Swap Permutation


Problem Statement :


You are given an array A = [1, 2, 3, ..., n]:

1.How many sequences (S1) can you get after exact k adjacent swaps on A?

2.How many sequences (S2) can you get after at most k swaps on A?

An adjacent swap can be made between two elements of the Array A, A[i] and A[i+1] or A[i] and A[i-1].
A swap otherwise can be between any two elements of the array A[i] and A[j] ∀ 1 ≤ i, j ≤ N, i ≠ j.

Input Format

First and only line contains n and k separated by space.

Constraints

1 ≤ n ≤ 2500
1 ≤ k ≤ 2500

Output Format

Output S1 % MOD and S2 % MOD in one line, where MOD = 1000000007.


Solution :



title-img


                            Solution in C :

In C++ :





/*
*/

#pragma comment(linker, "/STACK:16777216")
#include <fstream>
#include <iostream>
#include <string>
#include <complex>
#include <math.h>
#include <set>
#include <vector>
#include <map>
#include <queue>
#include <stdio.h>
#include <stack>
#include <algorithm>
#include <list>
#include <ctime>
#include <memory.h>

#define y0 sdkfaslhagaklsldk
#define y1 aasdfasdfasdf
#define yn askfhwqriuperikldjk
#define j1 assdgsdgasghsf
#define tm sdfjahlfasfh
#define lr asgasgash

#define eps 1e-11
//#define M_PI 3.141592653589793
#define bs 1000000007
#define bsize 256

using namespace std;

long n,k,s[2600][2600],dp[2600][2600],lwr,upr,ts;
long long ans1,ans2;

int main(){
//freopen("trade.in","r",stdin);
//freopen("trade.out","w",stdout);
//freopen("C:/input.txt","r",stdin);
//freopen("C:/output.txt","w",stdout);
ios_base::sync_with_stdio(0);
//cin.tie(0);

cin>>n>>k;
dp[0][0]=1;
for (int i=0;i<=2500;i++)
 s[0][i]=1;
 
for (int i=0;i<n;i++)
{
 for (int j=0;j<=k;j++)
 {
  upr=j;
  lwr=j-(n-i-1);
  if (lwr<0)lwr=0;
  ts=s[i][upr];
  if (lwr>0)ts-=s[i][lwr-1];
  ts+=bs;
  ts%=bs;
  dp[i+1][j]=ts;
 }   
 for (int j=0;j<=k;j++)
 {
  s[i+1][j]=dp[i+1][j];
  if (j){s[i+1][j]=(s[i+1][j-1]+s[i+1][j])%bs;}
 }
}

ans1=0;
for (int i=k;i>=0;i-=2)
{ans1=(ans1+dp[n][i])%bs;}cout<<ans1;//<<endl;

for (int i=0;i<=n;i++)
 for (int j=0;j<=k;j++)
  dp[i][j]=s[i][j]=0;


dp[0][0]=1;
for (int i=0;i<=2500;i++)
 s[0][i]=1;
 
for (int i=0;i<n;i++)
{
 for (int j=0;j<=k;j++)
 {
  lwr=n-i-1;
  
  dp[i+1][j]=dp[i][j];
  if (j)dp[i+1][j]+=(1ll*dp[i][j-1]*lwr)%bs;
  dp[i+1][j]%=bs;
 }   
 for (int j=0;j<=k;j++)
 {
  s[i+1][j]=dp[i+1][j];
  if (j){s[i+1][j]=(s[i+1][j-1]+s[i+1][j])%bs;}
 }
}

ans1=0;
for (int i=0;i<=k;i++)
 ans1=(ans1+dp[n][i])%bs;
 cout<<" "<<ans1<<endl;
 
cin.get();cin.get();
return 0;}








In Java :





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

public class Solution {

    final static long mod = 1000000007;

    public static void solve(Input in, PrintWriter out) throws IOException {
        int n = in.nextInt();
        int k = in.nextInt();
        long[] d = new long[k + 1];
        d[0] = 1;
        for (int it = 0; it < n; ++it) {
            long[] dd = new long[k + 1];
            long sum = 0;
            for (int i = 0; i <= k; ++i) {
                sum = (sum + d[i]) % mod;
                dd[i] = sum;
                if (i >= it) {
                    sum = (sum + mod - d[i - it]) % mod;
                }
            }
            d = dd;
        }
//        System.err.println(Arrays.toString(d));
        long ans1 = 0;
        for (int i = 0; i <= k; ++i) {
            if (i % 2 == k % 2) {
                ans1 = (ans1 + d[i]) % mod;
            }
        }
        long[] d1 = new long[n + 1];
        d1[0] = 1;
        for (int it = 0; it < n; ++it) {
            for (int i = n; i >= 1; --i) {
                d1[i] = (d1[i] * it + d1[i - 1]) % mod;
            }
            d1[0] = 0;
        }
//        System.err.println(Arrays.toString(d1));
        long ans2 = 0;
        for (int i = 1; i <= n; ++i) {
            if (n - i <= k) {
                ans2 = (ans2 + d1[i]) % mod;
            }
        }
        out.println(ans1 + " " + ans2);
    }

