Given an array arr[] of size N, the task is to find the smallest possible remaining array element by repeatedly removing a pair, say (arr[i], arr[j]) from the array and inserting the Ceil value of their average.
Examples:
Input: arr[] = { 1, 2, 3 }
Output: 2
Explanation:
Removing the pair (arr[1], arr[2]) from arr[] and inserting (arr[1] + arr[2] + 1) / 2 into arr[] modifies arr[] to { 1, 2 }.
Removing the pair (arr[0], arr[1]) from arr[] and inserting (arr[0] + arr[1] + 1) / 2 into arr[] modifies arr[] to { 2 }.
Therefore, the required output is 2.Input: arr[] = { 30, 16, 40 }
Output: 26
Explanation:
Removing the pair (arr[0], arr[2]) from arr[] and inserting (arr[0] + arr[2] + 1) / 2 into arr[] modifies arr[] to { 16, 35 } .
Removing the pair (arr[0], arr[1]) from arr[] and inserting (arr[0] + arr[1] + 1) / 2 into arr[] modifies arr[] to { 26 } .
Therefore, the required output is 26.
Approach: The problem can be solved using Greedy technique. The idea is to repeatedly remove the maximum and the second-maximum array element and insert their average. Finally, print the smallest element left in the array.
- Initialize a priority_queue, say PQ, to store the array elements such that the largest element is always present at the top of PQ.
- Traverse the array and store all the array elements in PQ.
- Iterate over the elements of the priority_queue while count of elements in the priority_queue is greater than 1. In every iteration, pop the top two elements from PQ and insert the Ceil value of their average.
- Finally, print the element left in PQ.
Below is the implementation of the above approach:
// C++ program to implement // the above approach #include <bits/stdc++.h> using namespace std;
// Function to find the smallest element // left in the array by removing pairs // and inserting their average int findSmallestNumLeft( int arr[], int N)
{ // Stores array elements such that the
// largest element present at top of PQ
priority_queue< int > PQ;
// Traverse the array
for ( int i = 0; i < N; i++) {
// Insert arr[i] into PQ
PQ.push(arr[i]);
}
// Iterate over elements of PQ while count
// of elements in PQ greater than 1
while (PQ.size() > 1) {
// Stores largest element of PQ
int top1 = PQ.top();
// Pop the largest element of PQ
PQ.pop();
// Stores largest element of PQ
int top2 = PQ.top();
// Pop the largest element of PQ
PQ.pop();
// Insert the ceil value of average
// of top1 and top2
PQ.push((top1 + top2 + 1) / 2);
}
return PQ.top();
} // Driver Code int main()
{ int arr[] = { 30, 16, 40 };
int N = sizeof (arr)
/ sizeof (arr[0]);
cout << findSmallestNumLeft(
arr, N);
return 0;
} |
// Java program to implement // the above approach import java.util.PriorityQueue;
class GFG{
// Function to find the smallest element // left in the array by removing pairs // and inserting their average static int findSmallestNumLeft( int arr[], int N)
{ // Stores array elements such that the
// largest element present at top of PQ
PriorityQueue<Integer> PQ = new PriorityQueue<Integer>((a,b)->b-a);
// Traverse the array
for ( int i = 0 ; i < N; i++)
{
// Insert arr[i] into PQ
PQ.add(arr[i]);
}
// Iterate over elements of PQ while count
// of elements in PQ greater than 1
while (PQ.size() > 1 )
{
// Stores largest element of PQ
int top1 = PQ.peek();
// Pop the largest element of PQ
PQ.remove();
// Stores largest element of PQ
int top2 = PQ.peek();
// Pop the largest element of PQ
PQ.remove();
// Insert the ceil value of average
// of top1 and top2
PQ.add((top1 + top2 + 1 ) / 2 );
}
return PQ.peek();
