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Number of ways to remove elements to maximize arithmetic mean
  • Last Updated : 03 Jun, 2021

Given an array arr[], the task is to find the number of ways to remove elements from the array so as to maximize the arithmetic mean of the remaining array.
Examples: 
 

Input: arr[] = { 1, 2, 1, 2 } 
Output: 3
Remove elements at indices: 
{ 0, 1, 2 } 
{ 0, 2, 3 } 
{ 0, 2 } 
Input: arr[] = { 1, 2, 3 } 
Output:
 

 

Approach: The arithmetic mean of the array is maximized when only the maximum element(s) remains in the array. 
Now consider the array arr[] = { 3, 3, 3, 3 } 
We just need to make sure that at least one instance of the maximum element remains in the array after removing the other elements. This will guarantee maximization of the arithmetic mean. Hence we need to remove at most 3 elements from the above array. The number of ways to remove at most 3 elements:
 

  1. Zero elements removed. Number of ways = 1.
  2. One element removed. Number of ways = 4.
  3. Two elements removed. Number of ways = 6.
  4. Three elements removed. Number of ways = 4.

Hence total = 1 + 4 + 6 + 4 = 15 = 24 – 1.
Now consider the array = { 1, 4, 3, 2, 3, 4, 4 } 
On sorting the array becomes = { 1, 2, 3, 3, 4, 4, 4 }. In this case, there are elements other than 4. We can remove at most 2 instances of 4 and when those instances are removed, the other elements (which are not 4) should always be removed with them. Hence the number of ways will remain the same as the number of ways to remove at most 2 instances of 4.
The various ways of removing elements: 
{ 1, 2, 3, 3 } 
{ 1, 2, 3, 3, 4 } 
{ 1, 2, 3, 3, 4 } 
{ 1, 2, 3, 3, 4 } 
{ 1, 2, 3, 3, 4, 4 } 
{ 1, 2, 3, 3, 4, 4 } 
{ 1, 2, 3, 3, 4, 4 }
Therefore the answer is 2count of max element – 1.
Below is the implementation of the above approach. 
 



C++




// C++ implementation of the above approach
#include <bits/stdc++.h>
 
#define ll long long
 
using namespace std;
const int mod = 1000000007;
 
// Function to compute a^n
ll power(ll a, ll n)
{
    if (n == 0)
        return 1;
 
    ll p = power(a, n / 2) % mod;
    p = (p * p) % mod;
    if (n & 1)
        p = (p * a) % mod;
 
    return p;
}
 
// Function to return number of ways to maximize arithmetic mean
ll numberOfWays(int* arr, int n)
{
 
    int max_count = 0;
    int max_value = *max_element(arr, arr + n);
    for (int i = 0; i < n; i++) {
        if (arr[i] == max_value)
            max_count++;
    }
    return (power(2, max_count) - 1 + mod) % mod;
}
 
// Driver code
int main()
{
    int arr[] = { 1, 2, 1, 2 };
    int n = sizeof(arr) / sizeof(arr[0]);
    cout << numberOfWays(arr, n);
    return 0;
}

Java




// Java implementation of the above approach
import java.util.Arrays;
 
class GFG
{
     
static int mod = 1000000007;
 
// Function to compute a^n
static int power(int a, int n)
{
    if (n == 0)
        return 1;
 
    int p = power(a, n / 2) % mod;
    p = (p * p) % mod;
    if ((n & 1) > 0)
        p = (p * a) % mod;
 
    return p;
}
 
// Function to return number of
// ways to maximize arithmetic mean
static int numberOfWays(int []arr, int n)
{
 
    int max_count = 0;
    int max_value = Arrays.stream(arr).max().getAsInt();
    for (int i = 0; i < n; i++)
    {
        if (arr[i] == max_value)
            max_count++;
    }
    return (power(2, max_count) - 1 + mod) % mod;
}
 
// Driver code
public static void main (String[] args)
{
    int []arr = { 1, 2, 1, 2 };
    int n = arr.length;
    System.out.println(numberOfWays(arr, n));
}
}
 
