Minimum operations required to make all elements in an array of first N odd numbers equal

Given an array consisting of first N odd numbers, the task is to find the minimum number of operations required to make all the array elements equal by repeatedly selecting a pair and incrementing one element and decrementing the other element in the pair by 1.

Examples:

Input: N = 3
Output: 2
Explanation:
Initially, the array is {1, 3, 5}. Following operations are performed:
Operation 1: Decrement arr[2] by 1 and increment arr[0] by 1. The array modifies to {2, 3, 4}.
Operation 2: Decrement arr[2] by 1 and increment arr[0] by 1. The array modifies to {3, 3, 3}.
Therefore, the minimum number of operations required is 2.

Input: N = 6
Output: 9

Naive Approach: The given problem can be solved based on the following observations:

• It can be observed that making all the array elements equal to the middle element of the array requires minimum number of operations.
• Therefore, the idea is to traverse the array using a variable i and select the element at index i and (N – i – 1) in each operation and make them equal to the middle element.

Follow the steps below to solve the problem:

• Initialize a variable, say mid, that stores the middle element of the array.
• Initialize an array arr[] that stores the array element as arr[i] = 2 * i + 1.
• Find the sum of the array arr[] and store it in a variable say sum.
• If N is odd, then update the value of mid to arr[N / 2]. Otherwise, update the value of mid to sum / N.
• Initialize a variable, say ans as 0 that stores the minimum number of operations.
• Traverse the given array over the range [0, N/2] and increment the value of ans by the value (mid – arr[i]).
• After completing the above steps, print the value of ans as the result.

Below is the implementation of the above approach:

C++14

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;

// Function to find the minimum number
// of operations required to make the
// array elements equal
int minOperations(int N)
{

    // Stores the array elements
    int arr[N];

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

    // Iterate over the range [0, N]
    for(int i = 0; i < N; i++)
    {

        // Update the value arr[i]
        arr[i] = (2 * i) + 1;

        // Increment the sum by
        // the value arr[i]
        sum = sum + arr[i];
    }

    // Stores the middle element
    int mid = 0;

    // If N is even
    if (N % 2 == 0)
    {
        mid = sum / N;
    }

    // Otherwise
    else {
        mid = arr[N / 2];
    }

    // Stores the result
    int ans = 0;

    // Traverse the range [0, N / 2]
    for(int i = 0; i < N / 2; i++)
    {

        // Update the value of ans
        ans += mid - arr[i];
    }

    // Return the result
    return ans;
}

// Driver Code
int main()
{
    int N = 6;
    cout << minOperations(N);

    return 0;
}

// This code is contributed by susmitakundugoaldanga

Java

// Java program for the above approach

class GFG {

    // Function to find the minimum number
    // of operations required to make the
    // array elements equal
    public static int minOperations(int N)
    {
        // Stores the array elements
        int[] arr = new int[N];

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

        // Iterate over the range [0, N]
        for (int i = 0; i < N; i++) {

            // Update the value arr[i]
            arr[i] = (2 * i) + 1;

            // Increment the sum by
            // the value arr[i]
            sum = sum + arr[i];
        }

        // Stores the middle element
        int mid = 0;

        // If N is even
        if (N % 2 == 0) {
            mid = sum / N;
        }

        // Otherwise
        else {
            mid = arr[N / 2];
        }

        // Stores the result
        int ans = 0;

        // Traverse the range [0, N / 2]
        for (int i = 0; i < N / 2; i++) {

            // Update the value of ans
            ans += mid - arr[i];
        }

        // Return the result
        return ans;
    }

    // Driver Code
    public static void main(String[] args)
    {
        int N = 6;
        System.out.println(
            minOperations(N));
    }
}

Python3

# Python3 program for the above approach

# Function to find the minimum number
# of operations required to make the
# array elements equal
def minOperations(N):

    # Stores the array elements
    arr = [0] * N

    # Stores the sum of the array
    sum = 0

    # Iterate over the range [0, N]
    for i in range(N) :

        # Update the value arr[i]
        arr[i] = (2 * i) + 1

        # Increment the sum by
        # the value arr[i]
        sum = sum + arr[i]

    # Stores the middle element
    mid = 0

    # If N is even
    if N % 2 == 0 :
        mid = sum / N

    # Otherwise
    else:
        mid = arr[int(N / 2)]

    # Stores the result
    ans = 0

    # Traverse the range [0, N / 2]
    for i in range(int(N / 2)):

        # Update the value of ans
        ans += mid - arr[i]

    # Return the result
    return int(ans)

# Driver Code
N = 6
print(minOperations(N))

# This code is contributed by Dharanendra L V.

C#

// C# program for the above approach
using System;

class GFG{

// Function to find the minimum number
// of operations required to make the
// array elements equal
public static int minOperations(int N)
{

    // Stores the array elements
    int[] arr = new int[N];

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

    // Iterate over the range [0, N]
    for(int i = 0; i < N; i++)
    {

        // Update the value arr[i]
        arr[i] = (2 * i) + 1;

