最小化将数组转换为前 N 个自然数的排列所需的数字总和

给定大小为N的数组A[] ，任务是找到将数组转换为1到N的排列所需添加到数组元素的最小数字总和。如果数组无法转换为所需的排列，则打印-1 。

例子：

Input: A[] = {1, 1, 1, 1, 1}
Output: 10
Explanation: Increment A[1] by 1, A[2] by 2, A[3] by 3, A[4] by 4, thus A[] becomes {1, 2, 3, 4, 5}.
Minimum additions required = 1 + 2 + 3 + 4 = 10

Input: A[] = {2, 2, 3}
Output: -1

方法：想法是使用排序。请按照以下步骤解决此问题：

• 初始化一个变量ans来存储所需的结果。
• 按升序对给定数组A[]进行排序。
• 使用变量i遍历数组A[]
  • 如果A[i] > i+1的值，将ans更新为-1并跳出循环。
  • 否则通过i+1和A[i]的差异增加ans 。
• 打印ans的值作为结果。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
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

// 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

# 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#

// 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

Javascript

// 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

输出：

10

时间复杂度： O(N* log(N))
辅助空间： O(1)

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