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.
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++ Program to implement // the above approach #include <bits/stdc++.h> #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<ll>& a, vector<ll>& b) { int n = a.size();
// Stores (i - 1) * (n - i) * a[i]
// for every i-th element
vector<ll> 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<ll> a = { 1, 2, 3 };
vector<ll> b = { 2, 3, 2 };
// Function call
cout << findMinValue(a, b) << endl;
} |
// 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 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 |
// C# program to implement // the above approach using System;
using System.Linq;
using System.Collections.Generic;
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] *= (a[i]);
;
}
// Sort the array in reverse order
List< int > input = new List< int >(b);
input = input.OrderBy(inp => -inp).ToList();
b = input.ToArray();
// Sort the products
Array.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
Console.WriteLine(findMinValue(a, b) + "\n" );
}
} // This code is contributed by phasing17 |
<script> // Javascript Program to implement // the above approach var mod = 1000000007;
// Function to rearrange the second array such // that the sum of its product of same indexed // elements from both the arrays is minimized function findMinValue(a, b)
{ var n = a.length;
// Stores (i - 1) * (n - i) * a[i]
// for every i-th element
var pro = Array(n);
for ( var i = 0; i < n; ++i)
{
// Updating the value of pro
// according to the function
pro[i] = ((i + 1) * (n - i));
pro[i] *= (1 * a[i]);
;
}
// Sort the array in reverse order
b.sort((a, b) => b - a)
// Sort the products
pro.sort((a, b) => a - b)
var ans = 0;
for ( var i = 0; i < n; ++i)
{
// Updating the ans
ans += (pro[i] % mod * b[i]) % mod;
ans %= mod;
}
// Return the ans
return ans;
} // Driver code var a = [ 1, 2, 3 ];
var b = [ 2, 3, 2 ];
// Function call document.write( findMinValue(a, b)); // This code is contributed by itsok </script> |
43
Time Complexity: O(N log N)
Auxiliary Space: O(N)