Open In App

# Maximum sum of values in a given range of an Array for Q queries when shuffling is allowed

Given an array arr[] and K subarrays in the form (Li, Ri), the task is to find the maximum possible value of

. It is allowed to shuffle the array before calculating this value to get the maximum sum.

Examples:

Input: arr[] = {1, 1, 2, 2, 4}, queries = {{1, 2}, {2, 4}, {1, 3}, {5, 5}, {3, 5}}
Output: 26
Explanation:
Shuffled Array to get the maximum sum – {2, 4, 2, 1, 1}
Subarray Sum = arr[1:2] + arr[2:4] + arr[1:3] + arr[5:5] + arr[3:5]
=> 6 + 7 + 8 + 1 + 4 = 26

Input: arr[] = {4, 1, 2, 1, 9, 2}, queries = {{1, 2}, {1, 3}, {1, 4}, {3, 4}}
Output: 49
Explanation:
Shuffled Array to get the maximum sum – {2, 4, 9, 2, 1, 1}
Subarray Sum = arr[1:2] + arr[1:3] + arr[1:4] + arr[3:4]
=> 6 + 15 + 17 + 11 = 49

Naive Approach: A simple solution is to compute the maximum sum of all possible permutations of the given array and check which sequence gives us the maximum summation.

Efficient Approach: The idea is to use Prefix Arrays to find out the frequency of indices over all subarrays. We can do this as follows:

//Initialize an array
prefSum[n] = {0, 0, ...0}
for each Li, Ri:
prefSum[Li]++
prefSum[Ri+1]--

// Find Prefix sum
for i=1 to N:
prefSum[i] += prefSum[i-1]

// prefSum contains frequency of
// each index over all subarrays

Finally, greedily choose the index with the highest frequency and put the largest element of the array at that index. This way we will get the largest possible sum.

Below is the implementation of the above approach:

## C++

 // C++ implementation to find the// maximum sum of K subarrays// when shuffling is allowed #include using namespace std; // Function to find the// maximum sum of all subarraysint maximumSubarraySum(    int a[], int n,    vector >& subarrays){    // Initialize maxsum    // and prefixArray    int i, maxsum = 0;    int prefixArray[n] = { 0 };     // Find the frequency    // using prefix Array    for (i = 0; i < subarrays.size(); ++i) {        prefixArray[subarrays[i].first - 1]++;        prefixArray[subarrays[i].second]--;    }     // Perform prefix sum    for (i = 1; i < n; i++) {        prefixArray[i] += prefixArray[i - 1];    }     // Sort both arrays to get a greedy result    sort(prefixArray,        prefixArray + n,        greater<int>());    sort(a, a + n, greater<int>());     // Finally multiply largest frequency with    // largest array element.    for (i = 0; i < n; i++)        maxsum += a[i] * prefixArray[i];     // Return the answer    return maxsum;} // Driver Codeint main(){    int n = 6;     // Initial Array    int a[] = { 4, 1, 2, 1, 9, 2 };     // Subarrays    vector > subarrays;    subarrays.push_back({ 1, 2 });    subarrays.push_back({ 1, 3 });    subarrays.push_back({ 1, 4 });    subarrays.push_back({ 3, 4 });     // Function Call    cout << maximumSubarraySum(a, n,                            subarrays);}

## Java

 // Java implementation to find the// maximum sum of K subarrays// when shuffling is allowedimport java.util.*;import java.lang.*; class GFG{     // Function to find the// maximum sum of all subarraysstatic int maximumSubarraySum(int a[], int n,           ArrayList> subarrays){         // Initialize maxsum    // and prefixArray    int i, maxsum = 0;    int[] prefixArray = new int[n];      // Find the frequency    // using prefix Array    for(i = 0; i < subarrays.size(); ++i)    {        prefixArray[subarrays.get(i).get(0) - 1]++;        prefixArray[subarrays.get(i).get(1)]--;    }      // Perform prefix sum    for(i = 1; i < n; i++)    {        prefixArray[i] += prefixArray[i - 1];    }      // Sort both arrays to get a    // greedy result    Arrays.sort(prefixArray);    Arrays.sort(a);      // Finally multiply largest    // frequency with largest    // array element.    for(i = 0; i < n; i++)        maxsum += a[i] * prefixArray[i];             // Return the answer    return maxsum;}  // Driver codepublic static void main (String[] args){    int n = 6;         // Initial Array    int a[] = { 4, 1, 2, 1, 9, 2 };         // Subarrays    ArrayList> subarrays = new ArrayList<>();    subarrays.add(Arrays.asList(1, 2));    subarrays.add(Arrays.asList(1, 3));    subarrays.add(Arrays.asList(1, 4));    subarrays.add(Arrays.asList(3, 4));         // Function call    System.out.println(maximumSubarraySum(a, n,                                          subarrays));}} // This code is contributed by offbeat

