Find Maximum Difference Between any Two Pairs By Following Operations Optimally

Given an array X[] of length N along with A and B. You can apply below type of operations:

• Operation 1: Choose two different indices, delete elements at both indices and insert the sum of them in X[]. This operation decrements A by 1.
• Operation 2: Choose two different indices, delete elements at both indices and insert the difference of them in X[]. This operation decrements B by 1.

Considered that after making optimal number of operations, we get array Y[].

Then your task is to output the maximum difference between the maximum and minimum element among all the possible Y[] that can be formed using given operation under the cost of A and B.

Note: It is not necessary to use all cost A and B.

Examples:

Input: N = 6, A = 1, B = 2, X[] = {8, -1, -4, 2, 6, -3}
Output: 23
Explanation: The operations are performed as:

• First Operation: Choose i = 4 and j = 6, So that A₄ = 2 and A₆ = -3. Difference between them is: 2 – (-3) = 5. Add difference into X[] and delete both elements. So updated X[] = {8, -1, -4, 6, 5}. This operation decrements B by 1. Now, A = 1 and B = 2 – 1 = 1.
• Second Operation: Choose i = 1 and j = 5, So that A₁ = 8 and A₅ = 5. Sum of them is: 8 + 5 = 13. Add sum into X[] and delete both elements. So updated X[] = {-1, -4, 6, 13}. This operation decrements A by 1. Now, A = 0 and B = 1.
• Third Operation: Choose i = 2 and j = 3, So that A₂ = -4 and A₃ = 6. Difference between them is: -4 – (6) = -10. Add difference into X[] and delete both elements. So updated X[] = {-1, -10, 13}. This costs decrements B by 1. Now, A = 0 and B = 0.

Now, In X[] max element and minimum element of X[] are 13 and -10 respectively. The difference between them is 13 – (-10) = 23. Which is maximum among all the possible arrays formed by given operation. Thus, output is 23.

Input: N = 3, A = 0, B = 0, X[] = {3, -1, 0}
Output: 4
Explanation: As A and B are initially zero. We can’t make any type of given operation, As the value of A or B must be greater than or equal to 1. Thus, the maximum possible difference between maximum and minimum value will be: 3 – (-1) = 4.

Approach: Follow below idea to solve the above problem:

Main logic: To maximize the difference, we have two straightforward options:

• Either increase the maximum element in the X[] by adding a number (raising the maximum), or decrease the minimum element by subtracting a number (lowering the minimum).

Before any operations, the result (denoted as Res) can be expressed as the difference between the maximum (Max_elem) and minimum (Min_elem) elements:

• To increase the maximum element, we add X: (Max_elem + X) – Min_elem = Res + X.
• To decrease the minimum element, we subtract X: Max_elem – (Min_elem – X) = Max_elem – Min_elem + X = Res + X. If X is negative, we add it to the minimum element to lower it, which simplifies to: Max_elem – (Min_elem + (-X)) = Max_elem – (Min_elem – x) = Res + X.
• This leads us to the third observation: the number’s parity is inconsequential. In every scenario, we add the absolute value of X to the result. This process continues for (A + B) iterations or until the array is exhausted, capped at min(A+B, N-2) iterations. We use N-2 because X[0] and X[n-1] are the minimum and maximum elements initially used to calculate Res.

Example:

X[] = {-5, -4, 3, 7}

At current, the difference is 7- (-5) = 12. Let A = 1, B = 0, obviously you could have just added 3 to 7 thinking its a positive number and move on . But when you grow serious, you realize that instead of adding 3 to 7, (which makes net difference ((7+3) – (-5) = 15), you could have added -4 to -5 to get a more difference (7 – (-5-4) = 16). So, what we realized, we need to deal with absolute value of numbers instead of numbers itself.

Steps were taken to solve the problem:

• Sort X[].
• Declare a variable let say Res and store the initial difference between max and min element of X[].
• If (A == 0 and B == 0)
  • Output value store in Res.
• Declare a variable let say Ops and initialize it with the total available cost. Formally, A+B
• Run a loop for i = 1 to i < N-1 and make all the elements positive.
• Sort X[], except first and last element.
• Run a loop for i = N – 2 to i>0 and follow below mentioned steps under the scope of loop:
  • If (Ops == 0)
    • Break
  • Else
    • Res += X[i]
    • Ops—
• Output the value stored in Res.

Below is the implementation of the above idea:

C++

// code by flutterfly
#include <iostream>
#include <algorithm>
#include <vector>
using namespace std;

// Function to output maximum difference
void Max_diff(long long N, long long A, long long B, vector<long long>& X)
{
    // Sorting X[] using inbuilt function
    sort(X.begin(), X.end());

    // Max difference in the initial array
    long long res = X[N - 1] - X[0];

    // Variable to hold the cumulative sum of both costs
    long long ops = A + B;

    // If both costs are zero
    // Then output the initial difference
    if (A == 0 && B == 0) {
        cout << res << endl;
    }

    // Else implementing the discussed approach
    else {

        // Looping over each element and changing them
        // into positive elements
        for (int i = 1; i < N - 1; i++) {
            X[i] = abs(X[i]);
        }

        // Sorting X
        sort(X.begin() + 1, X.end() - 1);

        // Calculating the difference
        for (int i = N - 2; i > 0; i--) {
            if (ops == 0) {
                break;
            }
            else {
                res += X[i];
                ops--;
            }
        }