    public static void main(String[] args) throws IOException {
        PrintWriter out = new PrintWriter(System.out);
        solve(new Input(new BufferedReader(
        new InputStreamReader(System.in))), out);
        out.close();
    }

    static class Input {
        BufferedReader in;
        StringBuilder sb = new StringBuilder();

        public Input(BufferedReader in) {
            this.in = in;
        }

        public Input(String s) {
            this.in = new BufferedReader(new StringReader(s));
        }

        public String next() throws IOException {
            sb.setLength(0);
            while (true) {
                int c = in.read();
                if (c == -1) {
                    return null;
                }
                if (" \n\r\t".indexOf(c) == -1) {
                    sb.append((char)c);
                    break;
                }
            }
            while (true) {
                int c = in.read();
                if (c == -1 || " \n\r\t".indexOf(c) != -1) {
                    break;
                }
                sb.append((char)c);
            }
            return sb.toString();
        }

        public int nextInt() throws IOException {
            return Integer.parseInt(next());
        }

        public long nextLong() throws IOException {
            return Long.parseLong(next());
        }

        public double nextDouble() throws IOException {
            return Double.parseDouble(next());
        }
    }
}








In C :




#include<stdio.h>

int i,j,k,n,mod;
long long ans1,ans2;
long long f[2600][2600],g[2600][2600];

main()
{
	mod=1000000007;
	scanf("%d%d",&n,&k);
	f[1][0]=1;
	for (i=2; i<=n; i++)
	{
		f[i][0]=1;
		for (j=1; j<=k && j<=i*(i-1)/2; j++)
		{
			f[i][j]=f[i][j-1];
			if (j<=(i-1)*(i-2)/2) f[i][j]=(f[i][j]+f[i-1][j])%mod;
			if (j>=i) f[i][j]=(f[i][j]-f[i-1][j-i]+mod)%mod;
		}
	}
	ans1=0;
	for (i=0; i<=k; i++)
		if ((k-i)%2==0) ans1=(ans1+f[n][i])%mod;
	g[1][1]=1;
	for (i=2; i<=n; i++)
	{
		g[i][1]=g[i-1][1]*(i-1)%mod;
		for (j=2; j<i; j++)
			g[i][j]=(g[i-1][j-1]+g[i-1][j]*(i-1)%mod)%mod;
		g[i][i]=1;
	}
	ans2=0;

	for (i=n; i>=n-k && i>=1; i--)
		ans2=(ans2+g[n][i])%mod;
	printf("%lld %lld\n",ans1,ans2);
}








In Python3 :





import sys
import math

def adjSwaps(n,k):
    if k==0:
        return 1
    else:
        #n==2:
        cntksteps = [1]*(k+1)
        #n>2:
        tmp = [0]*(k+1)
        for i in range(3,n+1):
            #0...i-1
            prefixSum = 0
            for j in range(min(i,k+1)):
                prefixSum+=cntksteps[j]
                prefixSum%=1000000007
                tmp[j] = prefixSum
            #i...k
            for j in range(i,k+1):
                prefixSum+=cntksteps[j]
                prefixSum-=cntksteps[j-i]
                tmp[j] = prefixSum
            for j in range(k+1):
                cntksteps[j] = tmp[j]
            #print(cntksteps)
        return cntksteps[k]%1000000007

def anySwaps(n,k):
    if k==0:
        return 1
    elif n<=k:
        return math.factorial(n)%1000000007
    else:

        tmp = [0]*(k+1)
        #n==2:
        cntksteps = [2]*(k+1)
        cntksteps[0] = 1
        #n>2:
        for i in range(3,n+1):
            #if i<k:
                #cntksteps[i] = cntksteps[i-1]*i
            for j in range(1,k+1):
                tmp[j] = (cntksteps[j]+(i-1)*cntksteps[j-1])%1000000007
            for j in range(1,k+1):
                cntksteps[j] = tmp[j]
            #cntksteps[0] = 1
            #print(cntksteps)
        return cntksteps[k]%1000000007
    
f = sys.stdin
tmp = f.readline().split()
n = int(tmp[0])
k = int(tmp[1])
print(adjSwaps(n,k), anySwaps(n,k))
                        




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