Maximum sum after rearranging the array for K queries

Given two arrays arr[] containing N integers and Q[][] containing K queries where every query represents a range [L, R]. The task is to rearrange the array and find the maximum possible sum of all the subarrays where each subarray is defined by the elements of the array in the range [L, R] given by each query. 
Note: 1-based indexing is used in the Q[][] array to signify the ranges.
Examples:

Input: arr[] = { 2, 6, 10, 1, 5, 6 }, Q[][2] = {{1, 3}, {4, 6}, {3, 4}} 
Output: 46 
Explanation:
One possible way is to rearrange the array to arr[] = {2, 6, 10, 6, 5, 1}. 
In this arrangement: 
The sum of the subarray in the range [1, 3] = 2 + 6 + 10 = 18. 
The sum of the subarray in the range [4, 6] = 6 + 5 + 1 = 12. 
The sum of the subarray in the range [3, 4] = 10 + 6 = 16. 
The total sum of all the subarrays = 46 which is the maximum possible.
Input: arr[] = { 1, 2, 3, 4, 5, 6, 7, 8 }, Q[][2] = {{1, 4}, {5, 5}, {7, 8}, {8, 8}} 
Output: 43 
Explanation:
One possible way is to rearrange the array to arr[] = {2, 3, 4, 5, 6, 1, 7, 8}. 
In this arrangement: 
The sum of the subarray in the range [1, 4] = 2 + 3 + 4 + 5 = 14. 
The sum of the subarray in the range [5, 5] = 6 = 6. 
The sum of the subarray in the range [7, 8] = 7 + 8 = 15. 
The sum of the subarray in the range [8, 8] = 8 = 8. 
The total sum of all the subarrays = 43 which is the maximum possible. 

Approach: On observing clearly, one conclusion which can be made is that we get the maximum sum when the maximum elements are included in as many subarrays as possible. For this, we need to find the number of times every index is included by iterating all the queries. 
For example: Let the array be arr[] = {2, 6, 10, 6, 5, 1} and the queries be Q[][] = {{1, 3}, {4, 6}, {3, 4}}. 

1. Step 1: Create a count array C[] of size N. So, initially, the count array C[] = {0, 0, 0, 0, 0, 0}.
2. Step 2: For the query [1, 3], the elements at the index [1, 3] are incremented by 1. The count array after this query becomes {1, 1, 1, 0, 0, 0}.
3. Step 3: Similarly, for the next query, the count array becomes {1, 1, 1, 1, 1, 1} and finally, after the third query, the count array becomes {1, 1, 2, 2, 1, 1}.
4. Step 4: After obtaining the count array, the idea is to use sorting to get the maximum sum.
5. Step 5: After sorting, the array C[] = {1, 1, 1, 1, 2, 2} and arr[] = {1, 2, 5, 6, 6, 10}. The maximum possible sum is the weighted sum of both the arrays, i.e.: 
   

sum = ((1 * 1) + (1 * 2) + (1 * 5) + (1 * 6) + (2 * 6) + (2 * 10)) = 46 

Below is the implementation of the above approach: 

C++

// C++ program to find the maximum sum
// after rearranging the array for K queries

#include <bits/stdc++.h>
using namespace std;

// Function to find maximum sum after
// rearranging array elements
int maxSumArrangement(int A[], int R[][2],
                      int N, int M)
{

    // Auxiliary array to find the
    // count of each selected elements
    int count[N];

    // Initialize with 0
    memset(count, 0, sizeof count);

    // Finding count of every element
    // to be selected
    for (int i = 0; i < M; ++i) {

        int l = R[i][0], r = R[i][1] + 1;

        // Making it to 0-indexing
        l--;
        r--;

        // Prefix sum array concept is used
        // to obtain the count array
        count[l]++;

        if (r < N)
            count[r]--;
    }

    // Iterating over the count array
    // to get the final array
    for (int i = 1; i < N; ++i) {
        count[i] += count[i - 1];
    }

    // Variable to store the maximum sum
    int ans = 0;

    // Sorting both the arrays
    sort(count, count + N);
    sort(A, A + N);


    // Loop to find the maximum sum
    for (int i = N - 1; i >= 0; --i) {
        ans += A[i] * count[i];
    }

    return ans;
}

// Driver code
int main()
{

    int A[] = { 2, 6, 10, 1, 5, 6 };
    int R[][2]
        = { { 1, 3 }, { 4, 6 }, { 3, 4 } };

    int N = sizeof(A) / sizeof(A[0]);
    int M = sizeof(R) / sizeof(R[0]);

    cout << maxSumArrangement(A, R, N, M);

    return 0;
}

Java

// Java program to find the maximum sum
// after rearranging the array for K queries
import java.util.*;

class GFG
{

    // Function to find maximum sum after
    // rearranging array elements
    static int maxSumArrangement(int A[], int R[][],
                          int N, int M)
    {

        // Auxiliary array to find the
        // count of each selected elements
        int count[] = new int[N];
        int i;

        // Finding count of every element
        // to be selected
        for ( i = 0; i < M; ++i) {

            int l = R[i][0], r = R[i][1] + 1;

            // Making it to 0-indexing
            l--;
            r--;

            // Prefix sum array concept is used
            // to obtain the count array
            count[l]++;

            if (r < N)
                count[r]--;
        }

        // Iterating over the count array
        // to get the final array
        for (i = 1; i < N; ++i) {
            count[i] += count[i - 1];
        }

