Maximize the minimum Array value by changing elements with adjacent K times
Last Updated :
08 Oct, 2023
Given an array arr[] of N integers and an integer K, where K denotes the maximum number of operations which can be applied to the array, the task is to maximize the minimum value of arr[] by using the given operation at most K times.
- In one operation it is possible to select any element of the given arr[] and can change it with its adjacent element.
Examples:
Input: N = 7, K = 4, arr[] = {9, 7, 3, 5, 7, 8, 7}
Output: 7
Explanation: First operation: Change 3 at index 2 with 7 at index 1.
So the arr[] becomes: {9, 7, 7, 5, 7, 8, 7}
Second Operation: Change 5 at index 3 with 7 at index 2.
So the arr[] becomes: {9, 7, 7, 7, 7, 8, 7}
Third operation: Change 7 at index 6 with 8 at index 5.
So the arr[] becomes: {9, 7, 7, 7, 7, 8, 8}
Fourth Operation: Change 7 at index 1 with 9 at index 0.
So the arr[] becomes: {9, 9, 7, 7, 7, 8, 8}
The minimum value in arr[] after applying operation at most K times is: 7
Input: N = 4, K = 2, arr[] = {2, 5, 6, 8}
Output: 6
Explanation: First operation: Change 5 at index 1 with 6 at index 2.
So that the arr[] becomes: {2, 6, 6, 8}
Second operation: Change 2 at index 0 with 6 at index 1.
So that the arr[] becomes: {6, 6, 6, 8}
The minimum value of arr[] can be achieved by applying operations is: 6
Approach: To solve the problem follow the below idea:
Sort the arr[], if K is greater than or equal to length of arr[], simply return element at last index of arr[] else return element at Kth index of arr[].
Illustration with an Example:
Consider N = 6, K = 3, arr[] = {9, 7, 3, 1, 2, 5}
We can perform the following operations
Operation 1:- Change 2 at index 4 with 5 at index 5 . So the arr[] becomes: {9, 7, 3, 1, 5, 5}
Operation 2:- Change 1 at index 3 with 5 at index 4 . So the arr[] becomes: {9, 7, 3, 5, 5, 5}
Operation 3:- Change 3 at index 2 with 7 at index 1 . So the arr[] becomes: {9, 7, 7, 5, 5, 5}
Minimum element after applying operation at most 3 times is: 5
When you will sort the arr[] and return arr[K] you will get the same output :-
Sorted arr[]: {1, 2, 3, 5, 7, 9}
arr[K] = arr[3] = 5, which is out required answer.
Follow the steps to solve the problem:
- Sort the array.
- Check if K is greater than or equal to arr[] or not.
- If yes, then simply return the element at the last index of arr[].
- Else return the element at the Kth index of arr[].
- Print the output.
Below is the implementation for the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
int main() {
int N = 6, K = 3 ;
int arr[] = { 9, 1, 3, 7, 2, 5 } ;
sort( arr, arr+N ) ;
if( K >= N )
cout << arr[ N-1 ] ;
else
cout << arr[ K ] ;
return 0;
}
|
Java
import java.util.*;
class GFG {
public static void main(String[] args)
{
int N = 6, K = 3;
int[] arr = { 9, 7, 3, 1, 2, 5 };
System.out.println(Min_Value(N, K, arr));
}
static int Min_Value(int N, int K, int arr[])
{
Arrays.sort(arr);
if (K == arr.length || K > arr.length)
return (arr[arr.length - 1]);
else
return (arr[K]);
}
}
|
Python3
if __name__ == "__main__":
N = 6
K = 3
arr = [9, 1, 3, 7, 2, 5]
arr.sort()
if(K >= N):
print(arr[N-1])
else:
print(arr[K])
|
C#
using System;
public class GFG {
static public void Main()
{
int N = 6, K = 3;
int[] arr = { 9, 7, 3, 1, 2, 5 };
Console.WriteLine(Min_Value(N, K, arr));
}
static int Min_Value(int N, int K, int[] arr)
{
Array.Sort(arr);
if (K == arr.Length || K > arr.Length)
return (arr[arr.Length - 1]);
else
return (arr[K]);
}
}
|
Javascript
<script>
let N = 6, K = 3;
let arr = [9, 1, 3, 7, 2, 5];
arr.sort();
if( K >= N )
document.write(arr[N-1],"</br>");
else
document.write(arr[K],"</br>");
</script>
|
Time Complexity: O(N * logN), because sorting is performed.
Auxiliary Space: O(1), as no extra space is required.
Another Approach:
- Sort the input array in non-decreasing order. We can use any sorting algorithm like quicksort, heapsort, or mergesort to do this.
- Check if the value of K is greater than or equal to the length of the array. If it is, then we can simply return the last element of the sorted array as this would be the maximum possible value that can be achieved after applying K operations.
- If the value of K is less than the length of the array, then we need to determine the maximum value that can be achieved by applying K operations.
- We can achieve this by repeatedly swapping adjacent elements that are
C++
#include <bits/stdc++.h>
using namespace std;
int main() {
int N = 6, K = 3 ;
int arr[] = { 9, 1, 3, 7, 2, 5 } ;
nth_element(arr, arr+K, arr+N);
if( K >= N )
cout << arr[ N-1 ] ;
else
cout << arr[ K ] ;
return 0;
}
|
Java
import java.util.Arrays;
public class Main {
public static void main(String[] args) {
int N = 6, K = 3;
int[] arr = { 9, 1, 3, 7, 2, 5 };
Arrays.sort(arr);
if (K >= N) {
System.out.println(arr[N - 1]);
}
else {
System.out.println(arr[K]);
}
}
}
|
Python3
if __name__ == "__main__":
N = 6
K = 3
arr = [9, 1, 3, 7, 2, 5]
arr.sort()
if(K >= N):
print(arr[N-1])
else:
print(arr[K])
|
C#
using System;
public class Program {
public static void Main()
{
int N = 6, K = 3;
int[] arr = { 9, 1, 3, 7, 2, 5 };
Array.Sort(arr);
if (K >= N) {
Console.WriteLine(arr[N - 1]);
}
else {
Console.WriteLine(arr[K]);
}
}
}
|
Javascript
let N = 6, K = 3 ;
let arr = [ 9, 1, 3, 7, 2, 5 ];
arr.sort((a, b) => a - b);
if( K >= N )
console.log(arr[ N-1 ]) ;
else
console.log(arr[ K ]) ;
|
Time Complexity: O(N*logN)
Auxiliary Space: O(1)
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