# 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:
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(2N
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 * (2N – 1)

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

 // C++ program for the above approach #include 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 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 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# 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

Output:
35

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

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