Given two arrays arr[] and brr[] consisting of N and K elements respectively, the task is to find the maximum subarray sum possible from the array arr[] by swapping any element from the array arr[] with any element of the array brr[] any number of times.
Examples:
Input: N = 5, K = 4, arr[] = { 7, 2, -1, 4, 5 }, brr[] = { 1, 2, 3, 2 }
Output : 21
Explanation : Swapping arr[2] with brr[2] modifies arr[] to {7, 2, 3, 4, 5}
Maximum subarray sum of the array arr[] = 21Input : N = 2, K = 2, arr[] = { -4, -4 }, brr[] = { 8, 8 }
Output : 16
Explanation: Swap arr[0] with brr[0] and arr[1] with brr[1] modifies arr[] to {8, 8}
Maximum sum subarray of the array arr[] = 16
Approach: The idea to solve this problem is that by swapping elements of array arr and brr, the elements within arr can also be swapped in three swaps. Below are some observations:
- If two elements in the array arr[] having indices i and j are needed to be swapped, then take any temporary element from array brr[], say at index k, and perform the following operations:
- Swap arr[i] and brr[k].
- Swap brr[k] and arr[j].
- Swap arr[i] and brr[k].
- Now elements between array arr[] and brr[] can be swapped within the array arr[] as well. Therefore, greedily arrange elements in array arr[] such that it contains all the positive integers in a continuous manner.
Follow the steps below to solve the problem:
- Store all elements of array arr[] and brr[] in another array crr[].
- Sort the array crr[] in descending order.
- Calculate the sum till the last index (less than N) in the array crr[] which contains a positive element.
- Print the sum obtained.
Below is the implementation of the above approach.
// C++ program for the above approach #include <bits/stdc++.h> using namespace std;
// Function to find the maximum subarray sum // possible by swapping elements from array // arr[] with that from array brr[] void maxSum( int * arr, int * brr, int N, int K)
{ // Stores elements from the
// arrays arr[] and brr[]
vector< int > crr;
// Store elements of array arr[]
// and brr[] in the vector crr
for ( int i = 0; i < N; i++) {
crr.push_back(arr[i]);
}
for ( int i = 0; i < K; i++) {
crr.push_back(brr[i]);
}
// Sort the vector crr
// in descending order
sort(crr.begin(), crr.end(),
greater< int >());
// Stores maximum sum
int sum = 0;
// Calculate the sum till the last
// index in crr[] which is less than
// N which contains a positive element
for ( int i = 0; i < N; i++) {
if (crr[i] > 0) {
sum += crr[i];
}
else {
break ;
}
}
// Print the sum
cout << sum << endl;
} // Driver code int main()
{ // Given arrays and respective lengths
int arr[] = { 7, 2, -1, 4, 5 };
int N = sizeof (arr) / sizeof (arr[0]);
int brr[] = { 1, 2, 3, 2 };
int K = sizeof (brr) / sizeof (brr[0]);
// Calculate maximum subarray sum
maxSum(arr, brr, N, K);
} |
// Java program for the above approach import java.util.*;
class GFG
{ // Function to find the maximum subarray sum
// possible by swapping elements from array
// arr[] with that from array brr[]
static void maxSum( int arr[], int brr[], int N, int K)
{
// Stores elements from the
// arrays arr[] and brr[]
Vector<Integer> crr = new Vector<Integer>();
// Store elements of array arr[]
// and brr[] in the vector crr
for ( int i = 0 ; i < N; i++)
{
crr.add(arr[i]);
}
for ( int i = 0 ; i < K; i++)
{
crr.add(brr[i]);
}
// Sort the vector crr
// in descending order
Collections.sort(crr);
Collections.reverse(crr);
// Stores maximum sum
int sum = 0 ;
// Calculate the sum till the last
// index in crr[] which is less than
// N which contains a positive element
for ( int i = 0 ; i < N; i++)
{
if (crr.get(i) > 0 )
