Find minimum value of the expression by choosing K elements from given array
Last Updated :
15 Sep, 2022
Given an array of integers arr of size N, the task is to find the minimum possible of the expression by choosing exactly K(? N) integers form given array arr. Let say if chosen elements are stored in array B (B1, B2, B3.....Bk) then value of expression:
x = \sum_{i=1}^k\sum_{j=1}^k(B_i - B_j)^{2}
Examples:
Input : arr[] = {2, 0, 9, 5}, k = 2
Output : 8
Let say, chosen elements are {2, 0}, then x = 8, which is minimum possible
Input : arr[] = {4, 21, 5, 3, 8}, k = 3
Output : 200
Approach :
The above expression can be simplified as:
- \sum_{i=1}^k\sum_{j=1}^k(B_i^2 + B_j^2 - 2*B_i*B_j)
- \sum_{i=1}^k\sum_{j=1}^k{B_i^2} + \sum_{i=1}^k\sum_{j=1}^k{B_j^2} - 2\sum_{i=1}^k\sum_{j=1}^k{B_i*B_j}
- k*\sum_{i=1}^k{B_i^2} + k*\sum_{i=1}^k{B_j^2} - 2\sum_{i=1}^k{B_i}*\sum_{j=1}^k{B_j}
- 2*k*\sum_{i=1}^k{B_i^2} - 2\sum_{i=1}^k{B_i^2}
- (2*k-2)*\sum_{i=1}^k{B_i^2}
So, all we need to do is select the k smallest elements from the array and solve the expression.
Below is the implementation of the above approach:
C++
// CPP program to find the minimum possible of the expression
// by choosing exactly K(? N) integers form given array arr
#include <bits/stdc++.h>
using namespace std;
// Function to find the minimum possible of the expression
// by choosing exactly K(? N) integers form given array arr
int minimumValue(int arr[], int n, int k)
{
// Sorting the array for least k element selection
sort(arr, arr + n);
int answer = 0;
// Select first k elements from sorted array
for (int i = 0; i < k; i++)
answer += arr[i] * arr[i];
// Return value of solved expression
return answer * (2 * k - 2);
}
// Driver code
int main()
{
int arr[] = { 4, 21, 5, 3, 8 }, k = 3;
int n = sizeof(arr) / sizeof(arr[0]);
// Function call
cout << minimumValue(arr, n, k);
return 0;
}
// JAVA program to find the minimum possible of the expression
// by choosing exactly K(? N) integers form given array arr
import java.util.*;
class GFG{
// Function to find the minimum possible of the expression
// by choosing exactly K(? N) integers form given array arr
static int minimumValue(int arr[], int n, int k)
{
// Sorting the array for least k element selection
Arrays.sort(arr);
int answer = 0;
// Select first k elements from sorted array
for (int i = 0; i < k; i++)
answer += arr[i] * arr[i];
// Return value of solved expression
return answer * (2 * k - 2);
}
// Driver code
public static void main(String[] args)
{
int arr[] = { 4, 21, 5, 3, 8 }, k = 3;
int n = arr.length;
// Function call
System.out.print(minimumValue(arr, n, k));
}
}
// This code is contributed by Rajput-Ji
# Python program to find the minimum
# possible of the expression by choosing
# exactly K(? N) integers form given array arr
# Function to find the minimum
# possible of the expression by
# choosing exactly K(? N) integers
# form given array arr
def minimumValue(arr, n, k):
# Sorting the array for least k element selection
arr.sort();
answer = 0;
# Select first k elements from sorted array
for i in range(k):
answer += arr[i] * arr[i];
# Return value of solved expression
return answer * (2 * k - 2);
# Driver code
if __name__ == '__main__':
arr = [ 4, 21, 5, 3, 8 ];
k = 3;
n = len(arr);
# Function call
print(minimumValue(arr, n, k));
# This code is contributed by Rajput-Ji
// C# program to find the minimum possible of the expression
// by choosing exactly K(? N) integers form given array arr
using System;
class GFG{
// Function to find the minimum possible of the expression
// by choosing exactly K(? N) integers form given array arr
static int minimumValue(int []arr, int n, int k)
{
// Sorting the array for least k element selection
Array.Sort(arr);
int answer = 0;
// Select first k elements from sorted array
for (int i = 0; i < k; i++)
answer += arr[i] * arr[i];
// Return value of solved expression
return answer * (2 * k - 2);
}
// Driver code
public static void Main(String[] args)
{
int []arr = { 4, 21, 5, 3, 8 };
int k = 3;
int n = arr.Length;
// Function call
Console.Write(minimumValue(arr, n, k));
}
}
// This code is contributed by 29AjayKumar
<script>
// JavaScript program to find the minimum possible of the expression
// by choosing exactly K(? N) integers form given array arr
// Function to find the minimum possible of the expression
// by choosing exactly K(? N) integers form given array arr
function minimumValue(arr, n, k)
{
// Sorting the array for least k element selection
arr.sort((a, b) => a - b);
let answer = 0;
// Select first k elements from sorted array
for (let i = 0; i < k; i++)
answer += arr[i] * arr[i];
// Return value of solved expression
return answer * (2 * k - 2);
}
// Driver code
let arr = [ 4, 21, 5, 3, 8 ], k = 3;
let n = arr.length;
// Function call
document.write(minimumValue(arr, n, k));
// This code is contributed by Surbhi Tyagi.
</script>
Time Complexity: O(n * log n)
Auxiliary Space: O(1)
Another Efficient Method: This problem can be solved efficiently using a Priority Queue. As we have to find the k smallest elements from the array in order to solve the expression so we use a priority queue which will find the k smallest elements in O(n*log(k)) time complexity where n is the size of the array and k is the number of smallest elements needed. Once we get the k smallest elements in the priority queue we will use the simplified expression given above to find the minimum value.
