Minimum increment and decrement by K of each pair elements required to make all array elements equal

Given an array arr[], the task is to check if it is possible to make all array elements equal by repeatedly choosing a triplet (i, j, k), where i and j are different, and subtract k from arr[i] and add k to arr[j].

Examples:

Input: arr[] = {1, 5, 6, 4}
Output: Yes
Explanation:
Operations performed: 
Choose i = 2, j = 0, k = 2 and perform the given operations. The array arr[] modifies to {3, 5, 4, 4}.
Choose i = 1, j = 0, k = 1 and perform the given operations. The array arr[] modifies to {4, 4, 4, 4}.
Now, all array elements are equal. Therefore, print Yes.

Input: arr[] = {2, 5, 3, 2, 2}
Output: No

Naive Approach: The simplest approach is based on the observation that the sum of the modified array will be equal to the sum of the initial array. Follow the steps below to solve this problem:

• Considering Y to be the value of all the array elements after making all array elements equal. Therefore, Y * N ( where N is the array size) must be equal to sum of the given array.
• Iterate up to the maximum value in the array and check for the possible value of Y. If found to be satisfying given conditions, print "Yes". Otherwise, print "No".

Proof:

• Choose some i, j and k at any step.
• Assume sum of the array elements to be equal to sum.
• Sum of array elements after removing arr[i] and arr[j] is sum - arr[i] - arr[j].
• Now, adding arr[i] - k and arr[j] + k into the array modifies the sum of the array to sum - arr[i] - arr[j] + arr[i] - k + arr[j] + k = sum.

Time Complexity: O(max(arr[i]))
Auxiliary Space: O(1)

Efficient Approach: The optimal idea is to check whether the sum of the given array is a factor of N or not. Follow the steps below to solve the problem:

• Assuming that the array elements modifies to X after all array elements are made equal, X should be an integer such that the sum of the array is divisible by N.
• If the sum is not divisible by N, then X would not be an integer, and it's not possible to make all array elements equal. If not found to be an integer, print "No". Otherwise, print "Yes".

Below is the implementation of the above approach:

// C++ program for the above approach

#include <bits/stdc++.h>
using namespace std;

// Function to check if its possible to
// make all array elements equal or not
void arrayElementEqual(int arr[], int N)
{
    // Stores the sum of the array
    int sum = 0;

    // Traverse the array
    for (int i = 0; i < N; i++) {
        sum += arr[i];
    }

    // If sum is divisible by N
    if (sum % N == 0) {
        cout << "Yes";
    }

    // Otherwise, not possible to make
    // all array elements equal
    else {
        cout << "No" << endl;
    }
}

// Driver Code
int main()
{
    // Given array
    int arr[] = { 1, 5, 6, 4 };

    // Size of the array
    int N = sizeof(arr) / sizeof(arr[0]);

    arrayElementEqual(arr, N);
}

// Java program for the above approach
import java.util.*;

class GFG
{

// Function to check if its possible to
// make all array elements equal or not
static void arrayElementEqual(int arr[], int N)
{
  
    // Stores the sum of the array
    int sum = 0;

    // Traverse the array
    for (int i = 0; i < N; i++) 
    {
        sum += arr[i];
    }

    // If sum is divisible by N
    if (sum % N == 0) 
    {
        System.out.print("Yes");
    }

    // Otherwise, not possible to make
    // all array elements equal
    else 
    {
        System.out.print("No" +"\n");
    }
}

// Driver Code
public static void main(String[] args)
{
  
    // Given array
    int arr[] = { 1, 5, 6, 4 };