## Python3

 # Python3 implementation to find the# maximum sum of K subarrays# when shuffling is allowed # Function to find the# maximum sum of all subarraysdef maximumSubarraySum(a, n, subarrays):     # Initialize maxsum    # and prefixArray    maxsum = 0    prefixArray = [0] * n     # Find the frequency    # using prefix Array    for i in range(len(subarrays)):        prefixArray[subarrays[i][0] - 1] += 1        prefixArray[subarrays[i][1]] -= 1     # Perform prefix sum    for i in range(1, n):        prefixArray[i] += prefixArray[i - 1]     # Sort both arrays to get a greedy result    prefixArray.sort()    prefixArray.reverse()    a.sort()    a.reverse()     # Finally multiply largest frequency with    # largest array element.    for i in range(n):        maxsum += a[i] * prefixArray[i]     # Return the answer    return maxsum # Driver Coden = 6 # Initial Arraya = [ 4, 1, 2, 1, 9, 2 ] # Subarrayssubarrays = [ [ 1, 2 ], [ 1, 3 ],              [ 1, 4 ], [ 3, 4 ] ] # Function Callprint(maximumSubarraySum(a, n, subarrays)) # This code is contributed by divyeshrabadiya07

## C#

 // C# implementation to find the// maximum sum of K subarrays// when shuffling is allowedusing System;using System.Collections.Generic; class GFG{     // Function to find the// maximum sum of all subarraysstatic int maximumSubarraySum(int[] a, int n,                    List> subarrays){         // Initialize maxsum    // and prefixArray    int i, maxsum = 0;    int[] prefixArray = new int[n];       // Find the frequency    // using prefix Array    for(i = 0; i < subarrays.Count; ++i)    {        prefixArray[subarrays[i][0] - 1]++;        prefixArray[subarrays[i][1]]--;    }       // Perform prefix sum    for(i = 1; i < n; i++)    {        prefixArray[i] += prefixArray[i - 1];    }       // Sort both arrays to get a    // greedy result    Array.Sort(prefixArray);    Array.Sort(a);       // Finally multiply largest    // frequency with largest    // array element.    for(i = 0; i < n; i++)        maxsum += a[i] * prefixArray[i];              // Return the answer    return maxsum;} // Driver Codestatic void Main(){    int n = 6;         // Initial Array    int[] a = { 4, 1, 2, 1, 9, 2 };          // Subarrays    List> subarrays = new List>();    subarrays.Add(new List<int>{ 1, 2 });    subarrays.Add(new List<int>{ 1, 3 });    subarrays.Add(new List<int>{ 1, 4 });    subarrays.Add(new List<int>{ 3, 4 });          // Function call    Console.WriteLine(maximumSubarraySum(a, n,subarrays));}} // This code is contributed by divyesh072019

## Javascript

 

Output:

49

### Brute Force in python:

Approach:

One simple approach is to generate all possible permutations of the given array and then compute the sum of the subarray for each query. Finally, we can return the maximum sum obtained among all permutations.

the maximum_sum function takes in the input array arr and a list of query tuples queries, where each query tuple represents a range of indices to consider for summing the subarray. The function uses the itertools.permutations method to generate all possible permutations of the input array arr, and for each permutation, it computes the sum of subarrays for all query tuples in queries. The maximum of all such sums is returned as the final output.

## Python3

 import itertools def maximum_sum(arr, queries):    max_sum = float('-inf')    for perm in itertools.permutations(arr):        curr_sum = 0        for q in queries:            curr_sum += sum(perm[q[0]-1:q[1]])        max_sum = max(max_sum, curr_sum)    return max_sum arr1 = [1, 1, 2, 2, 4]queries1 = [(1, 2), (2, 4), (1, 3), (5, 5), (3, 5)]print(maximum_sum(arr1, queries1))  # Output: 26 arr2 = [4, 1, 2, 1, 9, 2]queries2 = [(1, 2), (1, 3), (1, 4), (3, 4)]print(maximum_sum(arr2, queries2))  # Output: 49

Output

26
49


Time Complexity: O(n! * q * n), where n is the length of the array and q is the number of queries.
Space Complexity: O(n! * n)