Número de soluciones integrales no negativas de la ecuación de suma

Dado un número n (número de variables) y val (suma de las variables), averigüe cuántas soluciones integrales no negativas son posibles. 

Ejemplos: 

Input : n = 5, val = 1
Output : 5
Explanation:
x1 + x2 + x3 + x4 + x5 = 1
Number of possible solution are : 
(0 0 0 0 1), (0 0 0 1 0), (0 0 1 0 0),
(0 1 0 0 0), (1 0 0 0 0)
Total number of possible solutions are 5

Input : n = 5, val = 4
Output : 70
Explanation:
x1 + x2 + x3 + x4 + x5 = 4
Number of possible solution are: 
(1 1 1 1 0), (1 0 1 1 1), (0 1 1 1 1), 
(2 1 0 0 1), (2 2 0 0 0)........ so on......
Total numbers of possible solutions are 70

Preguntado en: Entrevista de Microsoft 

1. Realice una llamada de función recursiva a countSolutions(int n, int val) 
2. Llame a esta función de solución countSolutions(n-1, val-i) hasta que n = 1 y val >=0 y luego devuelva 1.

A continuación se muestra la implementación del enfoque anterior:

C++

// CPP program to find the numbers
// of non negative integral solutions
#include<iostream>
using namespace std;
 
// return number of non negative
// integral solutions
int countSolutions(int n, int val)
{
    // initialize total = 0
    int total = 0;
 
    // Base Case if n = 1 and val >= 0
    // then it should return 1
    if (n == 1 && val >=0)
        return 1;
 
    // iterate the loop till equal the val
    for (int i = 0; i <= val; i++){
         
        // total solution of equations
        // and again call the recursive
        // function Solutions(variable,value)
        total += countSolutions(n-1, val-i);
         
    }
     
    // return the total no possible solution
    return total;
}
 
// driver code
int main(){
     
    int n = 5;
    int val = 20;
     
    cout<<countSolutions(n, val);
}
 
//This code is contributed by Smitha Dinesh Semwal

Java

// Java program to find the numbers
// of non negative integral solutions
class GFG {
 
  // return number of non negative
  // integral solutions
  static int countSolutions(int n, int val) {
 
    // initialize total = 0
    int total = 0;
 
    // Base Case if n = 1 and val >= 0
    // then it should return 1
    if (n == 1 && val >= 0)
      return 1;
 
    // iterate the loop till equal the val
    for (int i = 0; i <= val; i++) {
 
      // total solution of equations
      // and again call the recursive
      // function Solutions(variable, value)
      total += countSolutions(n - 1, val - i);
    }
 
    // return the total no possible solution
    return total;
  }
 
  // Driver code
  public static void main(String[] args) {
    int n = 5;
    int val = 20;
 
    System.out.print(countSolutions(n, val));
  }
}
 
// This code is contributed by Anant Agarwal.

Python3

# Python3 program to find the numbers
# of non negative integral solutions
 
# return number of non negative
# integral solutions
def countSolutions(n, val):
 
    # initialize total = 0
    total = 0
 
    # Base Case if n = 1 and val >= 0
    # then it should return 1
    if n == 1 and val >=0:
        return 1
 
    # iterate the loop till equal the val
    for i in range(val+1):
     
        # total solution of equations
        # and again call the recursive
        # function Solutions(variable,value)
        total += countSolutions(n-1, val-i)
 
    # return the total no possible solution
    return total
 
# driver code
n = 5
val = 20
print(countSolutions(n, val))

C#

// C# program to find the numbers
// of non negative integral solutions
using System;
 
class GFG {
 
    // return number of non negative
    // integral solutions
    static int countSolutions(int n, int val) {
     
        // initialize total = 0
        int total = 0;
     
        // Base Case if n = 1 and val >= 0
        // then it should return 1
        if (n == 1 && val >= 0)
        return 1;
     
        // iterate the loop till equal the val
        for (int i = 0; i <= val; i++) {
         
            // total solution of equations
            // and again call the recursive
            // function Solutions(variable, value)
            total += countSolutions(n - 1, val - i);
        }
     
        // return the total no possible solution
        return total;
    }
     
    // Driver code
    public static void Main()
    {
        int n = 5;
        int val = 20;
     
        Console.WriteLine(countSolutions(n, val));
    }
}
 
// This code is contributed by Anant vt_m.

