-1][ii][1])
    {
    g++;
    while(((1<<min)&ii) == 0) min++;

    for(jj=0;jj<l;jj++)
       // if(kk=(maz[jj]&ii)) makaj1(ind,ii,kk,0,0);

        if(b[kk=(ii&maz[jj])]!=g && (kk&(1<<min)))
           {
             b[kk]=g;
             a[ind][ii^kk][1] = 1;
             a[ind][ii^kk][0] = (a[ind][ii^kk][0] + a[ind-1][ii][0])%P;
           }


    }


//kon=1;

return;
}


int main()
{

scanf("%lld",&t);
for(tt=0;tt<t;tt++)
 {
 scanf("%lld",&n);
 for(i=0;i<n;i++) scanf("%lld %lld",&p[i][0],&p[i][1]);

 for(i=0;i<(1<<n);i++) {maz[i]=0;mmaz[i]=0;}


//  for(i=0;i<n;i++) {mmaz[(1<<i)]=1;maz[(1<<i)]=1;}
 if(n==1) {mmaz[1]=1;maz[1]=1;}



 for(i=0;i<n;i++)
  for(j=i+1;j<n;j++)
    {
    priamka(i,j);
 //   printf("%lld %lld -> %lld\n",i,j,priamka(i,j));
    }

for(i=0;i<(1<<n);i++) maz[i]=mmaz[i];

l=0;
  for(i=1;i<(1<<n);i++)
     if(maz[i]) maz[l++] = i;

//printf("%lld\n",l);

//for(i=0;i<l;i++) printf("%lld..\n",maz[i]);


 k=0;
 for(i=0;i<(1<<n);i++) a[0][i][1]=0;
 a[0][(1<<n)-1][1] = 1;
 a[0][(1<<n)-1][0] = 1;


 kon=0;
 i=0;

 while(kon==0)
   {
   i++;
   pocitaj(i);

//  for(j=0;j<(1<<n);j++) printf("%lld %lld---> %lld\n",i,j,a[i][j][0]);

   }

 v = a[i][0][0];
 for(j=2;j<=i;j++) v= (v*j)%P;

  printf("%lld %lld\n",i,v);
 }


return 0;
}

 

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