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Rencontres Number (Counting partial derangements)

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  • Difficulty Level : Hard

Given two numbers, n >= 0 and 0 <= k <= n, count the number of derangements with k fixed points.
Examples: 
 

Input : n = 3, k = 0
Output : 2
Since k = 0, no point needs to be on its
original position. So derangements
are {3, 1, 2} and {2, 3, 1}

Input : n = 3, k = 1
Output : 3
Since k = 1, one point needs to be on its
original position. So partial derangements
are {1, 3, 2}, {3, 2, 1} and {2, 1, 3}

Input : n = 7, k = 2
Output : 924

 

In combinatorial mathematics, the rencontres number< or D(n, k) represents count of partial derangements.
The recurrence relation to find Rencontres Number Dn, k:
 

D(0, 0) = 1 
D(0, 1) = 0 
D(n+2, 0) = (n+1) * (D(n+1, 0) + D(n, 0)
D(n, k) = nCk * D(n-k, 0))

Given the two positive integer n and k. The task is find rencontres number D(n, k) for giver n and k.
Below is Recursive solution of this approach:
 

C++




// Recursive CPP program to find n-th Rencontres
// Number
#include <bits/stdc++.h>
using namespace std;
 
// Returns value of Binomial Coefficient C(n, k)
int binomialCoeff(int n, int k)
{
    // Base Cases
    if (k == 0 || k == n)
        return 1;
 
    // Recurrence relation
    return binomialCoeff(n - 1, k - 1) +
           binomialCoeff(n - 1, k);
}
 
// Return Recontres number D(n, m)
int RencontresNumber(int n, int m)
{
    // base condition
    if (n == 0 && m == 0)
        return 1;
 
    // base condition
    if (n == 1 && m == 0)
        return 0;
 
    // base condition
    if (m == 0)
        return (n - 1) * (RencontresNumber(n - 1, 0) +
                          RencontresNumber(n - 2, 0));
 
    return binomialCoeff(n, m) * RencontresNumber(n - m, 0);
}
 
// Driver Program
int main()
{
    int n = 7, m = 2;
    cout << RencontresNumber(n, m) << endl;
    return 0;
}

Java




// Recursive Java program to find n-th Rencontres
// Number
import java.io.*;
 
class GFG {
     
    // Returns value of Binomial Coefficient
    // C(n, k)
    static int binomialCoeff(int n, int k)
    {
         
        // Base Cases
        if (k == 0 || k == n)
            return 1;
 
        // Recurrence relation
        return binomialCoeff(n - 1, k - 1) +
                         binomialCoeff(n - 1, k);
    }
 
    // Return Recontres number D(n, m)
    static int RencontresNumber(int n, int m)
    {
         
        // base condition
        if (n == 0 && m == 0)
            return 1;
 
        // base condition
        if (n == 1 && m == 0)
            return 0;
 
        // base condition
        if (m == 0)
            return (n - 1) * (RencontresNumber(n - 1, 0)
                          + RencontresNumber(n - 2, 0));
 
        return binomialCoeff(n, m) *
                             RencontresNumber(n - m, 0);
    }
 
    // Driver Program
    public static void main(String[] args)
    {
        int n = 7, m = 2;
        System.out.println(RencontresNumber(n, m));
    }
}
 
// This code is contributed by vt_m.

Python3




# Recursive CPP program to find
# n-th Rencontres Number
 
# Returns value of Binomial Coefficient C(n, k)
def binomialCoeff(n, k):
 
    # Base Cases
    if (k == 0 or k == n):
        return 1
 
    # Recurrence relation
    return (binomialCoeff(n - 1, k - 1)
          + binomialCoeff(n - 1, k))
 
# Return Recontres number D(n, m)
def RencontresNumber(n, m):
 
    # base condition
    if (n == 0 and m == 0):
        return 1
 
    # base condition
    if (n == 1 and m == 0):
        return 0
 
    # base condition
    if (m == 0):
        return ((n - 1) * (RencontresNumber(n - 1, 0)
                         + RencontresNumber(n - 2, 0)))
 
    return (binomialCoeff(n, m) *
            RencontresNumber(n - m, 0))
 
# Driver Program
n = 7; m = 2
print(RencontresNumber(n, m))
 
# This code is contributed by Smitha Dinesh Semwal.

C#




// Recursive C# program to find n-th Rencontres
// Number
using System;
 
class GFG {
     
    // Returns value of Binomial Coefficient
    // C(n, k)
    static int binomialCoeff(int n, int k)
    {
         
        // Base Cases
        if (k == 0 || k == n)
            return 1;
 
        // Recurrence relation
        return binomialCoeff(n - 1, k - 1) +
                     binomialCoeff(n - 1, k);
    }
 
