Maximum possible score that can be obtained by constructing a Binary Tree based on given conditions

Given an array arr[] of (N – 1) integers and each value arr[i](1-based indexing) is the score of the nodes having degree i. The task is to determine the maximum score of any tree of N nodes that can be constructed.

Examples:

Input: arr[] = {1, 3, 0}
Output: 8
Explanation:
One possible way to construct tree is:
1
/ \
2 3
\
4
Node 1 have degree 2. Therefore, its score is 3.
Node 2 have degree 1. Therefore, its score is 1.
Node 3 have degree 2. Therefore, its score is 3.
Node 4 have degree 1. Therefore, its score is 1.
Therefore, the total score = 3 + 1 + 3 + 1 = 8.

Input: arr[] = {0, 1}
Output: 1
Explanation:
One possible way to construct tree is:
1
/ \
2 3
Node 1 have degree 2. Therefore, its score is 1.
Node 2 have degree 1. Therefore, its score is 0.
Node 3 have degree 1. Therefore, its score is 0.
Therefore, total score = 1 + 0 + 0 = 1.

Naive Approach: The simplest approach is to generate all possible combinations of constructing a tree having N nodes and find the total score for each of them. Then, print the maximum of all the scores obtained.

Time Complexity: (N!) where N is the number of nodes in the tree.
Auxiliary Space: O(N)

Efficient Approach: To optimize the above approach, the idea is to use Dynamic Programming by creating a dp[][] table where dp[i][j] represents the maximum score using i nodes having the sum of degrees of the nodes as j. Follow the below steps to solve the problem:

Initialize an array dp[N + 1][2*(N – 1) + 1] where N is the number of nodes and (2*(N – 1)) is the maximum sum of degrees.
Initialize dp[0][0] with 0.
Iterate two nested loops, one over the range [1, N], and another for till the possible maximum score 2*(N – 1) from 1 and for each score s in the range [1, N] traverse the given array of scores arr[] and updating dp[i][s] as:

dp[i][s] = max(dp[i][s], scores[j-1] dp[i-1][s-j])
where dp[i][s] represents the maximum score of tree having i nodes and sum of degrees as s.

For a tree with N vertices and (N – 1) edges, the sum of all degrees should be 2 * (N – 1). Therefore, print the value of dp[N][2*(N – 1)] as the maximum score for a tree with N nodes.

Below is the implementation of the above approach:

C++

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

using namespace std;
 
// Function to find the maximum score
// for one possible tree having N nodes
// N - 1 Edges

int maxScore(vector<int>& arr)
{

    int N = arr.size();
 
    // Number of nodes

    N++;
 
    // Initialize dp[][]

    vector<vector<int> >

        dp(N + 1, vector<int>(2 * N,

                              -100000));
 
    // Score with 0 vertices is 0

    dp[0][0] = 0;
 
    // Traverse the nodes from 1 to N

    for (int i = 1; i <= N; i++) {
 
        // Find maximum scores for

        // each sum

        for (int s = 1;

             s <= 2 * (N - 1); s++) {
 
            // Iterate over degree of

            // new node

            for (int j = 1; j <= N - 1

                            and j <= s;

                 j++) {
 
                // Update the current

                // state

                dp[i][s]

                    = max(dp[i][s],

                          arr[j - 1]

                              + dp[i - 1][s - j]);

            }

        }

    }
 
    // Return maximum score for N node

    // tree having 2(N - 1) sum of degree

    return dp[N][2 * (N - 1)];
}
 
// Driver Code

int main()
{

    // Given array of scores

    vector<int> arr = { 1, 3, 0 };
 
    // Function Call

    cout << maxScore(arr);
 
    return 0;
}

Java

// Java program for the above approach

import java.util.*;
 
class GFG{
 
// Function to find the maximum score
// for one possible tree having N nodes
// N - 1 Edges

static int maxScore(int[] arr)
{

    int N = arr.length;
 
    // Number of nodes

    N++;
 
    // Initialize dp[][]

    int [][] dp = new int[N + 1][2 * (N - 1) + 1];

    // Score with 0 vertices is 0

    dp[0][0] = 0;
 