Below is the implementation of the above approach:
C++
// C++ code to implement the approach
#include <bits/stdc++.h>
using namespace std;
// Function to find the minimum possible of the expression
// by choosing exactly K(<= N) integers form given array arr
int minimumValue(int arr[], int n, int k)
{
// Using a Priority Queue
// to find k smallest elements
priority_queue<int> heap1;
for (int i = 0; i < n; ++i) {
// Insert elements into
// the priority queue
heap1.push(arr[i]);
// If size of the priority
// queue exceeds k then remove
// the largest element from
// the priority queue
if (heap1.size() > k) {
heap1.pop();
}
}
int answer = 0;
// Using first k elements from priority queue
// to find the minimum value
for (int i = 0; i < k; i++)
{
answer += heap1.top() * heap1.top();
heap1.pop();
}
// Return value of solved expression
return answer * (2 * k - 2);
}
// Driver code
int main()
{
int arr[] = { 4, 21, 5, 3, 8 }, k = 3;
int n = sizeof(arr) / sizeof(arr[0]);
// Function call
cout << minimumValue(arr, n, k);
return 0;
}
// This code is contributed by Pushpesh Raj
// Java code to implement the approach
import java.util.*;
class GFG {
// Function to find the minimum possible of the
// expression by choosing exactly K(<= N) integers form
// given array arr
static int minimumValue(int[] arr, int n, int k)
{
// Using a Priority Queue
// to find k smallest elements
PriorityQueue<Integer> heap1
= new PriorityQueue<Integer>(
Collections.reverseOrder());
for (int i = 0; i < n; ++i) {
// Insert elements into
// the priority queue
heap1.add(arr[i]);
// If size of the priority
// queue exceeds k then remove
// the largest element from
// the priority queue
if (heap1.size() > k) {
heap1.poll();
}
}
int answer = 0;
// Using first k elements from priority queue
// to find the minimum value
for (var i = 0; i < k; i++) {
answer += heap1.peek() * heap1.peek();
heap1.poll();
}
// Return value of solved expression
return answer * (2 * k - 2);
}
// Driver code
public static void main(String[] args)
{
int[] arr = { 4, 21, 5, 3, 8 };
int k = 3;
int n = arr.length;
// Function call
System.out.println(minimumValue(arr, n, k));
}
}
// This code is contributed by phasing17
# Python3 code to implement the approach
# Function to find the minimum possible of the expression
# by choosing exactly K(<= N) integers form given array arr
def minimumValue(arr, n, k):
# Using a Priority Queue
# to find k smallest elements
heap1 = [];
for i in range(n):
# Insert elements into
# the priority queue
heap1.append(arr[i]);
heap1.sort(reverse = True)
# If size of the priority
# queue exceeds k then remove
# the largest element from
# the priority queue
if (len(heap1) > k):
heap1.pop(0);
answer = 0;
# Using first k elements from priority queue
# to find the minimum value
for i in range(k):
answer += heap1[0] * heap1[0];
heap1.pop(0);
# Return value of solved expression
return answer * (2 * k - 2);
# Driver code
arr = [4, 21, 5, 3, 8 ]
k = 3;
n = len(arr);
# Function call
print(minimumValue(arr, n, k));
# This code is contributed by phasing17
// C# code to implement the approach
using System;
using System.Collections.Generic;
class GFG
{
// Function to find the minimum possible of the
// expression by choosing exactly K(<= N) integers form
// given array arr
static int minimumValue(int[] arr, int n, int k)
{
// Using a Priority Queue
// to find k smallest elements
List<int> heap1 = new List<int>();
for (int i = 0; i < n; ++i) {
// Insert elements into
// the priority queue
heap1.Add(arr[i]);
heap1.Sort();
heap1.Reverse();
// If size of the priority
// queue exceeds k then remove
// the largest element from
// the priority queue
if (heap1.Count > k) {
heap1.RemoveAt(0);
}
}
int answer = 0;
// Using first k elements from priority queue
// to find the minimum value
for (var i = 0; i < k; i++) {
answer += heap1[0] * heap1[0];
heap1.RemoveAt(0);
}
// Return value of solved expression
return answer * (2 * k - 2);
}
// Driver code
public static void Main(string[] args)
{
int[] arr = { 4, 21, 5, 3, 8 };
int k = 3;
int n = arr.Length;
// Function call
Console.WriteLine(minimumValue(arr, n, k));
}
}
// This code is contributed by phasing17
// JS code to implement the approach
// Function to find the minimum possible of the expression
// by choosing exactly K(<= N) integers form given array arr
function minimumValue(arr, n, k)
{
// Using a Priority Queue
// to find k smallest elements
let heap1 = [];
for (var i = 0; i < n; ++i) {
// Insert elements into
// the priority queue
heap1.push(arr[i]);
heap1.sort(function(a, b) { return a < b})
// If size of the priority
// queue exceeds k then remove
// the largest element from
// the priority queue
if (heap1.length > k) {
heap1.shift();
}
}
let answer = 0;
// Using first k elements from priority queue
// to find the minimum value
for (var i = 0; i < k; i++)
{
answer += heap1[0] * heap1[0];
heap1.shift();
}
// Return value of solved expression
return answer * (2 * k - 2);
}
// Driver code
let arr = [4, 21, 5, 3, 8 ]
let k = 3;
let n = arr.length;
// Function call
console.log(minimumValue(arr, n, k));
// This code is contributed by phasing17
Time Complexity: O(k+nlog(k)) where n is the size of the array and k is the given number of elements to choose from.
Auxiliary Space: O(k) since the priority queue at any time holds k elements.
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