Conteo de strings que se pueden formar usando a, b y c bajo restricciones dadas

Dada una longitud n, cuente el número de strings de longitud n que se pueden hacer usando ‘a’, ‘b’ y ‘c’ con como máximo una ‘b’ y dos ‘c’ permitidas.

Ejemplos: 

Input : n = 3 
Output : 19 
Below strings follow given constraints:
aaa aab aac aba abc aca acb acc baa
bac bca bcc caa cab cac cba cbc cca ccb 

Input  : n = 4
Output : 39

Preguntado en la entrevista de Google

Una solución simple es contar recursivamente todas las combinaciones posibles de strings que pueden ser moda hasta el último ‘a’, ‘b’ y ‘c’. 

A continuación se muestra la implementación de la idea anterior. 

// C++ program to count number of strings
// of n characters with
#include<bits/stdc++.h>
using namespace std;
 
// n is total number of characters.
// bCount and cCount are counts of 'b'
// and 'c' respectively.
int countStr(int n, int bCount, int cCount)
{
    // Base cases
    if (bCount < 0 || cCount < 0) return 0;
    if (n == 0) return 1;
    if (bCount == 0 && cCount == 0) return 1;
 
    // Three cases, we choose, a or b or c
    // In all three cases n decreases by 1.
    int res = countStr(n-1, bCount, cCount);
    res += countStr(n-1, bCount-1, cCount);
    res += countStr(n-1, bCount, cCount-1);
 
    return res;
}
 
// Driver code
int main()
{
    int n = 3;  // Total number of characters
    cout << countStr(n, 1, 2);
    return 0;
}

// Java program to count number
// of strings of n characters with
import java.io.*;
 
class GFG
{
     
// n is total number of characters.
// bCount and cCount are counts of 'b'
// and 'c' respectively.
static int countStr(int n,
                    int bCount,
                    int cCount)
{
    // Base cases
    if (bCount < 0 || cCount < 0) return 0;
    if (n == 0) return 1;
    if (bCount == 0 && cCount == 0) return 1;
 
    // Three cases, we choose, a or b or c
    // In all three cases n decreases by 1.
    int res = countStr(n - 1, bCount, cCount);
    res += countStr(n - 1, bCount - 1, cCount);
    res += countStr(n - 1, bCount, cCount - 1);
 
    return res;
}
 
// Driver code
public static void main (String[] args)
{
    int n = 3; // Total number of characters
    System.out.println(countStr(n, 1, 2));
}
}
 
// This code is contributed by akt_mit

# Python 3 program to
# count number of strings
# of n characters with
 
# n is total number of characters.
# bCount and cCount are counts
# of 'b' and 'c' respectively.
def countStr(n, bCount, cCount):
 
    # Base cases
    if (bCount < 0 or cCount < 0):
        return 0
    if (n == 0) :
        return 1
    if (bCount == 0 and cCount == 0):
        return 1
 
    # Three cases, we choose, a or b or c
    # In all three cases n decreases by 1.
    res = countStr(n - 1, bCount, cCount)
    res += countStr(n - 1, bCount - 1, cCount)
    res += countStr(n - 1, bCount, cCount - 1)
 
    return res
 
# Driver code
if __name__ =="__main__":
    n = 3 # Total number of characters
    print(countStr(n, 1, 2))
 
# This code is contributed
# by ChitraNayal

// C# program to count number
// of strings of n characters
// with a, b and c under given
// constraints
using System;
 
class GFG
{
     
// n is total number of
// characters. bCount and
// cCount are counts of
// 'b' and 'c' respectively.
static int countStr(int n,
                    int bCount,
                    int cCount)
{
    // Base cases
    if (bCount < 0 || cCount < 0)
        return 0;
    if (n == 0) return 1;
    if (bCount == 0 && cCount == 0)
        return 1;
 
    // Three cases, we choose,
    // a or b or c. In all three
    // cases n decreases by 1.
    int res = countStr(n - 1,
                       bCount, cCount);
    res += countStr(n - 1,
                    bCount - 1, cCount);
    res += countStr(n - 1,
                    bCount, cCount - 1);
 
    return res;
}
 
// Driver code
static public void Main ()
{
    // Total number
    // of characters
    int n = 3;
    Console.WriteLine(countStr(n, 1, 2));
}
}
 
// This code is contributed by aj_36

<?php
// PHP program to count number of
// strings of n characters with
 
// n is total number of characters.
// bCount and cCount are counts
// of 'b' and 'c' respectively.
function countStr($n, $bCount,
                      $cCount)
{
    // Base cases
    if ($bCount < 0 ||
        $cCount < 0)
        return 0;
    if ($n == 0)
    return 1;
    if ($bCount == 0 &&
        $cCount == 0)
        return 1;
 