    // Traverse the nodes from 1 to N

    for(int i = 1; i <= N; i++)

    {

        // Find maximum scores for

        // each sum

        for(int s = 1; s <= 2 * (N - 1); s++)

        {

            // Iterate over degree of

            // new node

            for(int j = 1; j <= N - 1 && j <= s; j++)

            {

                // Update the current

                // state

                dp[i][s] = Math.max(dp[i][s],

                                   arr[j - 1] + 

                                    dp[i - 1][s - j]);

            }

        }

    }
 
    // Return maximum score for N node

    // tree having 2(N - 1) sum of degree

    return dp[N][2 * (N - 1)] - 1;
}
 
// Driver Code

public static void main(String[] args)
{

    // Given array of scores

    int [] arr = { 1, 3, 0 };
 
    // Function Call

    System.out.print(maxScore(arr));
}
}
 
// This code is contributed by Amit Katiyar

Python3

# Python3 program for the above approach
 
# Function to find the maximum score
# for one possible tree having N nodes
# N - 1 Edges

def maxScore(arr):

    N = len(arr)
 
    # Number of nodes

    N += 1
 
    # Initialize dp[][]

    dp = [[-100000 for i in range(2 * N)] 

                   for i in range(N + 1)]
 
    # Score with 0 vertices is 0

    dp[0][0] = 0
 
    # Traverse the nodes from 1 to N

    for i in range(1, N + 1):

        # Find maximum scores for

        # each sum

        for s in range(1, 2 * (N - 1) + 1):

            # Iterate over degree of

            # new node

            j = 1

            while j <= N - 1 and j <= s:

                # Update the current

                # state

                dp[i][s] = max(dp[i][s], arr[j - 1] +

                               dp[i - 1][s - j])

                j += 1

    # Return maximum score for N node

    # tree having 2(N - 1) sum of degree

    return dp[N][2 * (N - 1)]
 
# Driver Code

if __name__ == '__main__':

    # Given array of scores

    arr = [ 1, 3, 0 ]
 
    # Function Call

    print(maxScore(arr))
 
# This code is contributed by mohit kumar 29

// C# program for the 
// above approach

using System;

class GFG{
 
// Function to find the 
// maximum score for one 
// possible tree having N 
// nodes N - 1 Edges

static int maxScore(int[] arr)
{

  int N = arr.Length;
 
  // Number of nodes

  N++;
 
  // Initialize [,]dp

  int [,] dp = new int[N + 1,

                       2 * (N - 

                       1) + 1];
 
  // Score with 0 vertices 

  // is 0

  dp[0, 0] = 0;
 
  // Traverse the nodes from 

  // 1 to N

  for(int i = 1; i <= N; i++)

  {

    // Find maximum scores for

    // each sum

    for(int s = 1; 

            s <= 2 * (N - 1); s++)

    {

      // Iterate over degree of

      // new node

      for(int j = 1; 

              j <= N - 1 && j <= s; j++)

      {

        // Update the current

        // state

        dp[i, s] = Math.Max(dp[i, s],

                            arr[j - 1] + 

                            dp[i - 1,

                               s - j]);

      }

    }

  }
 
  // Return maximum score for

  // N node tree having 2(N - 1) 

  // sum of degree

  return dp[N, 2 * (N - 

            1)] - 1;
}
 
// Driver Code

public static void Main(String[] args)
{    

  // Given array of scores

  int [] arr = {1, 3, 0};
 
  // Function Call

  Console.Write(maxScore(arr));
}
}
 
// This code is contributed by Princi Singh

Javascript

<script>

// Javascript program for the above approach
 
// Function to find the maximum score
// for one possible tree having N nodes
// N - 1 Edges

function maxScore(arr)
{

    var N = arr.length;
 
    // Number of nodes

    N++;
 
    // Initialize dp[][]

    var dp = Array.from(Array(N+1), ()=> Array(2*N).fill(-10000000));
 
    // Score with 0 vertices is 0

    dp[0][0] = 0;
 