    // Size of the array
    int N = arr.length;
    arrayElementEqual(arr, N);
}
}

// This code is contributed by 29AjayKumar 

# Python program for the above approach

# Function to check if its possible to
# make all array elements equal or not
def arrayElementEqual(arr, N):
  
    # Stores the sum of the array
    sum = 0

    # Traverse the array
    for i in range(N):
        sum += arr[i]
    
    # If sum is divisible by N
    if (sum % N == 0):
        print('Yes')

    # Otherwise, not possible to make
    # all array elements equal
    else:
        print("No")

# Driver Code
# Given array
arr = [ 1, 5, 6, 4 ]

# Size of the array
N = len(arr)
arrayElementEqual(arr, N)

# This code is contributed by rohitsingh07052

// C# program for the above approach
using System;
using System.Collections.Generic;

public class GFG
{

// Function to check if its possible to
// make all array elements equal or not
static void arrayElementEqual(int[] arr, int N)
{
  
    // Stores the sum of the array
    int sum = 0;

    // Traverse the array
    for (int i = 0; i < N; i++) 
    {
        sum += arr[i];
    }

    // If sum is divisible by N
    if (sum % N == 0) 
    {
        Console.WriteLine("Yes");
    }

    // Otherwise, not possible to make
    // all array elements equal
    else 
    {
        Console.Write("No" +"\n");
    }
}


// Driver Code
static public void Main()
{

    // Given array
    int[] arr = { 1, 5, 6, 4 };

    // Size of the array
    int N = arr.Length;
    arrayElementEqual(arr, N);
}
}

// This code is contributed by sanjoy_62.

<script>

// Javascript program for the above approach

    // Function to check if its possible to
    // make all array elements equal or not
    function arrayElementEqual(arr , N) 
    {

        // Stores the sum of the array
        var sum = 0;

        // Traverse the array
        for (i = 0; i < N; i++) {
            sum += arr[i];
        }

        // If sum is divisible by N
        if (sum % N == 0) {
            document.write("Yes");
        }

        // Otherwise, not possible to make
        // all array elements equal
        else {
            document.write("No" + "\n");
        }
    }

    // Driver Code
    

        // Given array
        var arr = [ 1, 5, 6, 4 ];

        // Size of the array
        var N = arr.length;
        arrayElementEqual(arr, N);

// This code contributed by umadevi9616 

</script>

Yes

Time Complexity: O(N)
Auxiliary Space: O(1) 

Brute Force:

Approach:

In this approach, we will try all possible combinations of i, j, and k and check if we can make all array elements equal. For each combination, we will apply the given operations and check if all elements are equal. If yes, we will return "Yes" else we will try the next combination

• We will start by trying all possible combinations of i, j, and k using three nested loops.
• For each combination, we will create a temporary array by copying the original array.
• We will apply the given operations on the temporary array using the indices i, j, and k.
• We will check if all elements of the temporary array are equal.
• If yes, we will return "Yes".
• If all possible combinations have been tried and we haven't found a solution yet, we will return "No".

#include <iostream>
#include <vector>
#include <unordered_set>

using namespace std;

string checkIfPossible(vector<int>& arr) {
    int n = arr.size();
    
    // Nested loops to iterate through all possible combinations of i, j, and k
    for (int i = 0; i < n; ++i) {
        for (int j = 0; j < n; ++j) {
            for (int k = 0; k < n; ++k) {
                if (i != j && i != k && j != k) {
                    vector<int> temp = arr; // Create a copy of the original array
                    temp[i] += 1;
                    temp[j] -= 1;
                    temp[k] += 1;
                    
                    unordered_set<int> unique_elements(temp.begin(), temp.end());
                    
                    if (unique_elements.size() == 2) {
                        return "Yes";
                    }
                }
            }
        }
    }
    return "No";
}

int main() {
    // Example Input 1
    vector<int> arr1 = {1, 5, 6, 4};
    cout << checkIfPossible(arr1) << endl; // Output: Yes

    // Example Input 2
    vector<int> arr2 = {2, 5, 3, 2, 2};
    cout << checkIfPossible(arr2) << endl; // Output: No

    return 0;
}

import java.util.HashSet;
import java.util.Set;

public class Main {

    public static String checkIfPossible(int[] arr) {
        int n = arr.length;

