检查是否可以重新分配阵列

给定一个包含N个整数的数组arr[] ，任务是检查是否可以重新分配数组，使得对于每个1 ≤ i ≤ N （基于 1 的索引） arr[i] = i 。重新分配数组意味着数组的所有元素都可以更改为任何其他元素，但结果数组的总和必须等于原始数组总和。
例子：

Input: arr[] = {7, 4, 1, 1, 2}
Output: Yes
7 + 4 + 1 + 1 + 2 = 15
1 + 2 + 3 + 4 + 5 = 15
Input: arr[] = {1, 1, 1, 1}
Output: No
1 + 1 + 1 + 1 = 4
1 + 2 + 3 + 4 = 10

编程需要懂一点英语

处理方法：给定数组的和在修改后一定不能改变。因此，计算给定数组的总和，为了使数组的形式为1, 2, 3, ..., N ，数组元素的总和必须为(N * (N + 1)) / 2 。否则，这是不可能的。
下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
 
// Function that returns true if the
// array can be redistributed to
// the form 1, 2, 3, ..., N
bool canRedistribute(int* a, int n)
{
 
    // Calculate the sum of the array elements
    int sum = 0;
    for (int i = 0; i < n; i++)
        sum += a[i];
 
    // If can be redistributed
    if (sum == (n * (n + 1)) / 2)
        return true;
 
    return false;
}
 
// Driver code
int main()
{
    int a[] = { 7, 4, 1, 1, 2 };
    int n = sizeof(a) / sizeof(int);
 
    if (canRedistribute(a, n))
        cout << "Yes";
    else
        cout << "No";
 
    return 0;
}

Java

// Java implementation of the approach
import java.io.*;
 
class GFG
{
 
// Function that returns true if the
// array can be redistributed to
// the form 1, 2, 3, ..., N
static boolean canRedistribute(int []a, int n)
{
 
    // Calculate the sum of the array elements
    int sum = 0;
    for (int i = 0; i < n; i++)
        sum += a[i];
 
    // If can be redistributed
    if (sum == (n * (n + 1)) / 2)
        return true;
 
    return false;
}
 
// Driver code
public static void main (String[] args)
{
    int a[] = { 7, 4, 1, 1, 2 };
    int n = a.length;
 
    if (canRedistribute(a, n))
        System.out.print( "Yes");
    else
        System.out.print("No");
}
}
 
// This code is contributed by anuj_67..

Python3

# Python implementation of the approach
 
# Function that returns true if the
# array can be redistributed to
# the form 1, 2, 3, ..., N
def canRedistribute(a, n):
 
    # Calculate the sum of the array elements
    sum = 0;
    for i in range(n):
        sum += a[i];
 
    # If can be redistributed
    if (sum == (n * (n + 1)) / 2):
        return True;
 
    return False;
 
# Driver code
 
a = [7, 4, 1, 1, 2 ];
n = len(a);
 
if (canRedistribute(a, n)):
    print("Yes");
else:
    print("No");
 
# This code is contributed by 29AjayKumar

C#

// C# implementation of the approach
using System;
     
class GFG
{
 
// Function that returns true if the
// array can be redistributed to
// the form 1, 2, 3, ..., N
static Boolean canRedistribute(int []a, int n)
{
 
    // Calculate the sum of the array elements
    int sum = 0;
    for (int i = 0; i < n; i++)
        sum += a[i];
 
    // If can be redistributed
    if (sum == (n * (n + 1)) / 2)
        return true;
 
    return false;
}
 
// Driver code
public static void Main (String[] args)
{
    int []a = { 7, 4, 1, 1, 2 };
    int n = a.Length;
 
    if (canRedistribute(a, n))
        Console.WriteLine( "Yes");
    else
        Console.WriteLine("No");
}
}
 
/* This code is contributed by PrinciRaj1992 */

Javascript

输出：

Yes