    // Three cases, we choose,
    // a or b or c. In all three
    // cases n decreases by 1.
    $res = countStr($n - 1,
                    $bCount,
                    $cCount);
    $res += countStr($n - 1,
                     $bCount - 1,
                     $cCount);
    $res += countStr($n - 1,
                     $bCount,
                     $cCount - 1);
 
    return $res;
}
 
// Driver code
$n = 3; // Total number
        // of characters
echo countStr($n, 1, 2);
     
// This code is contributed by ajit
?>

<script>
 
// JavaScript program for the above approach
   
// n is total number of characters.
// bCount and cCount are counts of 'b'
// and 'c' respectively.
function countStr(n, bCount, cCount)
{
 
    // Base cases
    if (bCount < 0 || cCount < 0) return 0;
    if (n == 0) return 1;
    if (bCount == 0 && cCount == 0) return 1;
   
    // Three cases, we choose, a or b or c
    // In all three cases n decreases by 1.
    let res = countStr(n - 1, bCount, cCount);
    res += countStr(n - 1, bCount - 1, cCount);
    res += countStr(n - 1, bCount, cCount - 1);
   
    return res;
}
 
// Driver Code
    let n = 3; // Total number of characters
    document.write(countStr(n, 1, 2));
         
        // This code is contributed by splevel62.
</script>

19

La complejidad temporal de la solución anterior es exponencial.

Solución eficiente:

Si ahogamos un árbol de recurrencia del código anterior, podemos notar que los mismos valores aparecen varias veces. Entonces almacenamos resultados que se usan más tarde si se repiten.

// C++ program to count number of strings
// of n characters with
#include<bits/stdc++.h>
using namespace std;
 
// n is total number of characters.
// bCount and cCount are counts of 'b'
// and 'c' respectively.
int countStrUtil(int dp[][2][3], int n, int bCount=1,
                 int cCount=2)
{
    // Base cases
    if (bCount < 0 || cCount < 0) return 0;
    if (n == 0) return 1;
    if (bCount == 0 && cCount == 0) return 1;
 
    // if we had saw this combination previously
    if (dp[n][bCount][cCount] != -1)
        return dp[n][bCount][cCount];
 
    // Three cases, we choose, a or b or c
    // In all three cases n decreases by 1.
    int res = countStrUtil(dp, n-1, bCount, cCount);
    res += countStrUtil(dp, n-1, bCount-1, cCount);
    res += countStrUtil(dp, n-1, bCount, cCount-1);
 
    return (dp[n][bCount][cCount] = res);
}
 
// A wrapper over countStrUtil()
int countStr(int n)
{
    int dp[n+1][2][3];
    memset(dp, -1, sizeof(dp));
    return countStrUtil(dp, n);
}
 
// Driver code
int main()
{
    int n = 3; // Total number of characters
    cout << countStr(n);
    return 0;
}

// Java program to count number of strings
// of n characters with
 
class GFG
{
    // n is total number of characters.
    // bCount and cCount are counts of 'b'
    // and 'c' respectively.
 
    static int countStrUtil(int[][][] dp, int n,
                            int bCount, int cCount)
    {
 
        // Base cases
        if (bCount < 0 || cCount < 0)
        {
            return 0;
        }
        if (n == 0)
        {
            return 1;
        }
        if (bCount == 0 && cCount == 0)
        {
            return 1;
        }
 
        // if we had saw this combination previously
        if (dp[n][bCount][cCount] != -1)
        {
            return dp[n][bCount][cCount];
        }
 
        // Three cases, we choose, a or b or c
        // In all three cases n decreases by 1.
        int res = countStrUtil(dp, n - 1, bCount, cCount);
        res += countStrUtil(dp, n - 1, bCount - 1, cCount);
        res += countStrUtil(dp, n - 1, bCount, cCount - 1);
 
        return (dp[n][bCount][cCount] = res);
    }
 
    // A wrapper over countStrUtil()
    static int countStr(int n, int bCount, int cCount)
    {
        int[][][] dp = new int[n + 1][2][3];
        for (int i = 0; i < n + 1; i++)
        {
            for (int j = 0; j < 2; j++)
            {
                for (int k = 0; k < 3; k++)
                {
                    dp[i][j][k] = -1;
                }
            }
        }
        return countStrUtil(dp, n,bCount,cCount);
    }
 
    // Driver code
    public static void main(String[] args)
    {
        int n = 3; // Total number of characters
        int bCount = 1, cCount = 2;
        System.out.println(countStr(n,bCount,cCount));
    }
}
 