    // Traverse the nodes from 1 to N

    for (var i = 1; i <= N; i++) {
 
        // Find maximum scores for

        // each sum

        for (var s = 1;

             s <= 2 * (N - 1); s++) {
 
            // Iterate over degree of

            // new node

            for (var j = 1; j <= N - 1

                            && j <= s;

                 j++) {
 
                // Update the current

                // state

                dp[i][s]

                    = Math.max(dp[i][s],

                          arr[j - 1]

                              + dp[i - 1][s - j]);

            }

        }

    }
 
    // Return maximum score for N node

    // tree having 2(N - 1) sum of degree

    return dp[N][2 * (N - 1)];
}
 
// Driver Code
 
// Given array of scores

var arr = [1, 3, 0];
 
// Function Call
document.write( maxScore(arr));
 
</script>

Output

Time Complexity: O(N³)
Auxiliary Space: O(N²)

Efficient Approach : Space optimization

we have eliminated the use of unnecessary rows in the dp array and used only two rows (one for the current iteration and one for the previous iteration) to store the values. We have also replaced the dp[i][s] with dp[curr][s] and dp[i-1][s-j] with dp[prev][s-j] to further reduce memory usage.

Implementation Steps:

Create a DP vector table of 2 rows to only determine current and previous computations.
Initialize DP with -100000 because we are finding a maximum value to update DP.
Now Iterate over subproblems through nested loops and get the computation of current problem from previous and current row of DP.
At last return answer store in dp[N % 2][2 * (N – 1)].

Implementation:

C++

// C++ program for above approach
 
#include <bits/stdc++.h>

using namespace std;
 
// Function to find the maximum score
// for one possible tree having N nodes
// N - 1 Edges

int maxScore(vector<int>& arr) {

    int N = arr.size();

       // Number of nodes

    N++;

      // initialize Dp of Two rows only to keep track 

    // of current and previous row computations

    vector<vector<int>> dp(2, vector<int>(2 * N, -100000));

    // Base Case    

    dp[0][0] = 0;

      // iterate over subproblems to get the current value

    for (int i = 1; i <= N; i++) {

          // current value and previous value

        int curr = i % 2, prev = (i - 1) % 2;

        for (int s = 1; s <= 2 * (N - 1); s++) {

              // update current index of DP with minimum value

            dp[curr][s] = -100000;

            for (int j = 1; j <= N - 1 and j <= s; j++) {

                  // store answer in DP get from current and previous row 

                dp[curr][s] = max(dp[curr][s], arr[j - 1] + dp[prev][s - j]);

            }

        }

    }

      // return answer

    return dp[N % 2][2 * (N - 1)];
}
 
// Driver code
 
int main() {

    vector<int> arr = { 1, 3, 0 };

   // Function call

    cout << maxScore(arr);

    return 0;
}
 
// this code is contributed by bhardwajji

Java

import java.util.*;
 
public class Main {

    // Function to find the maximum score

    // for one possible tree having N nodes

    // N - 1 Edges

    public static int maxScore(List<Integer> arr) {

        int N = arr.size();

        // Number of nodes

        N++;

        // initialize Dp of Two rows only to keep track 

        // of current and previous row computations

        List<List<Integer>> dp = new ArrayList<>();

        for (int i = 0; i < 2; i++) {

            List<Integer> row = new ArrayList<>();

            for (int j = 0; j < 2 * N; j++) {

                row.add(-100000);

            }

            dp.add(row);

        }

        // Base Case    

        dp.get(0).set(0, 0);

        // iterate over subproblems to get the current value

        for (int i = 1; i <= N; i++) {

            // current value and previous value

            int curr = i % 2, prev = (i - 1) % 2;

            for (int s = 1; s <= 2 * (N - 1); s++) {

                // update current index of DP with minimum value

                dp.get(curr).set(s, -100000);

                for (int j = 1; j <= N - 1 && j <= s; j++) {

                    // store answer in DP get from current and previous row 

                    dp.get(curr).set(s, Math.max(dp.get(curr).get(s), arr.get(j - 1) + dp.get(prev).get(s - j)));

                }

            }

        }

        // return answer

        return dp.get(N % 2).get(2 * (N - 1));