    // Return Recontres number D(n, m)
    static int RencontresNumber(int n, int m)
    {
         
        // base condition
        if (n == 0 && m == 0)
            return 1;
 
        // base condition
        if (n == 1 && m == 0)
            return 0;
 
        // base condition
        if (m == 0)
            return (n - 1) *
                (RencontresNumber(n - 1, 0)
                + RencontresNumber(n - 2, 0));
 
        return binomialCoeff(n, m) *
                 RencontresNumber(n - m, 0);
    }
 
    // Driver Program
    public static void Main()
    {
        int n = 7, m = 2;
         
        Console.Write(RencontresNumber(n, m));
    }
}
 
// This code is contributed by
// Smitha Dinesh Semwal

PHP




<?php
// Recursive PHP program to
// find n-th Rencontres
// Number
 
// Returns value of Binomial
// Coefficient C(n, k)
function binomialCoeff($n, $k)
{
     
    // Base Cases
    if ($k == 0 || $k == $n)
        return 1;
 
    // Recurrence relation
    return binomialCoeff($n - 1,$k - 1) +
              binomialCoeff($n - 1, $k);
}
 
// Return Recontres number D(n, m)
function RencontresNumber($n, $m)
{
     
    // base condition
    if ($n == 0 && $m == 0)
        return 1;
 
    // base condition
    if ($n == 1 && $m == 0)
        return 0;
 
    // base condition
    if ($m == 0)
        return ($n - 1) * (RencontresNumber($n - 1, 0) +
                           RencontresNumber($n - 2, 0));
 
    return binomialCoeff($n, $m) *
           RencontresNumber($n - $m, 0);
}
 
    // Driver Code
    $n = 7;
    $m = 2;
    echo RencontresNumber($n, $m),"\n";
     
// This code is contributed by ajit.
?>

Javascript




<script>
// Recursive Javascript program to
// find n-th Rencontres
// Number
 
// Returns value of Binomial
// Coefficient C(n, k)
function binomialCoeff(n, k)
{
     
    // Base Cases
    if (k == 0 || k == n)
        return 1;
 
    // Recurrence relation
    return binomialCoeff(n - 1,k - 1) +
              binomialCoeff(n - 1, k);
}
 
// Return Recontres number D(n, m)
function RencontresNumber(n, m)
{
     
    // base condition
    if (n == 0 && m == 0)
        return 1;
 
    // base condition
    if (n == 1 && m == 0)
        return 0;
 
    // base condition
    if (m == 0)
        return (n - 1) * (RencontresNumber(n - 1, 0) +
                           RencontresNumber(n - 2, 0));
 
    return binomialCoeff(n, m) *
           RencontresNumber(n - m, 0);
}
 
    // Driver Code
    let n = 7;
    let m = 2;
    document.write(RencontresNumber(n, m) + "<br>");
     
// This code is contributed by _saurabh_jaiswal.
</script>

Output: 

924

 

Time Complexity: O(n * m), where n and m represents the given integers.
Auxiliary Space: O(n*m), due to recursive stack space.

Below is the implementation using Dynamic Programming: 
 

C++




// DP based CPP program to find n-th Rencontres
// Number
#include <bits/stdc++.h>
using namespace std;
#define MAX 100
 
// Fills table C[n+1][k+1] such that C[i][j]
// represents table of binomial coefficient
// iCj
int binomialCoeff(int C[][MAX], int n, int k)
{
    // Calculate value of Binomial Coefficient
    // in bottom up manner
    for (int i = 0; i <= n; i++) {
        for (int j = 0; j <= min(i, k); j++) {
 
            // Base Cases
            if (j == 0 || j == i)
                C[i][j] = 1;
 
            // Calculate value using previously
            // stored values
            else
                C[i][j] = C[i - 1][j - 1] +
                          C[i - 1][j];
        }
    }
}
 
// Return Recontres number D(n, m)
int RencontresNumber(int C[][MAX], int n, int m)
{
    int dp[n+1][m+1] = { 0 };
 
    for (int i = 0; i <= n; i++) {
        for (int j = 0; j <= m; j++) {
            if (j <= i) {
 
                // base case
                if (i == 0 && j == 0)
                    dp[i][j] = 1;
 
                // base case
                else if (i == 1 && j == 0)
                    dp[i][j] = 0;
 
                else if (j == 0)
                    dp[i][j] = (i - 1) * (dp[i - 1][0] +
                                          dp[i - 2][0]);
                else
                    dp[i][j] = C[i][j] * dp[i - j][0];
            }
        }
    }
 
    return dp[n][m];
}
 
// Driver Program
int main()
{
    int n = 7, m = 2;
 
    int C[MAX][MAX];
    binomialCoeff(C, n, m);
 
    cout << RencontresNumber(C, n, m) << endl;
    return 0;
}

Java




// DP based Java program to find n-th Rencontres
// Number
 
import java.io.*;
 
class GFG {
 
    static int MAX = 100;
 
    // Fills table C[n+1][k+1] such that C[i][j]
    // represents table of binomial coefficient
    // iCj
    static void binomialCoeff(int C[][], int n, int k)
    {
 
        // Calculate value of Binomial Coefficient
        // in bottom up manner
        for (int i = 0; i <= n; i++) {
            for (int j = 0; j <= Math.min(i, k); j++)
            {
 
                // Base Cases
                if (j == 0 || j == i)
                    C[i][j] = 1;
 
                // Calculate value using previously
                // stored values
                else
                    C[i][j] = C[i - 1][j - 1] +
                                         C[i - 1][j];
            }
        }
    }
 
    // Return Recontres number D(n, m)
    static int RencontresNumber(int C[][], int n, int m)
    {
        int dp[][] = new int[n + 1][m + 1];
 
        for (int i = 0; i <= n; i++) {
            for (int j = 0; j <= m; j++) {
                if (j <= i) {
 
                    // base case
                    if (i == 0 && j == 0)
                        dp[i][j] = 1;
 
                    // base case
                    else if (i == 1 && j == 0)
                        dp[i][j] = 0;
 
                    else if (j == 0)
                        dp[i][j] = (i - 1) * (dp[i - 1][0]
                                           + dp[i - 2][0]);
                    else
                        dp[i][j] = C[i][j] * dp[i - j][0];
                }
            }
        }
 
        return dp[n][m];
    }
 
    // Driver Program
    public static void main(String[] args)
    {
        int n = 7, m = 2;
 
        int C[][] = new int[MAX][MAX];
        binomialCoeff(C, n, m);
 
        System.out.println(RencontresNumber(C, n, m));
    }
}
 
// This code is contributed by vt_m.

Python 3




# DP based Python 3 program to find n-th
# Rencontres Number
 
MAX = 100
 
# Fills table C[n+1][k+1] such that C[i][j]
# represents table of binomial coefficient
# iCj
def binomialCoeff(C, n, k) :
     
    # Calculate value of Binomial Coefficient
    # in bottom up manner
    for i in range(0, n + 1) :
        for j in range(0, min(i, k) + 1) :
             
            # Base Cases
            if (j == 0 or j == i) :
                C[i][j] = 1
 
            # Calculate value using previously
            # stored values
            else :
                C[i][j] = (C[i - 1][j - 1]
                               + C[i - 1][j])
                 
 
# Return Recontres number D(n, m)
def RencontresNumber(C, n, m) :
    w, h = m+1, n+1
    dp= [[0 for x in range(w)] for y in range(h)]
     
 
    for i in range(0, n+1) :
        for j in range(0, m+1) :
            if (j <= i) :
                 
                # base case
                if (i == 0 and j == 0) :
                    dp[i][j] = 1
 
                # base case
                elif (i == 1 and j == 0) :
                    dp[i][j] = 0
 
                elif (j == 0) :
                    dp[i][j] = ((i - 1) *
                     (dp[i - 1][0] + dp[i - 2][0]))
                else :
                    dp[i][j] = C[i][j] * dp[i - j][0]
                     
    return dp[n][m]
 
 
# Driver Program
n = 7
m = 2
C = [[0 for x in range(MAX)] for y in range(MAX)]
 
binomialCoeff(C, n, m)
 
print(RencontresNumber(C, n, m))
 
# This code is contributed by Nikita Tiwari.

C#




// DP based C# program
// to find n-th Rencontres
// Number
using System;
 
class GFG
{
    static int MAX = 100;
 
    // Fills table C[n+1][k+1]
    // such that C[i][j]
    // represents table of
    // binomial coefficient iCj
    static void binomialCoeff(int [,]C,
                              int n, int k)
    {
 
        // Calculate value of
        // Binomial Coefficient
        // in bottom up manner
        for (int i = 0; i <= n; i++)
        {
            for (int j = 0;
                     j <= Math.Min(i, k); j++)
            {
 
                // Base Cases
                if (j == 0 || j == i)
                    C[i,j] = 1;
 
                // Calculate value using
                // previously stored values
                else
                    C[i, j] = C[i - 1, j - 1] +
                              C[i - 1, j];
            }
        }
    }
 
    // Return Recontres
    // number D(n, m)
    static int RencontresNumber(int [,]C,
                                int n, int m)
    {
        int [,]dp = new int[n + 1,
                            m + 1];
 
        for (int i = 0; i <= n; i++)
        {
            for (int j = 0; j <= m; j++)
            {
                if (j <= i)
                {
 