        // Printing out the max difference
        cout << res << endl;
    }
}

// Driver Function
int main()
{
    // Inputs
    long long N = 7;
    long long A = 6;
    long long B = 6;
    vector<long long> X = { -2, -4, 2, -2, -3, -1, -1 };

    // Function call
    Max_diff(N, A, B, X);

    return 0;
}

Java

// Java code to implement the approach

import java.util.*;

// Driver Class
public class Main {

    // Driver Function
    public static void main(String[] args)
    {

        // Inputs
        long N = 7;
        long A = 6;
        long B = 6;
        long[] X = { -2, -4, 2, -2, -3, -1, -1 };

        // Function_call
        Max_diff(N, A, B, X);
    }

    // Method to output maximum difference
    public static void Max_diff(long N, long A, long B,
                                long[] X)
    {

        // Sorting X[] using inbuilt function
        Arrays.sort(X);

        // Max difference in initial array
        long res = X[(int)N - 1] - X[0];

        // Variable to hold the cumulative sum of
        // both costs
        long ops = A + B;

        // If both costs are zero
        // Then output initial difference
        if (A == 0 && B == 0) {
            System.out.println(res);
        }

        // Else implementing the discussed approach
        else {

            // Looping over each element and changing them
            // into positive elements
            for (int i = 1; i < N - 1; i++) {
                X[i] = Math.abs(X[i]);
            }

            // Sorting X
            Arrays.sort(X, 1, (int)N - 1);

            // Calculating the difference
            for (int i = (int)N - 2; i > 0; i--) {
                if (ops == 0) {
                    break;
                }
                else {
                    res += X[i];
                    ops--;
                }
            }

            // Printing out the max difference
            System.out.println(res);
        }
    }
}

Python

# code by flutterfly
# Python code to implement the approach

# Method to output maximum difference
def max_diff(N, A, B, X):
    # Sorting X[] using inbuilt function
    X.sort()

    # Max difference in the initial array
    res = X[N - 1] - X[0]

    # Variable to hold the cumulative sum of both costs
    ops = A + B

    # If both costs are zero, then output initial difference
    if A == 0 and B == 0:
        print(res)
    else:
        # Looping over each element and changing them into positive elements
        for i in range(1, N - 1):
            X[i] = abs(X[i])

        # Sorting X
        X[1:N - 1] = sorted(X[1:N - 1])

        # Calculating the difference
        for i in range(N - 2, 0, -1):
            if ops == 0:
                break
            else:
                res += X[i]
                ops -= 1

        # Printing out the max difference
        print(res)


# Driver Function
if __name__ == "__main__":
    # Inputs
    N = 7
    A = 6
    B = 6
    X = [-2, -4, 2, -2, -3, -1, -1]

    # Function call
    max_diff(N, A, B, X)

C#

using System;
using System.Collections.Generic;
using System.Linq;

class Program {
    // Function to output maximum difference
    static void Max_diff(long N, long A, long B,
                         List<long> X)
    {
        // Sorting X[] using inbuilt function
        X.Sort();

        // Max difference in the initial array
        long res = X[(int)N - 1] - X[0];

        // Variable to hold the cumulative sum of both costs
        long ops = A + B;

        // If both costs are zero
        // Then output the initial difference
        if (A == 0 && B == 0) {
            Console.WriteLine(res);
        }
        // Else implementing the discussed approach
        else {
            // Looping over each element and changing them
            // into positive elements
            for (int i = 1; i < N - 1; i++) {
                X[i] = Math.Abs(X[i]);
            }

            // Sorting X
            X.Sort(1, (int)N - 2, Comparer<long>.Default);

            // Calculating the difference
            for (int i = (int)N - 2; i > 0; i--) {
                if (ops == 0) {
                    break;
                }
                else {
                    res += X[i];
                    ops--;
                }
            }

            // Printing out the max difference
            Console.WriteLine(res);
        }
    }

    // Driver Function
    static void Main()
    {
        // Inputs
        long N = 7;
        long A = 6;
        long B = 6;
        List<long> X
            = new List<long>{ -2, -4, 2, -2, -3, -1, -1 };

        // Function call
        Max_diff(N, A, B, X);
    }
}

Javascript

// Function to output maximum difference
function maxDiff(N, A, B, X) {
    // Sorting X[] using inbuilt function
    X.sort((a, b) => a - b);

    // Max difference in the initial array
    let res = X[N - 1] - X[0];

    // Variable to hold the cumulative sum of both costs
    let ops = A + B;

    // If both costs are zero, then output initial difference
    if (A === 0 && B === 0) {
        console.log(res);
    } else {
        // Looping over each element and changing them into positive elements
        for (let i = 1; i < N - 1; i++) {
            X[i] = Math.abs(X[i]);
        }

        // Sorting X
        X.slice(1, N - 1).sort((a, b) => a - b);

        // Calculating the difference
        for (let i = N - 2; i > 0; i--) {
            if (ops === 0) {
                break;
            } else {
                res += X[i];
                ops -= 1;
            }
        }

        // Printing out the max difference
        console.log(res);
    }
}

// Driver Function
// Inputs
const N = 7;
const A = 6;
const B = 6;
const X = [-2, -4, 2, -2, -3, -1, -1];

// Function call
maxDiff(N, A, B, X);

Output

15

Time Complexity: O(N*logN), As Sorting is performed.
Auxiliary Space: O(1)