PHP

<?php
// PHP program to find the numbers
// of non negative integral solutions
 
// return number of non negative
// integral solutions
function countSolutions($n, $val)
{
    // initialize total = 0
    $total = 0;
 
    // Base Case if n = 1 and
    // val >= 0 then it should
    // return 1
    if ($n == 1 && $val >=0)
        return 1;
 
    // iterate the loop
    // till equal the val
    for ($i = 0; $i <= $val; $i++)
    {
         
        // total solution of equations
        // and again call the recursive
        // function Solutions(variable,value)
        $total += countSolutions($n - 1,
                                 $val - $i);
         
    }
     
    // return the total
    // no possible solution
    return $total;
}
 
// Driver Code
$n = 5;
$val = 20;
 
echo countSolutions($n, $val);
 
// This code is contributed by nitin mittal.
?>

Javascript

<script>
 
// JavaScript program to find the numbers
// of non negative integral solutions
 
  // return number of non negative
  // integral solutions
  function countSolutions(n, val) {
   
    // initialize total = 0
    let total = 0;
   
    // Base Case if n = 1 and val >= 0
    // then it should return 1
    if (n == 1 && val >= 0)
      return 1;
   
    // iterate the loop till equal the val
    for (let i = 0; i <= val; i++) {
   
      // total solution of equations
      // and again call the recursive
      // function Solutions(variable, value)
      total += countSolutions(n - 1, val - i);
    }
   
    // return the total no possible solution
    return total;
  }
 
// Driver code
         
        let n = 5;
        let val = 20;
   
    document.write(countSolutions(n, val));
         
</script>

Producción : 

10626

Enfoque de programación dinámica:

(Usando Memoización)

C++

// CPP program to find the numbers
// of non negative integral solutions
#include<bits/stdc++.h>
using namespace std;
 
int dp[1001][1001];
 
// return number of non negative
// integral solutions
int countSolutions(int n, int val)
{
    // initialize total = 0
    int total = 0;
 
    // Base Case if n = 1 and val >= 0
    // then it should return 1
    if (n == 1 && val >=0) {
        return 1;
    }
   
    // If a value already present in dp,
    // return it
    if(dp[n][val] != -1) {
        return dp[n][val];
    }
     
    // iterate the loop till equal the val
    for (int i = 0; i <= val; i++){
         
        // total solution of equations
        // and again call the recursive
        // function Solutions(variable,value)
        total += countSolutions(n-1, val-i);
         
    }
     
    // Store the value in dp
    dp[n][val] = total;
     
    // Return dp
    return dp[n][val];
}
 
// driver code
int main(){
     
    int n = 5;
    int val = 20;
     
    memset(dp, -1, sizeof(dp));
     
    cout << countSolutions(n, val);
}

Java

// Java program to find the numbers
// of non negative integral solutions
import java.util.*;
public class GFG
{
  static int dp[][] = new int[1001][1001];
 
  // return number of non negative
  // integral solutions
  static int countSolutions(int n, int val)
  {
    // initialize total = 0
    int total = 0;
 
    // Base Case if n = 1 and val >= 0
    // then it should return 1
    if (n == 1 && val >=0) {
      return 1;
    }
 
    // If a value already present in dp,
    // return it
    if(dp[n][val] != -1) {
      return dp[n][val];
    }
 
    // iterate the loop till equal the val
    for (int i = 0; i <= val; i++){
 
      // total solution of equations
      // and again call the recursive
      // function Solutions(variable,value)
      total += countSolutions(n-1, val-i);
 
    }
 
    // Store the value in dp
    dp[n][val] = total;
 
    // Return dp
    return dp[n][val];
  }
 
  // driver code
  public static void main(String args[]){
 
    int n = 5;
    int val = 20;
 
    for(int i = 0; i < 1001; i++) {
      for(int j = 0; j < 1001; j++) {
        dp[i][j]=-1;
      }
    }
 
    System.out.println(countSolutions(n, val));
  }
}
 
// This code is contributed by Samim Hossain Mondal.