                    // base case
                    if (i == 0 && j == 0)
                        dp[i, j] = 1;
 
                    // base case
                    else if (i == 1 && j == 0)
                        dp[i, j] = 0;
 
                    else if (j == 0)
                        dp[i, j] = (i - 1) *
                                   (dp[i - 1, 0] +
                                    dp[i - 2, 0]);
                    else
                        dp[i, j] = C[i, j] *
                                  dp[i - j, 0];
                }
            }
        }
 
        return dp[n, m];
    }
 
    // Driver Code
    static public void Main ()
    {
        int n = 7, m = 2;
        int [,]C = new int[MAX, MAX];
        binomialCoeff(C, n, m);
     
        Console.WriteLine(RencontresNumber(C, n, m));
    }
}
 
// This code is contributed
// by akt_mit

PHP




<?php
// DP based PHP program to find n-th Rencontres
// Number
$MAX=100;
 
// Fills table C[n+1][k+1] such that C[i][j]
// represents table of binomial coefficient
// iCj
function binomialCoeff(&$C, $n, $k)
{
    // Calculate value of Binomial Coefficient
    // in bottom up manner
    for ($i = 0; $i <= $n; $i++) {
        for ($j = 0; $j <= min($i, $k); $j++) {
 
            // Base Cases
            if ($j == 0 || $j == $i)
                $C[$i][$j] = 1;
 
            // Calculate value using previously
            // stored values
            else
                $C[$i][$j] = $C[$i - 1][$j - 1] +
                        $C[$i - 1][$j];
        }
    }
}
 
// Return Recontres number D(n, m)
function RencontresNumber($C, $n, $m)
{
    $dp=array_fill(0,$n+1,array_fill(0,$m+1,0));
 
    for ($i = 0; $i <= $n; $i++) {
        for ($j = 0; $j <= $m; $j++) {
            if ($j <= $i) {
 
                // base case
                if ($i == 0 && $j == 0)
                    $dp[$i][$j] = 1;
 
                // base case
                else if ($i == 1 && $j == 0)
                    $dp[$i][$j] = 0;
 
                else if ($j == 0)
                    $dp[$i][$j] = ($i - 1) * ($dp[$i - 1][0] +
                                        $dp[$i - 2][0]);
                else
                    $dp[$i][$j] = $C[$i][$j] * $dp[$i - $j][0];
            }
        }
    }
 
    return $dp[$n][$m];
}
 
// Driver Program
 
    $n = 7;
    $m = 2;
 
    $C=array(array());
    binomialCoeff($C, $n, $m);
 
    echo RencontresNumber($C, $n, $m);
 
// This code is contributed
// by mits
?>

Javascript




<script>
 
// DP based JavaScript program to find n-th
// Rencontres Number
const MAX = 100
 
// Fills table C[n+1][k+1] such that C[i][j]
// represents table of binomial coefficient
// iCj
function binomialCoeff(C, n, k){
     
    // Calculate value of Binomial Coefficient
    // in bottom up manner
    for(let i=0;i<n+1;i++){
        for(let j=0;j< Math.min(i, k) + 1;j++){
             
            // Base Cases
            if (j == 0 || j == i)
                C[i][j] = 1
 
            // Calculate value using previously
            // stored values
            else
                C[i][j] = (C[i - 1][j - 1]
                            + C[i - 1][j])
        }
    }
}
 
// Return Recontres number D(n, m)
function RencontresNumber(C, n, m){
    let w = m+1,h = n+1
    let dp= new Array(h).fill(0).map(()=>new Array(w).fill(0))
     
 
    for(let i=0;i<n+1;i++){
        for(let j=0;j<m+1;j++){
            if (j <= i) {
             
                // base case
                if (i == 0 && j == 0)
                    dp[i][j] = 1
 
                // base case
                else if (i == 1 && j == 0)
                    dp[i][j] = 0
 
                else if (j == 0){
                    dp[i][j] = ((i - 1) *
                    (dp[i - 1][0] + dp[i - 2][0]))
                }
                else
                    dp[i][j] = C[i][j] * dp[i - j][0]
            }
        }
    }
                     
    return dp[n][m]
}
 
 
// Driver Program
let n = 7
let m = 2
let C = new Array(MAX).fill(0).map(()=>new Array(MAX).fill(0))
 
binomialCoeff(C, n, m)
 
document.write(RencontresNumber(C, n, m),"</br>")
 
// This code is contributed by shinjanpatra
 
</script>

Output: 

924

 

Time Complexity: O(n * m), where n and m represents the given integers.
Auxiliary Space: O(n * m), where n and m represents the given integers.


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