// This code has been contributed by 29AjayKumar

# Python 3 program to count number of strings
# of n characters with
 
# n is total number of characters.
# bCount and cCount are counts of 'b'
# and 'c' respectively.
def countStrUtil(dp, n, bCount=1,cCount=2):
 
    # Base cases
    if (bCount < 0 or cCount < 0):
        return 0
    if (n == 0):
        return 1
    if (bCount == 0 and cCount == 0):
        return 1
 
    # if we had saw this combination previously
    if (dp[n][bCount][cCount] != -1):
        return dp[n][bCount][cCount]
 
    # Three cases, we choose, a or b or c
    # In all three cases n decreases by 1.
    res = countStrUtil(dp, n-1, bCount, cCount)
    res += countStrUtil(dp, n-1, bCount-1, cCount)
    res += countStrUtil(dp, n-1, bCount, cCount-1)
 
    dp[n][bCount][cCount] = res
    return dp[n][bCount][cCount]
 
# A wrapper over countStrUtil()
def countStr(n):
 
    dp = [ [ [-1 for x in range(2+1)] for y in range(1+1)]for z in range(n+1)]
    return countStrUtil(dp, n)
 
# Driver code
if __name__ == "__main__":
     
    n = 3 # Total number of characters
    print(countStr(n))
     
# This code is contributed by chitranayal   

// C# program to count number of strings
// of n characters with
using System;
 
class GFG
{
    // n is total number of characters.
    // bCount and cCount are counts of 'b'
    // and 'c' respectively.
    static int countStrUtil(int[,,] dp, int n,
                    int bCount=1, int cCount=2)
    {
        // Base cases
        if (bCount < 0 || cCount < 0)
            return 0;
        if (n == 0)
            return 1;
        if (bCount == 0 && cCount == 0)
            return 1;
     
        // if we had saw this combination previously
        if (dp[n,bCount,cCount] != -1)
            return dp[n,bCount,cCount];
     
        // Three cases, we choose, a or b or c
        // In all three cases n decreases by 1.
        int res = countStrUtil(dp, n - 1, bCount, cCount);
        res += countStrUtil(dp, n - 1, bCount - 1, cCount);
        res += countStrUtil(dp, n - 1, bCount, cCount - 1);
     
        return (dp[n, bCount, cCount] = res);
    }
     
    // A wrapper over countStrUtil()
    static int countStr(int n)
    {
        int[,,] dp = new int[n + 1, 2, 3];
        for(int i = 0; i < n + 1; i++)
            for(int j = 0; j < 2; j++)
                for(int k = 0; k < 3; k++)
                    dp[i, j, k] = -1;
        return countStrUtil(dp, n);
    }
     
    // Driver code
    static void Main()
    {
        int n = 3; // Total number of characters
         
        Console.Write(countStr(n));
    }
}
 
// This code is contributed by DrRoot_

<script>
 
// javascript program to count number of strings
// of n characters with
 
    // n is total number of characters.
    // bCount and cCount are counts of 'b'
    // and 'c' respectively.
 
    function countStrUtil(dp , n, bCount , cCount)
    {
 
        // Base cases
        if (bCount < 0 || cCount < 0)
        {
            return 0;
        }
        if (n == 0)
        {
            return 1;
        }
        if (bCount == 0 && cCount == 0)
        {
            return 1;
        }
 
        // if we had saw this combination previously
        if (dp[n][bCount][cCount] != -1)
        {
            return dp[n][bCount][cCount];
        }
 
        // Three cases, we choose, a or b or c
        // In all three cases n decreases by 1.
        var res = countStrUtil(dp, n - 1, bCount, cCount);
        res += countStrUtil(dp, n - 1, bCount - 1, cCount);
        res += countStrUtil(dp, n - 1, bCount, cCount - 1);
 
        return (dp[n][bCount][cCount] = res);
    }
 
    // A wrapper over countStrUtil()
    function countStr(n , bCount , cCount)
    {
        dp = Array(n+1).fill(0).map
        (x => Array(2).fill(0).map
        (x => Array(3).fill(0)));
        for (i = 0; i < n + 1; i++)
        {
            for (j = 0; j < 2; j++)
            {
                for (k = 0; k < 3; k++)
                {
                    dp[i][j][k] = -1;
                }
            }
        }
        return countStrUtil(dp, n,bCount,cCount);
    }
 
// Driver code
var n = 3; // Total number of characters
var bCount = 1, cCount = 2;
document.write(countStr(n,bCount,cCount));
 
 
// This code contributed by shikhasingrajput
 
</script>

19

Tiempo Complejidad : O(n) 
Espacio Auxiliar : O(n)

Gracias al Sr. Lazy por sugerir las soluciones anteriores.