} // Driver Code public static void main(String[] args)
{ int arr[] = { 30 , 16 , 40 };
int N = arr.length;
System.out.print(findSmallestNumLeft(
arr, N));
} } // This code is contributed by Amit Katiyar |
# Python3 program to implement # the above approach # Function to find the smallest element # left in the array by removing pairs # and inserting their average def findSmallestNumLeft(arr, N):
# Stores array elements such that the
# largest element present at top of PQ
PQ = []
# Traverse the array
for i in range (N):
# Insert arr[i] into PQ
PQ.append(arr[i])
# Iterate over elements of PQ while count
# of elements in PQ greater than 1
PQ = sorted (PQ)
while ( len (PQ) > 1 ):
# Stores largest element of PQ
top1 = PQ[ - 1 ]
# Pop the largest element of PQ
del PQ[ - 1 ]
# Stores largest element of PQ
top2 = PQ[ - 1 ]
# Pop the largest element of PQ
del PQ[ - 1 ]
# Insert the ceil value of average
# of top1 and top2
PQ.append((top1 + top2 + 1 ) / / 2 )
PQ = sorted (PQ)
return PQ[ - 1 ]
# Driver Code if __name__ = = '__main__' :
arr = [ 30 , 16 , 40 ]
N = len (arr)
print (findSmallestNumLeft(arr, N))
# This code is contributed by mohit kumar 29 |
// C# program to implement // the above approach using System;
using System.Collections.Generic;
class GfG
{ // Function to find the smallest element
// left in the array by removing pairs
// and inserting their average
static int findSmallestNumLeft( int [] arr, int N)
{
// Stores array elements such that the
// largest element present at top of PQ
List< int > PQ = new List< int >();
// Traverse the array
for ( int i = 0; i < N; i++) {
// Insert arr[i] into PQ
PQ.Add(arr[i]);
}
PQ.Sort();
PQ.Reverse();
// Iterate over elements of PQ while count
// of elements in PQ greater than 1
while (PQ.Count > 1) {
// Stores largest element of PQ
int top1 = PQ[0];
// Pop the largest element of PQ
PQ.RemoveAt(0);
// Stores largest element of PQ
int top2 = PQ[0];
// Pop the largest element of PQ
PQ.RemoveAt(0);
// Insert the ceil value of average
// of top1 and top2
PQ.Add((top1 + top2 + 1) / 2);
PQ.Sort();
PQ.Reverse();
}
return PQ[0];
}
// Driver code
public static void Main()
{
int [] arr = { 30, 16, 40 };
int N = arr.Length;
Console.Write(findSmallestNumLeft(arr, N));
}
} // This code is contributed by divyeshrabadiya07. |
<script> // Javascript program to implement // the above approach // Function to find the smallest element // left in the array by removing pairs // and inserting their average function findSmallestNumLeft(arr, N)
{ // Stores array elements such that the
// largest element present at top of PQ
let PQ = [];
// Traverse the array
for (let i = 0; i < N; i++)
{
// Insert arr[i] into PQ
PQ.push(arr[i]);
}
PQ.sort( function (a,b){ return b-a;});
// Iterate over elements of PQ while count
// of elements in PQ greater than 1
while (PQ.length > 1)
{
// Stores largest element of PQ
let top1 = PQ[0];
// Pop the largest element of PQ
PQ.shift();
// Stores largest element of PQ
let top2 = PQ[0];
// Pop the largest element of PQ
PQ.shift();
// Insert the ceil value of average
// of top1 and top2
PQ.push(Math.floor((top1 + top2 + 1) / 2));
PQ.sort( function (a,b){ return b-a;});
}
return PQ[0];
} // Driver Code let arr = [ 30, 16, 40]; let N = arr.length; document.write(findSmallestNumLeft( arr, N));
// This code is contributed by unknown2108 </script> |
26
Time Complexity: O(N*logN), as we are using a loop to traverse N times and priority queue operation will take logN time.
Auxiliary Space: O(N), as we are using extra space for priority queue.