Maximize Array sum by swapping at most K elements with another array
Last Updated :
08 Aug, 2022
Given two arrays A and B of size N and an integer K, the task is to find the maximum possible sum of array A by swapping at most K elements with array B.
Examples:
Input: A[] = {2, 3, 4}, B[] = {6, 8, 5}, K = 1
Output: 15
Explanation: Swap A[0] and B[1]. Hence sum = 8 + 3 + 4 = 15.
Input: A[] = {9, 7}, B[] = {5, 1}, K = 2
Output: 16
Explanation: Since all the elements of array A are greater than the elements of array B, no swaps are required.
Maximize Array sum by swapping at most K elements with another array using Sorting
The idea is to replace all the smallest elements present in A with the largest elements of B, till the value of A[i] < B[i] and K swaps are not complete.
Follow the steps mentioned below to implement the idea:
- Sort the array A and B in non-decreasing order.
- Traverse array A from the beginning and array B from the end, so that we can swap the minimum element of array A with the maximum element of array B.
- If the element of array A is smaller than that of array B, swap them. Otherwise, break the loop.
- Do this for at most K elements, and break the loop after that.
- Find the sum of the resultant array A.
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
void maximumSum( int a[], int b[],
int k, int n)
{
int i, j;
sort(a, a + n);
sort(b, b + n);
for (i = 0, j = n - 1; i < k;
i++, j--) {
if (a[i] < b[j])
swap(a[i], b[j]);
else
break ;
}
int sum = 0;
for (i = 0; i < n; i++)
sum += a[i];
cout << sum << endl;
}
int main()
{
int K = 1;
int A[] = { 2, 3, 4 };
int B[] = { 6, 8, 5 };
int N = sizeof (A) / sizeof (A[0]);
maximumSum(A, B, K, N);
return 0;
}
|
Java
import java.util.*;
class GFG{
static void maximumSum( int a[], int b[],
int k, int n)
{
int i, j;
Arrays.sort(a);
Arrays.sort(b);
for (i = 0 , j = n - 1 ; i < k; i++, j--)
{
if (a[i] < b[j])
{
int temp = a[i];
a[i] = b[j];
b[j] = temp;
}
else
break ;
}
int sum = 0 ;
for (i = 0 ; i < n; i++)
sum += a[i];
System.out.print(sum + "\n" );
}
public static void main(String[] args)
{
int K = 1 ;
int A[] = { 2 , 3 , 4 };
int B[] = { 6 , 8 , 5 };
int N = A.length;
maximumSum(A, B, K, N);
}
}
|
Python3
def maximumSum(a, b, k, n):
a.sort()
b.sort()
i = 0
j = n - 1
while i < k:
if (a[i] < b[j]):
a[i], b[j] = b[j], a[i]
else :
break
i + = 1
j - = 1
sum = 0
for i in range (n):
sum + = a[i]
print ( sum )
if __name__ = = "__main__" :
K = 1
A = [ 2 , 3 , 4 ]
B = [ 6 , 8 , 5 ]
N = len (A)
maximumSum(A, B, K, N)
|
C#
using System;
class GFG{
static void maximumSum( int []a,
int []b,
int k, int n)
{
int i, j;
Array.Sort(a);
Array.Sort(b);
for (i = 0, j = n - 1; i < k; i++, j--)
{
if (a[i] < b[j])
{
int temp = a[i];
a[i] = b[j];
b[j] = temp;
}
else
break ;
}
int sum = 0;
for (i = 0; i < n; i++)
sum += a[i];
Console.Write(sum + "\n" );
}
public static void Main()
{
int K = 1;
int []A = { 2, 3, 4 };
int []B = { 6, 8, 5 };
int N = A.Length;
maximumSum(A, B, K, N);
}
}
|
Javascript
<script>
function maximumSum(a, b, k, n)
{
let i, j;
a.sort();
b.sort();
for (i = 0, j = n - 1; i < k;
i++, j--) {
if (a[i] < b[j])
{
let temp = a[i];
a[i] = b[j];
b[j] = temp;
}
else
break ;
}
let sum = 0;
for (i = 0; i < n; i++)
sum += a[i];
document.write(sum);
}
let K = 1;
let A = [2, 3, 4 ];
let B = [ 6, 8, 5 ];
let N = A.length;
maximumSum(A, B, K, N);
</script>
|
Time Complexity: O(N*log(N))
Auxiliary Space: O(1)
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