Minimice la suma de los números necesarios para convertir una array en una permutación de los primeros N números naturales

Dada una array A[] de tamaño N , la tarea es encontrar la suma mínima de números necesarios para agregar a los elementos de la array para convertir la array en una permutación de 1 a N. Si la array no se puede convertir a la permutación deseada, imprima -1 .

Ejemplos:

Entrada: A[] = {1, 1, 1, 1, 1}
Salida: 10
Explicación: Incrementar A[1] en 1, A[2] en 2, A[3] en 3, A[4] en 4 , por lo que A[] se convierte en {1, 2, 3, 4, 5}.
Adiciones mínimas requeridas = 1 + 2 + 3 + 4 = 10

Entrada: A[] = {2, 2, 3}
Salida: -1

Enfoque: La idea es usar sorting . Siga estos pasos para resolver este problema:

• Inicialice una variable y almacene el resultado requerido.
• Ordene la array dada A[] en orden creciente .
• Atraviesa el arreglo , A[] usando la variable i
  • Si el valor de A[i] > i+1 , actualice ans a -1 y salga del bucle .
  • De lo contrario, incremente ans por la diferencia de i+1 y A[i] .
• Imprime el valor de ans como resultado.

A continuación se muestra la implementación del enfoque anterior:

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the minimum additions required 
// to convert the array into a permutation of 1 to N
int minimumAdditions(int a[], int n)
{
    // Sort the array in increasing order
    sort(a, a + n);
    int ans = 0;
 
    // Traverse the array
    for (int i = 0; i < n; i++) {
 
        // If a[i] > i + 1, then return -1
        if ((i + 1) - a[i] < 0) {
            return -1;
        }
        if ((i + 1) - a[i] > 0) {
 
            // Update answer
            ans += (i + 1 - a[i]);
        }
    }
 
    // Return the required result
    return ans;
}
 
// Driver Code
int main()
{
    // Given Input
    int A[] = { 1, 1, 1, 1, 1 };
    int n = sizeof(A) / sizeof(A[0]);
 
    // Function Call
    cout << minimumAdditions(A, n);
 
    return 0;
}

// Java program for the above approach
import java.util.Arrays;
 
public class GFG {
 
    // Function to find the minimum additions required 
    // to convert the array into a permutation of 1 to N
    static int minimumAdditions(int a[], int n)
    {
        // Sort the array in increasing order
        Arrays.sort(a);
        int ans = 0;
 
        // Traverse the array
        for (int i = 0; i < n; i++) {
 
            // If a[i] > i + 1, then return -1
            if ((i + 1) - a[i] < 0) {
                return -1;
            }
            if ((i + 1) - a[i] > 0) {
 
                // Update answer
                ans += (i + 1 - a[i]);
            }
        }
 
        // Return the required result
        return ans;
    }
 
    // Driver code
    public static void main(String[] args)
    {
       
      // Given Input
        int A[] = { 1, 1, 1, 1, 1 };
        int n = A.length;
 
        // Function Call
        System.out.println(minimumAdditions(A, n));
    }
}
 
// This code is contributed by abhinavjain194

# Python3 program for the above approach
 
# Function to find the minimum additions
# required to convert the array into a
# permutation of 1 to N
def minimumAdditions(a, n):
     
    # Sort the array in increasing order
    a = sorted(a)
    ans = 0
 
    # Traverse the array
    for i in range(n):
 
        # If a[i] > i + 1, then return -1
        if ((i + 1) - a[i] < 0):
            return -1
 
        if ((i + 1) - a[i] > 0):
 
            # Update answer
            ans += (i + 1 - a[i])
 
    # Return the required result
    return ans
 
# Driver Code
if __name__ == '__main__':
     
    # Given Input
    A = [ 1, 1, 1, 1, 1 ]
    n = len(A)
 
    # Function Call
    print(minimumAdditions(A, n))
 
# This code is contributed by mohit kumar 29

// C# program for the above approach
using System;
 
class GFG{
 
// Function to find the minimum additions
// required to convert the array into a
// permutation of 1 to N
static int minimumAdditions(int []a, int n)
{
     
    // Sort the array in increasing order
    Array.Sort(a);
    int ans = 0;
 
    // Traverse the array
    for(int i = 0; i < n; i++)
    {
         
        // If a[i] > i + 1, then return -1
        if ((i + 1) - a[i] < 0)
        {
            return -1;
        }
         
        if ((i + 1) - a[i] > 0)
        {
             
            // Update answer
            ans += (i + 1 - a[i]);
        }
    }
 
    // Return the required result
    return ans;
}
 
// Driver code
static void Main()
{
     
    // Given Input
    int[] A = { 1, 1, 1, 1, 1 };
    int n = A.Length;
     
    // Function Call
    Console.Write(minimumAdditions(A, n));
}
}
 
// This code is contributed by SoumikMondal

<script>
// Javascript program for the above approach
 
// Function to find the minimum additions required
    // to convert the array into a permutation of 1 to N
function minimumAdditions(a, n)
{
 
    // Sort the array in increasing order
        a.sort(function(a, b){return a-b;});
        let ans = 0;
  
        // Traverse the array
        for (let i = 0; i < n; i++) {
  
            // If a[i] > i + 1, then return -1
            if ((i + 1) - a[i] < 0) {
                return -1;
            }
            if ((i + 1) - a[i] > 0) {
  
                // Update answer
                ans += (i + 1 - a[i]);
            }
        }
  
        // Return the required result
        return ans;
}
 
// Driver code
// Given Input
let A = [ 1, 1, 1, 1, 1 ];
let n = A.length;
 
// Function Call
document.write(minimumAdditions(A, n));
 
// This code is contributed by unknown2108
</script>

10

Complejidad de tiempo: O(N* log(N))
Espacio auxiliar: O(1)