Una solución que funciona en tiempo O(1): 

Podemos aplicar los conceptos de combinatoria para resolver este problema en tiempo constante. podemos recordar la fórmula de que el número de formas en que podemos organizar un total de n objetos, de los cuales p número de objetos son de un tipo, q objetos son de otro tipo y r objetos son de un tercer tipo es n!/( p! q! r!)

Avancemos hacia la solución paso a paso.

¿Cuántas strings podemos formar sin ‘b’ y ‘c’? La respuesta es 1 porque podemos organizar una string que consiste solo en ‘a’ de una sola manera y la string sería aaaa….(n veces) .

¿Cuántas cuerdas podemos formar con una ‘b’? ¡La respuesta es n porque podemos organizar una string que consta de (n-1) ‘a’s y 1 ‘b’ es n!/(n-1)! = norte Lo mismo ocurre con ‘c’.

¿Cuántas strings podemos formar con 2 lugares, llenados por ‘b’ y/o ‘c’? La respuesta es n*(n-1) + n*(n-1)/2 . Porque ese 2 lugar puede ser 1 ‘b’ y 1 ‘c’ o 2 ‘c’ de acuerdo con nuestras restricciones dadas. Para el primer caso, el número total de arreglos es n!/(n-2)! = n*(n-1) y para el segundo caso es n!/(2!(n-2)!) = n*(n-1)/2 .

Finalmente, ¿cuántas strings podemos formar con 3 lugares, llenados por ‘b’ y/o ‘c’? La respuesta es (n-2)*(n-1)*n/2 . Porque ese tercer lugar solo puede consistir en 1 ‘b’ y 2’c’ de acuerdo con nuestras restricciones dadas. Entonces, el número total de arreglos es n!/(2!(n-3)!) = (n-2)*(n-1)*n/2 .

Implementación:

// A O(1) CPP program to find number of strings
// that can be made under given constraints.
#include<bits/stdc++.h>
using namespace std;
int countStr(int n){
     
    int count = 0;
     
    if(n>=1){
        //aaa...
        count += 1;
        //b...aaa...
          count += n;
        //c...aaa...
        count += n;
         
        if(n>=2){
          //bc...aaa...
          count += n*(n-1);
          //cc...aaa...
          count += n*(n-1)/2;
           
          if(n>=3){
            //bcc...aaa...
            count += (n-2)*(n-1)*n/2;
          }
        }
     
    }
     
    return count;
     
}
 
// Driver code
int main()
{
  int n = 3;
  cout << countStr(n);
  return 0;
}

// A O(1) Java program to
// find number of strings
// that can be made under
// given constraints.
import java.io.*;
 
class GFG
{
    static int countStr(int n)
    {
    return 1 + (n * 2) +
           (n * ((n * n) - 1) / 2);
    }
 
// Driver code
public static void main (String[] args)
{
    int n = 3;
    System.out.println( countStr(n));
}
}
 
// This code is contributed by ajit

# A O(1) Python3 program to find
# number of strings that can be
# made under given constraints.
 
def countStr(n):
    return (1 + (n * 2) +
                (n * ((n * n) - 1) // 2))
 
# Driver code
if __name__ == "__main__":
    n = 3
    print(countStr(n))
 
# This code is contributed
# by ChitraNayal

// A O(1) C# program to
// find number of strings
// that can be made under
// given constraints.
using System;
 
class GFG
{
    static int countStr(int n)
    {
    return 1 + (n * 2) +
          (n * ((n * n) - 1) / 2);
    }
 
// Driver code
static public void Main ()
{
    int n = 3;
    Console.WriteLine(countStr(n));
}
}
 
// This code is contributed by m_kit

<?php
// A O(1) PHP program to find
// number of strings that can
// be made under given constraints.
function countStr($n)
{
    return 1 + ($n * 2) + ($n *
              (($n * $n) - 1) / 2);
}
 
// Driver code
$n = 3;
echo countStr($n);
 
// This code is contributed by aj_36
?>

<script>
// A O(1) javascript program to
// find number of strings
// that can be made under
// given constraints.
    function countStr(n) {
        return 1 + (n * 2) + (n * ((n * n) - 1) / 2);
    }
 
    // Driver code
     
        var n = 3;
        document.write(countStr(n));
 
// This code is contributed by Princi Singh
</script>

19

Complejidad de Tiempo : O(1) 
Espacio Auxiliar : O(1)

Gracias a Niharika Sahai por proporcionar la solución anterior.

Este artículo es una contribución de Nishant Singh . Si te gusta GeeksforGeeks y te gustaría contribuir, también puedes escribir un artículo usando write.geeksforgeeks.org o enviar tu artículo por correo a review-team@geeksforgeeks.org. Vea su artículo que aparece en la página principal de GeeksforGeeks y ayude a otros Geeks.