        // Variable to store the maximum sum
        int ans = 0;

        // Sorting both the arrays
        Arrays.sort( count);
        Arrays.sort(A);

        // Loop to find the maximum sum
        for (i = N - 1; i >= 0; --i) {
            ans += A[i] * count[i];
        }

        return ans;
    }

    // Driver code
    public static void main(String []args)
    {
        int A[] = { 2, 6, 10, 1, 5, 6 };
        int R[][]
            = { { 1, 3 }, { 4, 6 }, { 3, 4 } };

        int N = A.length;
        int M = R.length;

        System.out.print(maxSumArrangement(A, R, N, M));

    }
}

// This code is contributed by chitranayal

Python3

# Python3 program to find the maximum sum
# after rearranging the array for K queries

# Function to find maximum sum after
# rearranging array elements
def maxSumArrangement( A,  R,  N,  M):

    # Auxiliary array to find the
    # count of each selected elements
    # Initialize with 0
    count = [0 for i in range(N)]

    # Finding count of every element
    # to be selected
    for i in range(M):

        l = R[i][0]
        r = R[i][1] + 1

        # Making it to 0-indexing
        l = l - 1
        r = r - 1

        # Prefix sum array concept is used
        # to obtain the count array
        count[l] = count[l] + 1

        if (r < N):
            count[r] = count[r] - 1

    # Iterating over the count array
    # to get the final array
    for i in range(1, N):
        count[i] = count[i] + count[i - 1]

    # Variable to store the maximum sum
    ans = 0

    # Sorting both the arrays
    count.sort()
    A.sort()

    # Loop to find the maximum sum
    for i in range(N - 1, -1, -1):
        ans = ans + A[i] * count[i]

    return ans

# Driver code
A = [ 2, 6, 10, 1, 5, 6 ]
R = [ [ 1, 3 ], [ 4, 6 ], [ 3, 4 ] ]

N = len(A)
M = len(R)

print(maxSumArrangement(A, R, N, M))

# This code is contributed by Sanjit_Prasad

C#

// C# program to find the maximum sum
// after rearranging the array for K queries
using System;

class GFG
{

    // Function to find maximum sum after
    // rearranging array elements
    static int maxSumArrangement(int []A, int [,]R,
                          int N, int M)
    {

        // Auxiliary array to find the
        // count of each selected elements
        int []count = new int[N];
        int i;

        // Finding count of every element
        // to be selected
        for ( i = 0; i < M; ++i) {

            int l = R[i, 0], r = R[i, 1] + 1;

            // Making it to 0-indexing
            l--;
            r--;

            // Prefix sum array concept is used
            // to obtain the count array
            count[l]++;

            if (r < N)
                count[r]--;
        }

        // Iterating over the count array
        // to get the readonly array
        for (i = 1; i < N; ++i) {
            count[i] += count[i - 1];
        }

        // Variable to store the maximum sum
        int ans = 0;

        // Sorting both the arrays
        Array.Sort( count);
        Array.Sort(A);

        // Loop to find the maximum sum
        for (i = N - 1; i >= 0; --i) {
            ans += A[i] * count[i];
        }

        return ans;
    }

    // Driver code
    public static void Main(String []args)
    {
        int []A = { 2, 6, 10, 1, 5, 6 };
        int [,]R
            = { { 1, 3 }, { 4, 6 }, { 3, 4 } };

        int N = A.Length;
        int M = R.GetLength(0);

        Console.Write(maxSumArrangement(A, R, N, M));
    }
}

// This code is contributed by Princi Singh

Javascript

<script>
//Javascript program to find the maximum sum
// after rearranging the array for K queries



//function to sort a array
function arrSort(a, n) {
   var i, j, min, temp;
   for (i = 0; i < n - 1; i++) {
      min = i;
      for (j = i + 1; j < n; j++)
      if (a[j] < a[min])
      min = j;
      temp = a[i];
      a[i] = a[min];
      a[min] = temp;
   }
}

// Function to find maximum sum after
// rearranging array elements
function maxSumArrangement(A, R, N, M)
{

    // Auxiliary array to find the
    // count of each selected elements
    var count = new Array(N);

    // Initialize with 0
    count.fill(0);

    // Finding count of every element
    // to be selected
    for (var i = 0; i < M; i++) {

        var l = R[i][0], r = R[i][1] + 1;

        // Making it to 0-indexing
        l--;
        r--;

        // Prefix sum array concept is used
        // to obtain the count array
        count[l]++;

        if (r < N)
            count[r]--;
    }

    // Iterating over the count array
    // to get the final array
    for (var i = 1; i < N; ++i) {
        count[i] += count[i - 1];
    }

    // Variable to store the maximum sum
    var ans = 0;

    // Sorting both the arrays
    count.sort();
    arrSort(A,N);

    // Loop to find the maximum sum
    for (var i = N - 1; i >= 0; --i) {
        ans += A[i] * count[i];
    }

    return ans;
}

var A = [ 2, 6, 10, 1, 5, 6 ];
var R = [ [ 1, 3 ], [ 4, 6 ], [ 3, 4 ] ];
var N = A.length;
var M = R.length;
document.write( maxSumArrangement(A, R, N, M));

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

Output:

46

Time Complexity: O(N* log(N))
Auxiliary Space: O(N), where N is the size of the given array.