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++
// 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
// 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
# 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#
// 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
// 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
?>
Javascript
<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>
Producción
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++
// 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
// 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
Python3
# 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#
// 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_
Javascript
<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>
Producción
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:
C++
// 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;
}
Java
// 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
Python 3
# 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
C#
// 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
<?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
?>
Javascript
<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>
Producción
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.
Publicación traducida automáticamente
Artículo escrito por GeeksforGeeks-1 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA