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Maximum average sum partition of an array

Last Updated : 20 Dec, 2022
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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 . 

Implementation:

C++ 
// 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 partitioning size
    cout << largestSumOfAverages(A, K) << endl;
    return 0;
}
Java 
// 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
Python3 
# 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 partitioning size
    print(largestSumOfAverages(A, K))
    
# This code is contributed by
# Surendra_Gangwar
C# 
// 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
JavaScript 
<script>

// JavaScript program for maximum average sum partition


let MAX = 1000;

let memo = new Array(MAX).fill(0).map(() => new Array(MAX).fill(0));

// bottom up approach to calculate score
function score(n, A, k) {
    if (memo[n][k] > 0)
        return memo[n][k];
    let sum = 0;
    for (let 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];
}

function largestSumOfAverages(A, K) {
    let n = A.length;
    let sum = 0;

    for (let 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);
}


let A = [9, 1, 2, 3, 9];
let K = 3; // atmost partitioning size
document.write(largestSumOfAverages(A, K) + "<br>");

</script>

Output
20

Above problem can now be easily understood as dynamic programming.

Let dp(i, k) be the best score partitioning A[i:j] into at most K parts. If the first group we partition A[i:j] into ends before j, then our candidate partition has score average(i, j) + dp(j, k-1)). Recursion in the general case is dp(i, k) = max(average(i, N), (average(i, j) + dp(j, k-1))). We can precompute the prefix sums for fast execution of out code. 

Implementation:

C++ 
// CPP program for maximum average sum partition
#include <bits/stdc++.h>
using namespace std;

double largestSumOfAverages(vector<int>& A, int K)
{
    int n = A.size();

    // storing prefix sums
    double pre_sum[n+1]; 
    pre_sum[0] = 0;
    for (int i = 0; i < n; i++)
        pre_sum[i + 1] = pre_sum[i] + A[i];

    // for each i to n storing averages 
    double dp[n] = {0};
    double sum = 0;
    for (int i = 0; i < n; i++) 
        dp[i] = (pre_sum[n] - pre_sum[i]) / (n - i);
    
    for (int k = 0; k < K - 1; k++) 
        for (int i = 0; i < n; i++) 
            for (int j = i + 1; j < n; j++) 
                dp[i] = max(dp[i], (pre_sum[j] -
                         pre_sum[i]) / (j - i) + dp[j]);
    
    return dp[0];
}

// Driver code
int main()
{
    vector<int> A = { 9, 1, 2, 3, 9 };
    int K = 3; // atmost partitioning size
    cout << largestSumOfAverages(A, K) << endl;
    return 0;
}
Java 
// Java program for maximum average sum partition
import java.util.*;

class GFG 
{

static double largestSumOfAverages(int[] A, int K)
{
    int n = A.length;

    // storing prefix sums
    double []pre_sum = new double[n + 1]; 
    pre_sum[0] = 0;
    for (int i = 0; i < n; i++)
        pre_sum[i + 1] = pre_sum[i] + A[i];

    // for each i to n storing averages 
    double []dp = new double[n];
    double sum = 0;
    for (int i = 0; i < n; i++) 
        dp[i] = (pre_sum[n] - pre_sum[i]) / (n - i);
    
    for (int k = 0; k < K - 1; k++) 
        for (int i = 0; i < n; i++) 
            for (int j = i + 1; j < n; j++) 
                dp[i] = Math.max(dp[i], (pre_sum[j] -
                        pre_sum[i]) / (j - i) + dp[j]);
    
    return dp[0];
}

// Driver code
public static void main(String[] args) 
{
    int []A = { 9, 1, 2, 3, 9 };
    int K = 3; // atmost partitioning size
    System.out.println(largestSumOfAverages(A, K));
}
}

// This code is contributed by PrinciRaj1992
Python3 
# Python3 program for maximum average 
# sum partition
def largestSumOfAverages(A, K):

    n = len(A);

    # storing prefix sums
    pre_sum = [0] * (n + 1); 
    pre_sum[0] = 0;
    for i in range(n):
        pre_sum[i + 1] = pre_sum[i] + A[i];

    # for each i to n storing averages 
    dp = [0] * n;
    sum = 0;
    for i in range(n):
        dp[i] = (pre_sum[n] - 
                 pre_sum[i]) / (n - i);
    
    for k in range(K - 1):
        for i in range(n):
            for j in range(i + 1, n): 
                dp[i] = max(dp[i], (pre_sum[j] -
                                    pre_sum[i]) / 
                                    (j - i) + dp[j]);
    
    return int(dp[0]);

# Driver code
A = [ 9, 1, 2, 3, 9 ];
K = 3; # atmost partitioning size
print(largestSumOfAverages(A, K));

# This code is contributed by Rajput-Ji
C# 
// C# program for maximum average sum partition
using System;
using System.Collections.Generic;
    
class GFG 
{
static double largestSumOfAverages(int[] A,
                                   int K)
{
    int n = A.Length;

    // storing prefix sums
    double []pre_sum = new double[n + 1]; 
    pre_sum[0] = 0;
    for (int i = 0; i < n; i++)
        pre_sum[i + 1] = pre_sum[i] + A[i];

    // for each i to n storing averages 
    double []dp = new double[n];
    for (int i = 0; i < n; i++) 
        dp[i] = (pre_sum[n] - pre_sum[i]) / (n - i);
    
    for (int k = 0; k < K - 1; k++) 
        for (int i = 0; i < n; i++) 
            for (int j = i + 1; j < n; j++) 
                dp[i] = Math.Max(dp[i], (pre_sum[j] -
                        pre_sum[i]) / (j - i) + dp[j]);
    
    return dp[0];
}

// Driver code
public static void Main(String[] args) 
{
    int []A = { 9, 1, 2, 3, 9 };
    int K = 3; // atmost partitioning size
    Console.WriteLine(largestSumOfAverages(A, K));
}
}

// This code is contributed by PrinciRaj1992
JavaScript 
<script>
// javascript program for maximum average sum partition

    function largestSumOfAverages(A , K) {
        var n = A.length;

        // storing prefix sums
        var pre_sum = Array(n + 1).fill(-1);
        pre_sum[0] = 0;
        for (var i = 0; i < n; i++)
            pre_sum[i + 1] = pre_sum[i] + A[i];

        // for each i to n storing averages
        var dp = Array(n).fill(-1);
        var sum = 0;
        for (var i = 0; i < n; i++)
            dp[i] = (pre_sum[n] - pre_sum[i]) / (n - i);

        for (k = 0; k < K - 1; k++)
            for (i = 0; i < n; i++)
                for (j = i + 1; j < n; j++)
                    dp[i] = Math.max(dp[i], (pre_sum[j] - pre_sum[i]) / (j - i) + dp[j]);

        return dp[0];
    }

    // Driver code
        var A = [ 9, 1, 2, 3, 9 ];
        var K = 3; // atmost partitioning size
        document.write(largestSumOfAverages(A, K));

// This code is contributed by umadevi9616 
</script>

Output
20

Time Complexity: O(n2*K)
Auxiliary Space: O(n)
 


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