    }

    // Driver code

    public static void main(String[] args) {

        List<Integer> arr = new ArrayList<>(Arrays.asList(1, 3, 0));

        // Function call

        System.out.println(maxScore(arr));

    }
}

using System;
 
public class Program
{
 
  // Function to find the maximum score

  // for one possible tree having N nodes

  // N - 1 Edges

  public static int maxScore(int[] arr)

  {

    int N = arr.Length;
 
    // Number of nodes

    N++;
 
    // initialize Dp of Two rows only to keep track

    // of current and previous row computations

    int[][] dp = new int[2][];

    dp[0] = new int[2 * N];

    dp[1] = new int[2 * N];

    for (int i = 0; i < 2 * N; i++) {

      dp[0][i] = -100000;

      dp[1][i] = -100000;

    }
 
    // Base Case

    dp[0][0] = 0;
 
    // iterate over subproblems to get the current value

    for (int i = 1; i <= N; i++) 

    {
 
      // current value and previous value

      int curr = i % 2, prev = (i - 1) % 2;

      for (int s = 1; s <= 2 * (N - 1); s++) {
 
        // update current index of DP with minimum

        // value

        dp[curr][s] = -100000;
 
        for (int j = 1; j <= N - 1 && j <= s; j++) {
 
          // store answer in DP get from current

          // and previous row

          dp[curr][s] = Math.Max(

            dp[curr][s],

            arr[j - 1] + dp[prev][s - j]);

        }

      }

    }
 
    // return answer

    return dp[N % 2][2 * (N - 1)];

  }
 
  // Driver code

  public static void Main()

  {

    int[] arr = { 1, 3, 0 };
 
    // Function call

    Console.WriteLine(maxScore(arr));

  }
}
 
// This code is contributed by Prajwal Kandekar

Python3

def maxScore(arr):

    N = len(arr)

    # Number of nodes

    N += 1

    # initialize Dp of Two rows only to keep track 

    # of current and previous row computations

    dp = [[-100000] * (2 * N) for i in range(2)]

    # Base Case    

    dp[0][0] = 0

    # iterate over subproblems to get the current value

    for i in range(1, N + 1):

        # current value and previous value

        curr = i % 2

        prev = (i - 1) % 2

        for s in range(1, 2 * (N - 1) + 1):

            # update current index of DP with minimum value

            dp[curr][s] = -100000

            for j in range(1, N) :

                if j <= s:

                    # store answer in DP get from current and previous row 

                    dp[curr][s] = max(dp[curr][s], arr[j - 1] + dp[prev][s - j])

    # return answer

    return dp[N % 2][2 * (N - 1)]
 
# Driver code

arr = [1, 3, 0]
# Function call

print(maxScore(arr))

Javascript

// Javascript code addition 
 
function maxScore(arr) {

  let N = arr.length;

  // Number of nodes

  N++;
 
  // initialize Dp of Two rows only to keep track 

  // of current and previous row computations

  let dp = new Array(2).fill(null).map(() => new Array(2 * N).fill(-100000));
 
  // Base Case    

  dp[0][0] = 0;
 
  // iterate over subproblems to get the current value

  for (let i = 1; i <= N; i++) {

    // current value and previous value

    let curr = i % 2;

    let prev = (i - 1) % 2;
 
    for (let s = 1; s <= 2 * (N - 1); s++) {

      // update current index of DP with minimum value

      dp[curr][s] = -100000;
 
      for (let j = 1; j <= N - 1 && j <= s; j++) {

        // store answer in DP get from current and previous row 

        dp[curr][s] = Math.max(dp[curr][s], arr[j - 1] + dp[prev][s - j]);

      }

    }

  }
 
  // return answer

  return dp[N % 2][2 * (N - 1)];
}
 
// Driver code
let arr = [1, 3, 0];
// Function call
console.log(maxScore(arr));
 
// The code is contributed by Nidhi goel.

Output

Time Complexity: O(N*N*N)
Auxiliary Space: O(N) or O(N*2)

Article Tags :

Arrays

Combinatorial

DSA

Dynamic Programming

Tree