{
sum += crr.get(i);
}
else
{
break ;
}
}
// Print the sum
System.out.println(sum);
}
// Driver code
public static void main(String[] args)
{
// Given arrays and respective lengths
int arr[] = { 7 , 2 , - 1 , 4 , 5 };
int N = arr.length;
int brr[] = { 1 , 2 , 3 , 2 };
int K = brr.length;
// Calculate maximum subarray sum
maxSum(arr, brr, N, K);
}
} // This code is contributed by divyesh072019 |
# Python3 program for the above approach # Function to find the maximum subarray sum # possible by swapping elements from array # arr[] with that from array brr[] def maxSum(arr, brr, N, K):
# Stores elements from the
# arrays arr[] and brr[]
crr = []
# Store elements of array arr[]
# and brr[] in the vector crr
for i in range (N):
crr.append(arr[i])
for i in range (K):
crr.append(brr[i])
# Sort the vector crr
# in descending order
crr = sorted (crr)[:: - 1 ]
# Stores maximum sum
sum = 0
# Calculate the sum till the last
# index in crr[] which is less than
# N which contains a positive element
for i in range (N):
if (crr[i] > 0 ):
sum + = crr[i]
else :
break
# Print the sum
print ( sum )
# Driver code if __name__ = = '__main__' :
# Given arrays and respective lengths
arr = [ 7 , 2 , - 1 , 4 , 5 ]
N = len (arr)
brr = [ 1 , 2 , 3 , 2 ]
K = len (brr)
# Calculate maximum subarray sum
maxSum(arr, brr, N, K)
# This code is contributed by mohit kumar 29 |
// C# program for the above approach using System;
using System.Collections.Generic;
class GFG{
// Function to find the maximum subarray sum // possible by swapping elements from array // arr[] with that from array brr[] static void maxSum( int [] arr, int [] brr,
int N, int K)
{ // Stores elements from the
// arrays arr[] and brr[]
List< int > crr = new List< int >();
// Store elements of array arr[]
// and brr[] in the vector crr
for ( int i = 0; i < N; i++)
{
crr.Add(arr[i]);
}
for ( int i = 0; i < K; i++)
{
crr.Add(brr[i]);
}
// Sort the vector crr
// in descending order
crr.Sort();
crr.Reverse();
// Stores maximum sum
int sum = 0;
// Calculate the sum till the last
// index in crr[] which is less than
// N which contains a positive element
for ( int i = 0; i < N; i++)
{
if (crr[i] > 0)
{
sum += crr[i];
}
else
{
break ;
}
}
// Print the sum
Console.WriteLine(sum);
} // Driver Code static void Main()
{ // Given arrays and respective lengths
int [] arr = { 7, 2, -1, 4, 5 };
int N = arr.Length;
int [] brr = { 1, 2, 3, 2 };
int K = brr.Length;
// Calculate maximum subarray sum
maxSum(arr, brr, N, K);
} } // This code is contributed by divyeshrabadiya07 |
<script> // Javascript program for the above approach
// Function to find the maximum subarray sum
// possible by swapping elements from array
// arr[] with that from array brr[]
function maxSum(arr, brr, N, K)
{
// Stores elements from the
// arrays arr[] and brr[]
let crr = [];
// Store elements of array arr[]
// and brr[] in the vector crr
for (let i = 0; i < N; i++)
{
crr.push(arr[i]);
}
for (let i = 0; i < K; i++)
{
crr.push(brr[i]);
}
// Sort the vector crr
// in descending order
crr.sort( function (a, b){ return a - b});
crr.reverse();
// Stores maximum sum
let sum = 0;
// Calculate the sum till the last
// index in crr[] which is less than
// N which contains a positive element
for (let i = 0; i < N; i++)
{
if (crr[i] > 0)
{
sum += crr[i];
}
else
{
break ;
}
}
// Print the sum
document.write(sum);
}
// Given arrays and respective lengths
let arr = [ 7, 2, -1, 4, 5 ];
let N = arr.length;
let brr = [ 1, 2, 3, 2 ];
let K = brr.length;
// Calculate maximum subarray sum
maxSum(arr, brr, N, K);
</script> |
21
Time Complexity: O((N+K)*log(N+K))
Auxiliary Space: O(N+K)