# Minimize the Sum of all the subarrays made up of the products of same-indexed elements

• Last Updated : 22 Jun, 2021

Given two arrays arr[] and arr2[] of length N, the task is to find the minimum sum of all the subarrays made up of the products of the same indexed elements of both the arrays after rearranging the second array.

Note: Since the answer can be very large, print the answer modulo 109 + 7.

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Examples:

Input: arr[] = {1, 2}, arr2[] = {2, 3}
Output: 14
Explanation:
Rearrange the arr2[] to {3, 2}
Therefore, the product of same indexed elements of two arrays becomes {3, 4}.
Possible subarrays are {3}, {4}, {3, 4}
Sum of the subarrays = 3 + 4 + 7 = 14.
Input: arr[] = {1, 2, 3}, arr2[] = {2, 3, 2}
Output: 43
Explanation:
Rearrange arr2[] to {3, 2, 2}
Therefore, the product of thesame indexed elements of two arrays becomes {3, 4, 6}.
Therefore, sum of all the subarrays = 3 + 4 + 6 + 7 + 10 + 13 = 43

Approach:
It can be observed that, ith element occurs in (i + 1)*(n – i) subarrays. Therefore, the task is to maximize the sum of the (i + 1)*(n – i)* a[i] * b[i]. Follow the steps below to solve the problem:

• Since the elements of arr2[] can only be rearranged, so the value of (i + 1)*(n – i) * a[i] is constant for every ith element.
• Therefore, calculate the value of the (i + 1)*(n – i)* a[i] for all the indices and then sort the products.
• Sort the array arr2[] in descending order.
• For every ith index, calculate the sum of the product of the values of (i + 1)*(n – i)* a[i] in descending order and arr2[i] in ascending order.

Below is the implementation of the above approach:

## C++

 `// C++ Program to implement``// the above approach``#include ``#define ll long long``using` `namespace` `std;` `const` `int` `mod = (``int``)1e9 + 7;` `// Returns the greater of``// the two values``bool` `comp(ll a, ll b)``{``    ``if` `(a > b)``        ``return` `true``;``    ``else``        ``return` `false``;``}` `// Function to rearrange the second array such``// that the sum of its product of same indexed``// elements from both the arrays is minimized``ll findMinValue(vector& a, vector& b)``{` `    ``int` `n = a.size();` `    ``// Stores (i - 1) * (n - i) * a[i]``    ``// for every i-th element``    ``vector pro(n);` `    ``for` `(``int` `i = 0; i < n; ++i) {` `        ``// Updating the value of pro``        ``// according to the function``        ``pro[i] = ((ll)(i + 1) * (ll)(n - i));``        ``pro[i] *= (1LL * a[i]);``        ``;``    ``}` `    ``// Sort the array in reverse order``    ``sort(b.begin(), b.end(), comp);` `    ``// Sort the products``    ``sort(pro.begin(), pro.end());` `    ``ll ans = 0;``    ``for` `(``int` `i = 0; i < n; ++i) {` `        ``// Updating the ans``        ``ans += (pro[i] % mod * b[i]) % mod;``        ``ans %= mod;``    ``}` `    ``// Return the ans``    ``return` `ans;``}` `// Driver code``int` `main()``{` `    ``vector a = { 1, 2, 3 };``    ``vector b = { 2, 3, 2 };``    ``// Function call``    ``cout << findMinValue(a, b) << endl;``}`

## Java

 `// Java program to implement``// the above approach``import` `java.util.*;` `class` `GFG{` `static` `int` `mod = (``int``)1e9 + ``7``;` `// Function to rearrange the second array such``// that the sum of its product of same indexed``// elements from both the arrays is minimized``static` `int` `findMinValue(``int` `[] a, ``int` `[]b)``{``    ``int` `n = a.length;` `    ``// Stores (i - 1) * (n - i) * a[i]``    ``// for every i-th element``    ``int` `[] pro = ``new` `int``[n];` `    ``for``(``int` `i = ``0``; i < n; ++i)``    ``{``        ` `        ``// Updating the value of pro``        ``// according to the function``        ``pro[i] = ((i + ``1``) * (n - i));``        ``pro[i] *= (1L * a[i]);``        ``;``    ``}` `    ``// Sort the array in reverse order``    ``Integer[] input = Arrays.stream(b).boxed(``                      ``).toArray(Integer[]::``new``);``                      ` `    ``Arrays.sort(input, (x, y) -> y - x);``    ``b = Arrays.stream(input).mapToInt(``        ``Integer::intValue).toArray();``        ` `    ``// Sort the products``    ``Arrays.sort(pro);` `    ``int` `ans = ``0``;``    ``for``(``int` `i = ``0``; i < n; ++i)``    ``{``        ` `        ``// Updating the ans``        ``ans += (pro[i] % mod * b[i]) % mod;``        ``ans %= mod;``    ``}` `    ``// Return the ans``    ``return` `ans;``}` `// Driver code``public` `static` `void` `main(String[] args)``{``    ``int` `[]a = { ``1``, ``2``, ``3` `};``    ``int` `[]b = { ``2``, ``3``, ``2` `};``    ` `    ``// Function call``    ``System.out.print(findMinValue(a, b) + ``"\n"``);``}``}` `// This code is contributed by 29AjayKumar`

## Python3

 `# Python3 Program to implement``# the above approach``mod ``=` `1e9` `+``7` `# Function to rearrange``# the second array such``# that the sum of its``# product of same indexed``# elements from both``# the arrays is minimized``def` `findMinValue(a, b):` `    ``n ``=` `len``(a)` `    ``# Stores (i - 1) * (n - i) * a[i]``    ``# for every i-th element``    ``pro ``=` `[``0``] ``*` `(n)` `    ``for` `i ``in` `range` `(n):` `        ``# Updating the value of pro``        ``# according to the function``        ``pro[i] ``=` `((i ``+` `1``) ``*` `(n ``-` `i))``        ``pro[i] ``*``=` `(a[i])` `    ``# Sort the array in reverse order``    ``b.sort(reverse ``=` `True``)` `    ``# Sort the products``    ``pro.sort()` `    ``ans ``=` `0``    ``for` `i ``in` `range` `(n):` `        ``# Updating the ans``        ``ans ``+``=` `(pro[i] ``%` `mod ``*` `b[i]) ``%` `mod``        ``ans ``%``=` `mod``   ` `    ``# Return the ans``    ``return` `ans` `# Driver code``if` `__name__ ``=``=` `"__main__"``:` `    ``a ``=` `[``1``, ``2``, ``3``]``    ``b ``=` `[``2``, ``3``, ``2``]``    ` `    ``# Function call``    ``print` `(``int``(findMinValue(a, b)))` `# This code is contributed by Chitranayal`

## Javascript

 ``
Output:
`43`

Time Complexity: O(N log N)
Auxiliary Space: O(N)

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