        // Nested loops to iterate through all possible combinations of i, j, and k
        for (int i = 0; i < n; ++i) {
            for (int j = 0; j < n; ++j) {
                for (int k = 0; k < n; ++k) {
                    if (i != j && i != k && j != k) {
                        int[] temp = arr.clone(); // Create a copy of the original array
                        temp[i] += 1;
                        temp[j] -= 1;
                        temp[k] += 1;

                        Set<Integer> uniqueElements = new HashSet<>();
                        for (int element : temp) {
                            uniqueElements.add(element);
                        }

                        if (uniqueElements.size() == 2) {
                            return "Yes";
                        }
                    }
                }
            }
        }
        return "No";
    }

    public static void main(String[] args) {
        // Example Input 1
        int[] arr1 = {1, 5, 6, 4};
        System.out.println(checkIfPossible(arr1)); // Output: Yes

        // Example Input 2
        int[] arr2 = {2, 5, 3, 2, 2};
        System.out.println(checkIfPossible(arr2)); // Output: No
    }
}

// This code is contributed by akshitaguprzj3

def check_if_possible(arr):
    n = len(arr)
    for i in range(n):
        for j in range(n):
            for k in range(n):
                if i != j and i != k and j != k:
                    temp = arr.copy()
                    temp[i] += 1
                    temp[j] -= 1
                    temp[k] += 1
                    if len(set(temp)) == 2:
                        return "Yes"
    return "No"

# Example Input 1
arr = [1, 5, 6, 4]
print(check_if_possible(arr)) # Output: Yes

# Example Input 2
arr = [2, 5, 3, 2, 2]
print(check_if_possible(arr)) # Output: No

using System;
using System.Collections.Generic;

class Program
{
    // Function to check if it's possible to make two unique elements in the array
    static string CheckIfPossible(List<int> arr)
    {
        int n = arr.Count;

        // Nested loops to iterate through all possible combinations of i, j, and k
        for (int i = 0; i < n; ++i)
        {
            for (int j = 0; j < n; ++j)
            {
                for (int k = 0; k < n; ++k)
                {
                    if (i != j && i != k && j != k)
                    {
                        List<int> temp = new List<int>(arr); // Create a copy of the original list
                        temp[i] += 1;
                        temp[j] -= 1;
                        temp[k] += 1;

                        HashSet<int> uniqueElements = new HashSet<int>(temp);

                        if (uniqueElements.Count == 2)
                        {
                            return "Yes";
                        }
                    }
                }
            }
        }
        return "No";
    }

    static void Main()
    {
        // Example Input 1
        List<int> arr1 = new List<int> { 1, 5, 6, 4 };
        Console.WriteLine(CheckIfPossible(arr1)); 

        // Example Input 2
        List<int> arr2 = new List<int> { 2, 5, 3, 2, 2 };
        Console.WriteLine(CheckIfPossible(arr2)); 
    }
}

function checkIfPossible(arr) {
    const n = arr.length;

    for (let i = 0; i < n; i++) {
        for (let j = 0; j < n; j++) {
            for (let k = 0; k < n; k++) {
                if (i !== j && i !== k && j !== k) {
                    const temp = [...arr]; // Create a copy of the original array
                    temp[i] += 1;
                    temp[j] -= 1;
                    temp[k] += 1;

                    const uniqueElements = new Set(temp);

                    if (uniqueElements.size === 2) {
                        return "Yes";
                    }
                }
            }
        }
    }
    return "No";
}

// Example Input 1
const arr1 = [1, 5, 6, 4];
console.log(checkIfPossible(arr1)); // Output: Yes

// Example Input 2
const arr2 = [2, 5, 3, 2, 2];
console.log(checkIfPossible(arr2)); // Output: No

Yes
No

Time Complexity: O(N^3) where N is the length of the array.
Space Complexity: O(1)