        // Increment the sum by
        // the value arr[i]
        sum = sum + arr[i];
    }

    // Stores the middle element
    int mid = 0;

    // If N is even
    if (N % 2 == 0)
    {
        mid = sum / N;
    }

    // Otherwise
    else
    {
        mid = arr[N / 2];
    }

    // Stores the result
    int ans = 0;

    // Traverse the range [0, N / 2]
    for(int i = 0; i < N / 2; i++)
    {

        // Update the value of ans
        ans += mid - arr[i];
    }

    // Return the result
    return ans;
}

// Driver Code
public static void Main(string[] args)
{
    int N = 6;

    Console.WriteLine(minOperations(N));
}
}

// This code is contributed by ukasp

Javascript

<script>

// javascript program for the above approach

// Function to find the minimum number
// of operations required to make the
// array elements equal
function minOperations(N)
{

    // Stores the array elements
    let arr = Array.from({length: N}, (_, i) => 0);

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

    // Iterate over the range [0, N]
    for(let i = 0; i < N; i++)
    {

        // Update the value arr[i]
        arr[i] = (2 * i) + 1;

        // Increment the sum by
        // the value arr[i]
        sum = sum + arr[i];
    }

    // Stores the middle element
    let mid = 0;

    // If N is even
    if (N % 2 == 0)
    {
        mid = sum / N;
    }

    // Otherwise
    else
    {
        mid = arr[N / 2];
    }

    // Stores the result
    let ans = 0;

    // Traverse the range [0, N / 2]
    for(let i = 0; i < N / 2; i++)
    {

        // Update the value of ans
        ans += mid - arr[i];
    }

    // Return the result
    return ans;
}

// Driver Code

    let N = 6;
    document.write(minOperations(N));

</script>

Output:

9

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

Efficient Approach: The above approach can be optimized based on the following observation:

• If N is odd then the result is the sum of first N / 2 even numbers i.e., (K * (K + 1)) / 2, where K = (N / 2).
• Otherwise, the result is the sum of the first K odd numbers i.e., K * K, where K = ((N – 1) / 2).

Therefore, if the value of N is even, then print the value of (N / 2)². Otherwise, print the value of K * (K + 1) / 2, where K = ((N – 1) / 2) as the resultant operation.

Below is the implementation of the above approach:

C++

// C++ code for the above approach
#include <bits/stdc++.h>
using namespace std;

// Function to find the minimum number
    // of operations required to make the
    // array elements equal
    int minOperation(int N)
    {
        // If the value of N is even
        if (N % 2 == 0) {

            // Return the value
            return (N / 2) * (N / 2);
        }

        // Otherwise, N is odd
        int k = (N - 1) / 2;

        // Return the value
        return k * (k + 1);
    }

// Driver Code
int main()
{
     int N = 6;
     cout << minOperation(N);

    return 0;
}

// This code is contributed by code_hunt.

Java

// Java program for the above approach

class GFG {

    // Function to find the minimum number
    // of operations required to make the
    // array elements equal
    public static int minOperation(int N)
    {
        // If the value of N is even
        if (N % 2 == 0) {

            // Return the value
            return (N / 2) * (N / 2);
        }

        // Otherwise, N is odd
        int k = (N - 1) / 2;

        // Return the value
        return k * (k + 1);
    }

    // Driver Code
    public static void main(String[] args)
    {
        int N = 6;
        System.out.println(
            minOperation(N));
    }
}

C#

// C# program for the above approach
using System;
class GFG{

// Function to find the minimum number
// of operations required to make the
// array elements equal
public static int minOperation(int N)
{

    // If the value of N is even
    if (N % 2 == 0)
    {

        // Return the value
        return (N / 2) * (N / 2);
    }

    // Otherwise, N is odd
    int k = (N - 1) / 2;

    // Return the value
    return k * (k + 1);
}

// Driver code
static void Main()
{
    int N = 6;

    Console.WriteLine(minOperation(N));
}
}

// This code is contributed by abhinavjain194

Python3

# Python3 program for the above approach

# Function to find the minimum number
# of operations required to make the
# array elements equal
def minOperation(N) :

    # If the value of N is even
    if (N % 2 == 0) :

        # Return the value
        return (N / 2) * (N / 2)

    # Otherwise, N is odd
    k = (N - 1) / 2

    # Return the value
    return (k * (k + 1))

# Driver Code

N = 6
print(int(minOperation(N)))

# This code is contributed by souravghosh0416.

Javascript

<script>

// Javascript program implementation
// of the approach

  // Function to find the minimum number
    // of operations required to make the
    // array elements equal
    function minOperation(N)
    {
        // If the value of N is even
        if (N % 2 == 0) {

            // Return the value
            return (N / 2) * (N / 2);
        }

        // Otherwise, N is odd
        let k = (N - 1) / 2;

        // Return the value
        return k * (k + 1);
    }


// Driver Code

        let N = 6;
         document.write(
            minOperation(N));

 // This code is contributed by splevel62.
</script>

Output:

9

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