Maximize array sum by replacing equal adjacent pairs by their sum and X respectively

Given two integers N and X which denotes the size of an array arr[] and the initial value of all the array elements respectively, the task is to find the maximum sum possible from the given array after performing the following operation any number of times.  

Choose any valid index i for which arr[i] = arr[i + 1] and update arr[i] = arr[i] + arr[i + 1] and arr[i + 1] = X.

Examples:

Input: N = 3, X = 5 
Output: 35 
Explanation:
Initially arr[] = [5, 5, 5] 
Performing the given operation on i = 1, arr[] = [10, 5, 5] 
Performing the given operation on i = 2, arr[] = [10, 10, 5] 
Performing the given operation on i = 1, arr[] = [20, 5, 5] 
Performing the given operation on i = 2, arr[] = [20, 10, 5] 
No adjacent equal elements are present in the array. 
Therefore, the maximum possible sum from the array is 35.

Input: N = 2, X = 3 
Output: 9 
Explanation:
Initially arr[] = [3, 3] 
Performing the given operation on i = 1, arr[] = [6, 3] 
No adjacent equal elements are present in the array. 
Therefore, the maximum possible sum from the array is 9. 

Naive Approach: The idea is to perform the given operation on every valid index in the initial array and find the maximum possible sum form all possible array rearrangements.

Time Complexity: O(2^N) 
Auxiliary Space: O(1)

Efficient Approach: The above approach can be optimized by using the following observation: 

• From the aforementioned examples, it can be observed that the value of the element at index i in the final array is given by: 
  

X * 2^{(N – i – 1)}

• Therefore, the sum of the final array will be equal to the sum of the series X * 2^{(N – i – 1)} for every valid index i.
• The sum of the above series is given by: 
  

Sum of the series = X * (2^N – 1) 

Therefore, simply print X * (2^N – 1) as the required answer. 
 Below is the implementation of the above approach:

C++

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

// Function to calculate x ^ y
int power(int x, int y)
{
    int temp;

    // Base Case
    if (y == 0)
        return 1;

    // Find the value in temp
    temp = power(x, y / 2);

    // If y is even
    if (y % 2 == 0)
        return temp * temp;
    else
        return x * temp * temp;
}

// Function that find the maximum
// possible sum of the array
void maximumPossibleSum(int N, int X)
{

    // Print the result using
    // the formula
    cout << (X * (power(2, N) - 1)) << endl;
}

// Driver code
int main()
{
    int N = 3, X = 5;

    // Function call
    maximumPossibleSum(N, X);
}

// This code is contributed by rutvik_56

Java

// Java program for the above approach

import java.io.*;

class GFG {

    // Function to calculate x ^ y
    static int power(int x, int y)
    {
        int temp;

        // Base Case
        if (y == 0)
            return 1;

        // Find the value in temp
        temp = power(x, y / 2);

        // If y is even
        if (y % 2 == 0)
            return temp * temp;
        else
            return x * temp * temp;
    }

    // Function that find the maximum
    // possible sum of the array
    public static void
    maximumPossibleSum(int N, int X)
    {
        // Print the result using
        // the formula
        System.out.println(
            X * (power(2, N) - 1));
    }

    // Driver Code
    public static void
        main(String[] args)
    {
        int N = 3, X = 5;

        // Function Call
        maximumPossibleSum(N, X);
    }
}

Python3

# Python3 program for the above approach

# Function to calculate x ^ y
def power(x, y):

    # Base Case
    if(y == 0):
        return 1

    # Find the value in temp
    temp = power(x, y // 2)

    # If y is even
    if(y % 2 == 0):
        return temp * temp
    else:
        return x * temp * temp

# Function that find the maximum
# possible sum of the array
def maximumPossibleSum(N, X):

    # Print the result using
    # the formula
    print(X * (power(2, N) - 1))

# Driver Code
N = 3
X = 5

# Function call
maximumPossibleSum(N, X)

# This code is contributed by Shivam Singh

C#

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

// Function to calculate x ^ y
static int power(int x, int y)
{
  int temp;

  // Base Case
  if (y == 0)
    return 1;

  // Find the value in temp
  temp = power(x, y / 2);

  // If y is even
  if (y % 2 == 0)
    return temp * temp;
  else
    return x * temp * temp;
}

// Function that find the maximum
// possible sum of the array
public static void maximumPossibleSum(int N,
                                      int X)
{
  // Print the result using
  // the formula
  Console.WriteLine(X * (power(2, N) - 1));
}

// Driver Code
public static void Main(String[] args)
{
  int N = 3, X = 5;

  // Function Call
  maximumPossibleSum(N, X);
}
}

// This code is contributed by shikhasingrajput

Javascript

<script>

// Javascript program for the above approach

// Function to calculate x ^ y
    function power(x , y) {
        var temp;

        // Base Case
        if (y == 0)
            return 1;

        // Find the value in temp
        temp = power(x, parseInt(y / 2));

        // If y is even
        if (y % 2 == 0)
            return temp * temp;
        else
            return x * temp * temp;
    }

    // Function that find the maximum
    // possible sum of the array
    function maximumPossibleSum(N , X)
    {
        // Print the result using
        // the formula
        document.write(X * (power(2, N) - 1));
    }

    // Driver Code

        var N = 3, X = 5;

        // Function Call
        maximumPossibleSum(N, X);

// This code contributed by aashish1995

</script>

Output:

35

Time Complexity: O(log N), The time complexity of this code is O(log N) because the power function is called recursively with a value of y being halved at each recursive call. Therefore, the number of recursive calls required to reach the base case of y=0 is log base 2 of N.
Space Complexity: O(1)