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))








View More Similar Problems

Largest Rectangle

Skyline Real Estate Developers is planning to demolish a number of old, unoccupied buildings and construct a shopping mall in their place. Your task is to find the largest solid area in which the mall can be constructed. There are a number of buildings in a certain two-dimensional landscape. Each building has a height, given by . If you join adjacent buildings, they will form a solid rectangle

View Solution →

Simple Text Editor

In this challenge, you must implement a simple text editor. Initially, your editor contains an empty string, S. You must perform Q operations of the following 4 types: 1. append(W) - Append W string to the end of S. 2 . delete( k ) - Delete the last k characters of S. 3 .print( k ) - Print the kth character of S. 4 . undo( ) - Undo the last (not previously undone) operation of type 1 or 2,

View Solution →

Poisonous Plants

There are a number of plants in a garden. Each of the plants has been treated with some amount of pesticide. After each day, if any plant has more pesticide than the plant on its left, being weaker than the left one, it dies. You are given the initial values of the pesticide in each of the plants. Determine the number of days after which no plant dies, i.e. the time after which there is no plan

View Solution →

AND xor OR

Given an array of distinct elements. Let and be the smallest and the next smallest element in the interval where . . where , are the bitwise operators , and respectively. Your task is to find the maximum possible value of . Input Format First line contains integer N. Second line contains N integers, representing elements of the array A[] . Output Format Print the value

View Solution →

Waiter

You are a waiter at a party. There is a pile of numbered plates. Create an empty answers array. At each iteration, i, remove each plate from the top of the stack in order. Determine if the number on the plate is evenly divisible ith the prime number. If it is, stack it in pile Bi. Otherwise, stack it in stack Ai. Store the values Bi in from top to bottom in answers. In the next iteration, do the

View Solution →

Queue using Two Stacks

A queue is an abstract data type that maintains the order in which elements were added to it, allowing the oldest elements to be removed from the front and new elements to be added to the rear. This is called a First-In-First-Out (FIFO) data structure because the first element added to the queue (i.e., the one that has been waiting the longest) is always the first one to be removed. A basic que

View Solution →