Maximum possible middle element of the array after deleting exactly k elements
Last Updated :
09 Jun, 2022
Given an integer array of size n and a number k. If the indexing is 1 based then the middle element of the array is the element at index (n + 1) / 2, if n is odd otherwise n / 2. The task is to delete exactly k elements from the array in such a way that the middle element of the reduced array is as maximum as possible. Find the maximum possible middle element of the array after deleting exactly k elements.
Examples:
Input :
n = 5, k = 2
arr[] = {9, 5, 3, 7, 10};
Output : 7
Input :
n = 9, k = 3
arr[] = {2, 4, 3, 9, 5, 8, 7, 6, 10};
Output : 9
In the first input, if we delete 5 and 3 then the array becomes {9, 7, 10} and
the middle element will be 7.
In the second input, if we delete one element before 9 and two elements after 9
(for example 2, 5, 8) then the array becomes {4, 3, 9, 7, 6, 10} and middle
element will be 9 and it will be the optimum solution.
Naive Approach :
The naive approach is to check all possible solutions. There could be C(n, k) possible solutions. If we check all possible solutions to find an optimal solution, it will consume a lot of time.
Optimal Approach :
After deleting k elements, the array will be reduced to size n - k. Since we can delete any k numbers from the array to find the maximum possible middle elements. If we note, the index of the middle element after deleting k elements will lie in the range ( n + 1 - k ) / 2 and ( n + 1 - k ) / 2 + k. So in order to find the optimal solution, simply iterate the array from the index ( n + 1 - k ) / 2 to index ( n + 1 - k ) / 2 + k and select the maximum element in this range.
The is the implementation is given below.
C++
#include <bits/stdc++.h>
using namespace std;
// Function to calculate maximum possible middle
// value of the array after deleting exactly k
// elements
int maximum_middle_value(int n, int k, int arr[])
{
// Initialize answer as -1
int ans = -1;
// Calculate range of elements that can give
// maximum possible middle value of the array
// since index of maximum possible middle
// value after deleting exactly k elements from
// array will lie in between low and high
int low = (n + 1 - k) / 2;
int high = (n + 1 - k) / 2 + k;
// Find maximum element of the array in
// range low and high
for (int i = low; i <= high; i++) {
// since indexing is 1 based so
// check element at index i - 1
ans = max(ans, arr[i - 1]);
}
// Return the maximum possible middle value
// of the array after deleting exactly k
// elements from the array
return ans;
}
// Driver Code
int main()
{
int n = 5, k = 2;
int arr[] = { 9, 5, 3, 7, 10 };
cout << maximum_middle_value(n, k, arr) << endl;
n = 9;
k = 3;
int arr1[] = { 2, 4, 3, 9, 5, 8, 7, 6, 10 };
cout << maximum_middle_value(n, k, arr1) << endl;
return 0;
}
// Java implementation of the approach
import java.util.*;
class GFG
{
// Function to calculate maximum possible middle
// value of the array after deleting exactly k
// elements
static int maximum_middle_value(int n, int k, int arr[])
{
// Initialize answer as -1
int ans = -1;
// Calculate range of elements that can give
// maximum possible middle value of the array
// since index of maximum possible middle
// value after deleting exactly k elements from
// array will lie in between low and high
int low = (n + 1 - k) / 2;
int high = (n + 1 - k) / 2 + k;
// Find maximum element of the array in
// range low and high
for (int i = low; i <= high; i++)
{
// since indexing is 1 based so
// check element at index i - 1
ans = Math.max(ans, arr[i - 1]);
}
// Return the maximum possible middle value
// of the array after deleting exactly k
// elements from the array
return ans;
}
// Driver Code
public static void main(String args[])
{
int n = 5, k = 2;
int arr[] = { 9, 5, 3, 7, 10 };
System.out.println( maximum_middle_value(n, k, arr));
n = 9;
k = 3;
int arr1[] = { 2, 4, 3, 9, 5, 8, 7, 6, 10 };
System.out.println( maximum_middle_value(n, k, arr1));
}
}
// This code is contributed by Arnab Kundu
# Python3 implementation of the approach
# Function to calculate maximum possible
# middle value of the array after
# deleting exactly k elements
def maximum_middle_value(n, k, arr):
# Initialize answer as -1
ans = -1
# Calculate range of elements that can give
# maximum possible middle value of the array
# since index of maximum possible middle
# value after deleting exactly k elements
# from array will lie in between low and high
low = (n + 1 - k) // 2
high = (n + 1 - k) // 2 + k
# Find maximum element of the
# array in range low and high
for i in range(low, high+1):
# since indexing is 1 based so
# check element at index i - 1
ans = max(ans, arr[i - 1])
# Return the maximum possible middle
# value of the array after deleting
# exactly k elements from the array
return ans
# Driver Code
if __name__ == "__main__":
n, k = 5, 2
arr = [9, 5, 3, 7, 10]
print(maximum_middle_value(n, k, arr))
n, k = 9, 3
arr1 = [2, 4, 3, 9, 5, 8, 7, 6, 10]
print(maximum_middle_value(n, k, arr1))
# This code is contributed by Rituraj Jain
// C# implementation of the approach
using System;
class GFG
{
// Function to calculate maximum possible middle
// value of the array after deleting exactly k
// elements
static int maximum_middle_value(int n, int k, int []arr)
{
// Initialize answer as -1
int ans = -1;
// Calculate range of elements that can give
// maximum possible middle value of the array
// since index of maximum possible middle
// value after deleting exactly k elements from
// array will lie in between low and high
int low = (n + 1 - k) / 2;
int high = (n + 1 - k) / 2 + k;
// Find maximum element of the array in
// range low and high
for (int i = low; i <= high; i++)
{
// since indexing is 1 based so
// check element at index i - 1
ans = Math.Max(ans, arr[i - 1]);
}
// Return the maximum possible middle value
// of the array after deleting exactly k
// elements from the array
return ans;
}
// Driver Code
static public void Main ()
{
int n = 5, k = 2;
int []arr = { 9, 5, 3, 7, 10 };
Console.WriteLine( maximum_middle_value(n, k, arr));
n = 9;
k = 3;
int []arr1 = { 2, 4, 3, 9, 5, 8, 7, 6, 10 };
Console.WriteLine( maximum_middle_value(n, k, arr1));
}
}
// This code is contributed by ajit.
<script>
// Function to calculate maximum possible middle
// value of the array after deleting exactly k
// elements
function maximum_middle_value(n, k, arr)
{
// Initialize answer as -1
let ans = -1;
// Calculate range of elements that can give
// maximum possible middle value of the array
// since index of maximum possible middle
// value after deleting exactly k elements from
// array will lie in between low and high
let low = Math.floor((n + 1 - k) / 2);
let high = Math.floor(((n + 1 - k) / 2) + k);
// Find maximum element of the array in
// range low and high
for (let i = low; i <= high; i++) {
// since indexing is 1 based so
// check element at index i - 1
ans = Math.max(ans, arr[i - 1]);
}
// Return the maximum possible middle value
// of the array after deleting exactly k
// elements from the array
return ans;
}
// Driver Code
let n = 5, k = 2;
let arr = [ 9, 5, 3, 7, 10 ];
document.write(maximum_middle_value(n, k, arr) + "<br>");
n = 9;
k = 3;
let arr1 = [ 2, 4, 3, 9, 5, 8, 7, 6, 10 ];
document.write(maximum_middle_value(n, k, arr1) + "<br>");
// This code is contributed by Mayank Tyagi
</script>
Time Complexity: O(high-low), where high and low are the terms calculated
Auxiliary Space: O(1), as no extra space is used
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