Maximum average sum partition of an array
Last Updated :
20 Dec, 2022
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++
#include <bits/stdc++.h>
using namespace std;
#define MAX 1000
double memo[MAX][MAX];
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];
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;
cout << largestSumOfAverages(A, K) << endl;
return 0;
}
|
Java
import java.util.Arrays;
import java.util.Vector;
class GFG
{
static int MAX = 1000;
static double[][] memo = new double[MAX][MAX];
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);
memo[i + 1][1] = sum / (i + 1);
}
return score(n, A, K);
}
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));
}
}
|
Python3
MAX = 1000
memo = [[0.0 for i in range(MAX)]
for i in range(MAX)]
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]
memo[i + 1][1] = int(sum / (i + 1))
return score(n, A, K)
if __name__ == '__main__':
A = [9, 1, 2, 3, 9]
K = 3
print(largestSumOfAverages(A, K))
|
C#
using System;
using System.Collections.Generic;
class GFG
{
static int MAX = 1000;
static double[,] memo = new double[MAX, MAX];
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];
memo[i + 1, 1] = sum / (i + 1);
}
return score(n, A, K);
}
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));
}
}
|
Javascript
<script>
let MAX = 1000;
let memo = new Array(MAX).fill(0).map(() => new Array(MAX).fill(0));
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];
memo[i + 1][1] = sum / (i + 1);
}
return score(n, A, K);
}
let A = [9, 1, 2, 3, 9];
let K = 3;
document.write(largestSumOfAverages(A, K) + "<br>");
</script>
|
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++
#include <bits/stdc++.h>
using namespace std;
double largestSumOfAverages(vector<int>& A, int K)
{
int n = A.size();
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];
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];
}
int main()
{
vector<int> A = { 9, 1, 2, 3, 9 };
int K = 3;
cout << largestSumOfAverages(A, K) << endl;
return 0;
}
|
Java
import java.util.*;
class GFG
{
static double largestSumOfAverages(int[] A, int K)
{
int n = A.length;
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];
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];
}
public static void main(String[] args)
{
int []A = { 9, 1, 2, 3, 9 };
int K = 3;
System.out.println(largestSumOfAverages(A, K));
}
}
|
Python3
def largestSumOfAverages(A, K):
n = len(A);
pre_sum = [0] * (n + 1);
pre_sum[0] = 0;
for i in range(n):
pre_sum[i + 1] = pre_sum[i] + A[i];
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]);
A = [ 9, 1, 2, 3, 9 ];
K = 3;
print(largestSumOfAverages(A, K));
|
C#
using System;
using System.Collections.Generic;
class GFG
{
static double largestSumOfAverages(int[] A,
int K)
{
int n = A.Length;
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];
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];
}
public static void Main(String[] args)
{
int []A = { 9, 1, 2, 3, 9 };
int K = 3;
Console.WriteLine(largestSumOfAverages(A, K));
}
}
|
Javascript
<script>
function largestSumOfAverages(A , K) {
var n = A.length;
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];
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];
}
var A = [ 9, 1, 2, 3, 9 ];
var K = 3;
document.write(largestSumOfAverages(A, K));
</script>
|
Time Complexity: O(n2*K)
Auxiliary Space: O(n)
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