Python3

# Python3 program to find the numbers
# of non negative integral solutions
 
# Taking the matrix as globally
dp = [[-1 for i in range(1001)]
          for j in range(1001)]
 
# Return number of non negative
# integral solutions
def countSolutions(n, val):
 
    # Initialize total = 0
    total = 0
 
    # Base Case if n = 1 and val >= 0
    # then it should return 1
    if n == 1 and val >= 0:
        return 1
     
    # If a value is already present
    # in dp
    if (dp[n][val] != -1):
        return dp[n][val]
     
    # Iterate the loop till equal the val
    for i in range(val + 1):
     
        # total solution of equations
        # and again call the recursive
        # function Solutions(variable,value)
        total += countSolutions(n - 1, val - i)
 
    # Return the total no possible solution
    dp[n][val] = total
    return dp[n][val]
 
# Driver code
n = 5
val = 20
 
print(countSolutions(n, val))
 
# This code is contributed by Samim Hossain Mondal.

C#

// C# program to find the numbers
// of non negative integral solutions
using System;
class GFG
{
  static int [,]dp = new int[1001, 1001];
 
  // return number of non negative
  // integral solutions
  static int countSolutions(int n, int val)
  {
    // initialize total = 0
    int total = 0;
 
    // Base Case if n = 1 and val >= 0
    // then it should return 1
    if (n == 1 && val >=0) {
      return 1;
    }
 
    // If a value already present in dp,
    // return it
    if(dp[n, val] != -1) {
      return dp[n, val];
    }
 
    // iterate the loop till equal the val
    for (int i = 0; i <= val; i++){
 
      // total solution of equations
      // and again call the recursive
      // function Solutions(variable,value)
      total += countSolutions(n-1, val-i);
 
    }
 
    // Store the value in dp
    dp[n, val] = total;
 
    // Return dp
    return dp[n, val];
  }
 
  // driver code
  public static void Main(){
 
    int n = 5;
    int val = 20;
 
    for(int i = 0; i < 1001; i++) {
      for(int j = 0; j < 1001; j++) {
        dp[i, j] = -1;
      }
    }
 
    Console.Write(countSolutions(n, val));
  }
}
 
// This code is contributed by Samim Hossain Mondal.

Javascript

<script>
// JavaScript program to find the numbers
// of non negative integral solutions
  var dp = new Array(1001);
   
  // Loop to create 2D array using 1D array
  for (var i = 0; i < dp.length; i++) {
      dp[i] = new Array(1001);
  }
   
  // return number of non negative
  // integral solutions
  function countSolutions(n, val) {
   
    // initialize total = 0
    let total = 0;
   
    // Base Case if n = 1 and val >= 0
    // then it should return 1
    if (n == 1 && val >= 0)
      return 1;
     
    // if a value is already
    // present in dp
    if(dp[n][val] != -1)
        return dp[n][val];
     
    // iterate the loop till equal the val
    for (let i = 0; i <= val; i++) {
   
      // total solution of equations
      // and again call the recursive
      // function Solutions(variable, value)
      total += countSolutions(n - 1, val - i);
    }
   
    // return the total no possible solution
    return dp[n][val] = total;
  }
 
// Driver code
let n = 5;
let val = 20;
 
for(let i = 0; i < 1001; i++) {
    for(let j = 0; j < 1001; j++) {
        dp[i][j] = -1;
    }
}
document.write(countSolutions(n, val));
 
// This code is contributed by Samim Hossain Mondal.  
</script>
Producción

10626

Complejidad temporal: O(n * val)

Espacio Auxiliar: O(n * val)

Publicación traducida automáticamente

Artículo escrito por shrikanth13 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA

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