Distribution with i allocated to arr[i]

Given an integer array arr[] of n elements, where arr[i] represents the number of goodies currently held by the (i+1)th student. The goodies can be redistributed among the students. Determine whether it is possible to redistribute them so that the student at 1-based index i receives exactly i goodies, with no goodies lost or created. Return true if such a redistribution is possible; otherwise, return false.

Examples:

Input: arr[] = [7, 4, 1, 1, 2]
Output: true
Explanation: The total number of goodies is 15, which is equal to 1 + 2 + 3 + 4 + 5. Therefore, the goodies can be redistributed so that the ith student receives exactly i goodies.

Input: arr[] = [1, 1, 1, 1, 1]
Output: false
Explanation: The total number of goodies is 5, whereas 1 + 2 + 3 + 4 + 5 = 15. Hence, the required distribution is not possible.

[Naive Approach] Using Iterative Sum - O(n) Time and O(1) Space

The idea is to calculate the total number of goodies available and also compute the required number of goodies by adding numbers from 1 to n using a loop. If both totals are equal, the redistribution is possible; otherwise, it is not.

Working of Approach:

• Traverse the array and calculate the total number of goodies.
• Initialize the required sum as 0.
• Add numbers from 1 to n one by one.
• Compare the available and required totals.
• Return true if both are equal; otherwise return false.

#include <iostream>
#include <vector>
using namespace std;

bool isPossible(vector<int> &arr)
{

    int sum = 0;

    // Calculate the total number of goodies.
    for (int x : arr)
        sum += x;

    int need = 0;

    // Calculate the required goodies using a loop.
    for (int i = 1; i <= arr.size(); i++)
        need += i;

    return sum == need;
}

int main()
{

    vector<int> arr = {7, 4, 1, 1, 2};

    if (isPossible(arr))
        cout << "true";
    else
        cout << "false";

    return 0;
}

class GFG {
    public boolean isPossible(int[] arr)
    {

        int sum = 0;

        // Calculate the total number of goodies.
        for (int x : arr)
            sum += x;

        int need = 0;

        // Calculate the required goodies using a loop.
        for (int i = 1; i <= arr.length; i++)
            need += i;

        return sum == need;
    }

    public static void main(String[] args)
    {

        int[] arr = { 7, 4, 1, 1, 2 };

        GFG obj = new GFG();

        if (obj.isPossible(arr))
            System.out.print("true");
        else
            System.out.print("false");
    }
}

def isPossible(arr):

    sum = 0

    # Calculate the total number of goodies.
    for x in arr:
        sum += x

    need = 0

    # Calculate the required goodies using a loop.
    for i in range(1, len(arr) + 1):
        need += i

    return sum == need


if __name__ == "__main__":

    arr = [7, 4, 1, 1, 2]

    if isPossible(arr):
        print("true")
    else:
        print("false")

using System;

class GFG {
    public bool isPossible(int[] arr)
    {

        int sum = 0;

        // Calculate the total number of goodies.
        foreach(int x in arr) sum += x;

        int need = 0;

        // Calculate the required goodies using a loop.
        for (int i = 1; i <= arr.Length; i++)
            need += i;

        return sum == need;
    }

    static void Main()
    {

        int[] arr = { 7, 4, 1, 1, 2 };

        GFG obj = new GFG();

        if (obj.isPossible(arr))
            Console.Write("true");
        else
            Console.Write("false");
    }
}

function isPossible(arr)
{

    let sum = 0;

    // Calculate the total number of goodies.
    for (let x of arr)
        sum += x;

    let need = 0;

    // Calculate the required goodies using a loop.
    for (let i = 1; i <= arr.length; i++)
        need += i;

    return sum === need;
}

// Driver Code
let arr = [ 7, 4, 1, 1, 2 ];

if (isPossible(arr))
    console.log("true");
else
    console.log("false");

true

[Expected Approach] Sum Formula - O(n) Time and O(1) Space

The idea is to observe that goodies can be redistributed freely, so only the total number of goodies matters. Compute the required total using the formula n × (n + 1) / 2 and compare it with the sum of the array.

Working of Approach:

• Find the total number of goodies in the array.
• Compute the required total using n * (n + 1) / 2.
• Compare both totals.
• If they are equal, redistribution is possible.
• Otherwise, return false.

Let us understand with an example:
Input: arr[] = [7, 4, 1, 1, 2]

• Total goodies available = 7 + 4 + 1 + 1 + 2 = 15.
• Number of students = 5, so the required goodies are 1 + 2 + 3 + 4 + 5 = 15.
• Since the available total matches the required total, the goodies can be redistributed accordingly.
• Therefore, the answer is true.

#include <iostream>
#include <vector>
using namespace std;

bool isPossible(vector<int> &arr)
{
    int sum = 0;
    int need = arr.size() * (arr.size() + 1) / 2;
    for (int x : arr)
    {

        // Calculate the total number of goodies.
        sum += x;
    }
    return sum == need;
}

int main()
{

    vector<int> arr = {7, 4, 1, 1, 2};

    if (isPossible(arr))
        cout << "true";
    else
        cout << "false";

    return 0;
}

class GFG {
    public boolean isPossible(int[] arr)
    {
        int sum = 0;
        int need = arr.length * (arr.length + 1) / 2;
        for (int x : arr) {

            // Calculate the total number of goodies.
            sum += x;
        }
        return sum == need;
    }

    public static void main(String[] args)
    {

        int[] arr = { 7, 4, 1, 1, 2 };

        GFG obj = new GFG();

        if (obj.isPossible(arr))
            System.out.print("true");
        else
            System.out.print("false");
    }
}

def isPossible(arr):
    sum = 0
    need = len(arr) * (len(arr) + 1) // 2
    for x in arr:
        # Calculate the total number of goodies.
        sum += x
    return sum == need


arr = [7, 4, 1, 1, 2]

if __name__ == "__main__":

    arr = [7, 4, 1, 1, 2]

    if isPossible(arr):
        print("true")
    else:
        print("false")

using System;

class GFG {
    public bool isPossible(int[] arr)
    {
        int sum = 0;
        int need = arr.Length * (arr.Length + 1) / 2;

        foreach(int x in arr)
        {

            // Calculate the total number of goodies.
            sum += x;
        }

        return sum == need;
    }

    static void Main()
    {

        int[] arr = { 7, 4, 1, 1, 2 };

        GFG obj = new GFG();

        if (obj.isPossible(arr))
            Console.Write("true");
        else
            Console.Write("false");
    }
}

function isPossible(arr)
{
    let sum = 0;
    let need = arr.length * (arr.length + 1) / 2;
    for (let x of arr) {
        // Calculate the total number of goodies.
        sum += x;
    }
    return sum === need;
}

// Driver Code
let arr = [ 7, 4, 1, 1, 2 ];

if (isPossible(arr))
    console.log("true");
else
    console.log("false");

true