Given N elements, you can remove any two elements from the list, note their sum, and add the sum to the list. Repeat these steps while there is more than a single element in the list. The task is to minimize the sum of these chosen sums in the end.
Examples:
Input: arr[] = {1, 4, 7, 10}
Output: 39
Choose 1 and 4, Sum = 5, arr[] = {5, 7, 10}
Choose 5 and 7, Sum = 17, arr[] = {12, 10}
Choose 12 and 10, Sum = 39, arr[] = {22}
Input: arr[] = {1, 3, 7, 5, 6}
Output: 48
Approach: In order to minimize the sum, the elements that get chosen at every step must the minimum elements from the list. In order to do that efficiently, a priority queue can be used. At every step, while there is more than a single element in the list, choose the minimum and the second minimum, remove them from the list add their sum to the list after updating the running sum.
Steps to solve the problem:
- initialize two variable i and sum to store the minimized sum.
- initialize the priority queue with min heap.
- iterate through the array and push all elements in the queue.
- while size of queue is greater than one:
- initialize the min variable and store the top element in the queue.
- pop the top element from the queue.
- initialize the secondMin variable store the top element in the queue.
- update the sum variable with min and secondMin.
- push the sum of min and secondMin in the queue.
5. return the sum.
Below is the implementation of the above approach:
// C++ implementation of the approach #include<bits/stdc++.h> using namespace std;
// Function to return the minimized sum int getMinSum( int arr[], int n)
{ int i, sum = 0;
// Priority queue to store the elements of the array
// and retrieve the minimum element efficiently
priority_queue< int , vector< int >, greater< int > > pq;
// Add all the elements
// to the priority queue
for (i = 0; i < n; i++)
pq.push(arr[i]);
// While there are more than 1 elements
// left in the queue
while (pq.size() > 1)
{
// Remove and get the minimum
// element from the queue
int min = pq.top();
pq.pop();
// Remove and get the second minimum
// element (currently minimum)
int secondMin = pq.top();
pq.pop();
// Update the sum
sum += (min + secondMin);
// Add the sum of the minimum
// elements to the queue
pq.push(min + secondMin);
}
// Return the minimized sum
return sum;
} // Driver code int main()
{ int arr[] = { 1, 3, 7, 5, 6 };
int n = sizeof (arr)/ sizeof (arr[0]);
cout << (getMinSum(arr, n));
} // This code is contributed by mohit |
// Java implementation of the approach import java.util.PriorityQueue;
class GFG
{ // Function to return the minimized sum
static int getMinSum( int arr[], int n)
{
int i, sum = 0 ;
// Priority queue to store the elements of the array
// and retrieve the minimum element efficiently
PriorityQueue<Integer> pq = new PriorityQueue<>();
// Add all the elements
// to the priority queue
for (i = 0 ; i < n; i++)
pq.add(arr[i]);
// While there are more than 1 elements
// left in the queue
while (pq.size() > 1 )
{
// Remove and get the minimum
// element from the queue
int min = pq.poll();
// Remove and get the second minimum
// element (currently minimum)
int secondMin = pq.poll();
// Update the sum
sum += (min + secondMin);
// Add the sum of the minimum
// elements to the queue
pq.add(min + secondMin);
}
// Return the minimized sum
return sum;
}
// Driver code
public static void main(String[] args)
{
int arr[] = { 1 , 3 , 7 , 5 , 6 };
int n = arr.length;
System.out.print(getMinSum(arr, n));
}
} |
# Python3 implementation of the approach # importing heapq python module # for implementing min heap import heapq
# Function to return the minimized sum def getMinSum(arr, n):
summ = 0
# Heap to store the elements of the array
# and retrieve the minimum element efficiently
pq = arr
# creating min heap from array pq
heapq.heapify(pq)
# While there are more than 1 elements
# left in the queue
while ( len (pq) > 1 ):
# storing minimum element (root of min heap)
# into minn
minn = pq[ 0 ]
# replacing root with last element and
# deleting last element from min heap
# as per deleting procedure for HEAP
pq[ 0 ] = pq[ - 1 ]
pq.pop()
# maintaining the min heap property
heapq.heapify(pq)
# again storing minimum element (root of min heap)
# into secondMin
secondMin = pq[ 0 ]
# again replacing root with last element and
# deleting last element as per heap procedure
pq[ 0 ] = pq[ - 1 ]
pq.pop()
# Update the sum
summ + = (minn + secondMin)
# appending the summ as last element of min heap
pq.append(minn + secondMin)
# again maintaining the min heap property
heapq.heapify(pq)
return summ
# Driver Code if __name__ = = "__main__" :
arr = [ 1 , 3 , 7 , 5 , 6 ]
n = len (arr)
print (getMinSum(arr, n))
'''Code is written by RAJAT KUMAR [GLAU]''' |
// C# implementation of the approach using System;
using System.Collections.Generic;
class GFG
{ // Function to return the minimized sum
static int getMinSum( int [] arr, int n)
{
int i, sum = 0;
// Priority queue to store the elements of the array
// and retrieve the minimum element efficiently
List< int > pq = new List< int >();
// Add all the elements
// to the priority queue
for (i = 0; i < n; i++)
{
pq.Add(arr[i]);
}
// While there are more than 1 elements
// left in the queue
while (pq.Count > 1)
{
pq.Sort();
// Remove and get the minimum
// element from the queue
int min = pq[0];
pq.RemoveAt(0);
// Remove and get the second minimum
// element (currently minimum)
int secondMin = pq[0];
pq.RemoveAt(0);
// Update the sum
sum += (min + secondMin);
// Add the sum of the minimum
// elements to the queue
pq.Add(min + secondMin);
}
// Return the minimized sum
return sum;
}
// Driver code
static public void Main ()
{
int [] arr = { 1, 3, 7, 5, 6 };
int n = arr.Length;
Console.WriteLine(getMinSum(arr, n));
}
} // This code is contributed by avanitrachhadiya2155 |
<script> // JavaScript implementation of the approach // Function to return the minimized sum
function getMinSum(arr,n)
{
let i, sum = 0;
// Priority queue to store the elements of the array
// and retrieve the minimum element efficiently
let pq = [];
// Add all the elements
// to the priority queue
for (i = 0; i < n; i++)
pq.push(arr[i]);
// While there are more than 1 elements
// left in the queue
while (pq.length > 1)
{
// Remove and get the minimum
// element from the queue
let min = pq.shift();
// Remove and get the second minimum
// element (currently minimum)
let secondMin = pq.shift();
// Update the sum
sum += (min + secondMin);
// Add the sum of the minimum
// elements to the queue
pq.push(min + secondMin);
}
// Return the minimized sum
return sum;
}
// Driver code
let arr=[1, 3, 7, 5, 6];
let n = arr.length;
document.write(getMinSum(arr, n));
// This code is contributed by rag2127 </script> |
48
Time Complexity : O(N * log(N))
Auxiliary Space: O(N)