Maximum average sum partition of an array

Given an array, we partition a row of numbers A into at most K adjacent (non-empty) groups, then the score is the sum of the average of each group. What is the maximum score that can be scored?

Examples:

Input : A = { 9, 1, 2, 3, 9 }
K = 3
Output : 20
Explanation : We can partition A into [9], [1, 2, 3], [9]. The answer is 9 + (1 + 2 + 3) / 3 + 9 = 20.
We could have also partitioned A into [9, 1], [2], [3, 9]. That partition would lead to a score of 5 + 2 + 6 = 13, which is worse.

Input : A[] = { 1, 2, 3, 4, 5, 6, 7 }
K = 4
Output : 20.5
Explanation : We can partition A into [1, 2, 3, 4], [5], [6], [7]. The answer is 2.5 + 5 + 6 + 7 = 20.5.

A simple solution is to use recursion. An efficient solution is memorization where we keep the largest score upto k i.e. for 1, 2, 3… upto k;

Let memo[i][k] be the best score portioning A[i..n-1] into at most K parts. In the first group, we partition A[i..n-1] into A[i..j-1] and A[j..n-1], then our candidate partition has score average(i, j) + score(j, k-1)), where average(i, j) = (A[i] + A[i+1] + … + A[j-1]) / (j – i). We take the highest score of these.

In total, our recursion in the general case is :
memo[n][k] = max(memo[n][k], score(memo, i, A, k-1) + average(i, j))
for all i from n-1 to 1 .

C++

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// CPP program for maximum average sum partition
#include <bits/stdc++.h>
using namespace std;
  
#define MAX 1000
  
double memo[MAX][MAX];
  
// bottom up approach to calculate score
double score(int n, vector<int>& A, int k)
{
    if (memo[n][k] > 0)
        return memo[n][k];
    double sum = 0;
    for (int i = n - 1; i > 0; i--) {
        sum += A[i];
        memo[n][k] = max(memo[n][k], score(i, A, k - 1) +
                                          sum / (n - i));
    }
    return memo[n][k];
}
  
double largestSumOfAverages(vector<int>& A, int K)
{
    int n = A.size();
    double sum = 0;
    memset(memo, 0.0, sizeof(memo));
    for (int i = 0; i < n; i++) {
        sum += A[i];
  
        // storing averages from starting to each i ; 
        memo[i + 1][1] = sum / (i + 1);
    }
    return score(n, A, K);
}
  
int main()
{
    vector<int> A = { 9, 1, 2, 3, 9 };
    int K = 3; // atmost partioning size
    cout << largestSumOfAverages(A, K) << endl;
    return 0;
}

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Java

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// Java program for maximum average sum partition 
import java.util.Arrays;
import java.util.Vector;
  
class GFG 
{
  
    static int MAX = 1000;
    static double[][] memo = new double[MAX][MAX];
  
    // bottom up approach to calculate score
    public static double score(int n, Vector<Integer> A, int k)
    {
        if (memo[n][k] > 0)
            return memo[n][k];
  
        double sum = 0;
        for (int i = n - 1; i > 0; i--) 
        {
            sum += A.elementAt(i);
  
            memo[n][k] = Math.max(memo[n][k], 
                                  score(i, A, k - 1) +
                                  sum / (n - i));
        }
        return memo[n][k];
    }
  
    public static double largestSumOfAverages(Vector<Integer> A, int K)
    {
        int n = A.size();
        double sum = 0;
  
        for (int i = 0; i < memo.length; i++) 
        {
            for (int j = 0; j < memo[i].length; j++)
                memo[i][j] = 0.0;
        }
  
        for (int i = 0; i < n; i++) 
        {
            sum += A.elementAt(i);
  
            // storing averages from starting to each i ;
            memo[i + 1][1] = sum / (i + 1);
        }
  
        return score(n, A, K);
    }
  
    // Driver code
    public static void main(String[] args) 
    {
        Vector<Integer> A = new Vector<>(Arrays.asList(9, 1, 2, 3, 9));
        int K = 3;
        System.out.println(largestSumOfAverages(A, K));
  
    }
}
  
// This code is contributed by sanjeev2552

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Python3

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# Python3 program for maximum average sum partition
MAX = 1000
  
memo = [[0.0 for i in range(MAX)] 
             for i in range(MAX)]
  
# bottom up approach to calculate score
def score(n, A, k):
    if (memo[n][k] > 0):
        return memo[n][k]
    sum = 0
    i = n - 1
    while(i > 0):
        sum += A[i]
        memo[n][k] = max(memo[n][k], score(i, A, k - 1) + 
                                       int(sum / (n - i)))
  
        i -= 1
      
    return memo[n][k]
  
def largestSumOfAverages(A, K):
    n = len(A)
    sum = 0
    for i in range(n):
        sum += A[i]
  
        # storing averages from starting to each i ; 
        memo[i + 1][1] = int(sum / (i + 1))
      
    return score(n, A, K)
  
# Driver Code
if __name__ == '__main__':
    A = [9, 1, 2, 3, 9]
    K = 3 # atmost partioning size
    print(largestSumOfAverages(A, K))
      
# This code is contributed by
# Surendra_Gangwar

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C#

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// C# program for maximum average sum partition 
using System;
using System.Collections.Generic;
  
class GFG 
{
    static int MAX = 1000;
    static double[,] memo = new double[MAX, MAX];
  
    // bottom up approach to calculate score
    public static double score(int n, 
                               List<int> A, int k)
    {
        if (memo[n, k] > 0)
            return memo[n, k];
  
        double sum = 0;
        for (int i = n - 1; i > 0; i--) 
        {
            sum += A[i];
  
            memo[n, k] = Math.Max(memo[n, k], 
                                  score(i, A, k - 1) +
                                  sum / (n - i));
        }
        return memo[n, k];
    }
  
    public static double largestSumOfAverages(List<int> A, 
                                                   int K)
    {
        int n = A.Count;
        double sum = 0;
  
        for (int i = 0;
                 i < memo.GetLength(0); i++) 
        {
            for (int j = 0;
                     j < memo.GetLength(1); j++)
                memo[i, j] = 0.0;
        }
  
        for (int i = 0; i < n; i++) 
        {
            sum += A[i];
  
            // storing averages from
            // starting to each i;
            memo[i + 1, 1] = sum / (i + 1);
        }
  
        return score(n, A, K);
    }
  
    // Driver code
    public static void Main(String[] args) 
    {
        int [] arr = {9, 1, 2, 3, 9};
        List<int> A = new List<int>(arr);
        int K = 3;
        Console.WriteLine(largestSumOfAverages(A, K));
    }
}
  
// This code is contributed by Rajput-Ji

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