Check if it is possible to redistribute the Array

Given an array arr[] of N integers, the task is to check if it is possible to redistribute the array such that for every 1 ? i ? N (1-based indexing) arr[i] = i. Redistributing the array means all the elements of the array can be changed to any other element but the sum of the resultant array must be equal to the original array sum.

Examples:

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 

Approach: It is given that the sum of the array must not change after the modification. So, calculate the sum of the given array and in order for the array to be of the form 1, 2, 3, …, N, the sum of the array elements must be (N * (N + 1)) / 2. Else, it is impossible.

Below is the implementation of the above approach: 

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
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

<script>
 //Javascript implementation of the approach

// Function that returns true if the
// array can be redistributed to
// the form 1, 2, 3, ..., N
function canRedistribute(a, n)
{

    // Calculate the sum of the array elements
    var sum = 0;
    for (var i = 0; i < n; i++)
        sum += a[i];

    // If can be redistributed
    if (sum == (n * (n + 1)) / 2)
        return true;

    return false;
}

var a = [ 7, 4, 1, 1, 2 ];
var n = a.length;

    if (canRedistribute(a, n))
        document.write("Yes");
    else
       document.write( "No");


// This code is contributed by SoumikMondal
</script>

Output

Yes

Time Complexity: O(n), where n is the length of the given array.
Auxiliary Space: O(1)