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 Edgesint 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 Codeint main(){ // Given array of scores vector<int> arr = { 1, 3, 0 }; // Function Call cout << maxScore(arr); return 0;} |
Java
// Java program for the above approachimport java.util.*;class GFG{// Function to find the maximum score// for one possible tree having N nodes// N - 1 Edgesstatic 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 Codepublic 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 Edgesdef 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 Codeif __name__ == '__main__': # Given array of scores arr = [ 1, 3, 0 ] # Function Call print(maxScore(arr))# This code is contributed by mohit kumar 29 |
C#
// C# program for the // above approachusing System;class GFG{// Function to find the // maximum score for one // possible tree having N // nodes N - 1 Edgesstatic 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 Codepublic 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 Edgesfunction 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 scoresvar arr = [1, 3, 0];// Function Calldocument.write( maxScore(arr));</script> |
Output
8
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 Edgesint 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 codeint 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)); }} |
C#
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 codearr = [1, 3, 0]# Function callprint(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 codelet arr = [1, 3, 0];// Function callconsole.log(maxScore(arr));// The code is contributed by Nidhi goel. |
Output
8
Time Complexity: O(N*N*N)
Auxiliary Space: O(N) or O(N*2)
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