// This code is contributed by mits

Python3




# Python3 implementation of the
# above approach
 
mod = 1000000007;
 
# Function to compute a^n
def power(a, n) :
     
    if (n == 0) :
        return 1;
 
    p = power(a, n // 2) % mod;
    p = (p * p) % mod;
    if (n & 1) :
        p = (p * a) % mod;
 
    return p;
 
# Function to return number of ways
# to maximize arithmetic mean
def numberOfWays(arr, n) :
 
    max_count = 0;
    max_value = max(arr)
     
    for i in range(n) :
        if (arr[i] == max_value) :
            max_count += 1;
             
    return (power(2, max_count) - 1 + mod) % mod;
 
# Driver code
if __name__ == "__main__" :
 
    arr = [ 1, 2, 1, 2 ];
    n = len(arr) ;
     
    print(numberOfWays(arr, n));
 
# This code is contributed by Ryuga

C#




// C# implementation of the above approach
using System;
using System.Linq;
 
class GFG
{
     
static int mod = 1000000007;
 
// Function to compute a^n
static int power(int a, int n)
{
    if (n == 0)
        return 1;
 
    int p = power(a, n / 2) % mod;
    p = (p * p) % mod;
    if ((n & 1)>0)
        p = (p * a) % mod;
 
    return p;
}
 
// Function to return number of
// ways to maximize arithmetic mean
static int numberOfWays(int []arr, int n)
{
 
    int max_count = 0;
    int max_value = arr.Max();
    for (int i = 0; i < n; i++)
    {
        if (arr[i] == max_value)
            max_count++;
    }
    return (power(2, max_count) - 1 + mod) % mod;
}
 
// Driver code
static void Main()
{
    int []arr = { 1, 2, 1, 2 };
    int n = arr.Length;
    Console.WriteLine(numberOfWays(arr, n));
}
}
 
// This code is contributed by mits

PHP




<?php
// PHP implementation of the above approach
 
// Function to compute a^n
function power($x, $y, $p)
{
     
    // Initialize result
    $res = 1;
 
    // Update x if it is more
    // than or equal to p
    $x = $x % $p;
 
    while ($y > 0)
    {
        // If y is odd, multiply
        // x with result
        if ($y & 1)
            $res = ($res * $x) % $p;
 
        // y must be even now
         
        // y = $y/2
        $y = $y >> 1;
        $x = ($x * $x) % $p;
    }
    return $res;
}
 
// Function to return number of ways
// to maximize arithmetic mean
function numberOfWays($arr, $n)
{
    $mod = 1000000007;
    $max_count = 0;
    $max_value = $arr[0];
    for($i = 0; $i < $n; $i++)
    if($max_value < $arr[$i])
        $max_value = $arr[$i];
     
    for ($i = 0; $i < $n; $i++)
    {
        if ($arr[$i] == $max_value)
            $max_count++;
    }
    return (power(2, $max_count,
                     $mod) - 1 + $mod) % $mod;
}
 
// Driver code
$arr = array( 1, 2, 1, 2 );
$n = 4;
echo numberOfWays($arr, $n);
 
// This code is contributed
// by Arnab Kundu
?>

Javascript




<script>
    // Javascript implementation of the above approach
     
    let mod = 1000000007;
   
    // Function to compute a^n
    function power(a, n)
    {
        if (n == 0)
            return 1;
 
        let p = power(a, parseInt(n / 2, 10)) % mod;
        p = (p * p) % mod;
        if ((n & 1)>0)
            p = (p * a) % mod;
 
        return p;
    }
 
    // Function to return number of
    // ways to maximize arithmetic mean
    function numberOfWays(arr, n)
    {
 
        let max_count = 0;
        let max_value = Number.MIN_VALUE;
        for (let i = 0; i < n; i++)
        {
            max_value = Math.max(max_value, arr[i]);
        }
        for (let i = 0; i < n; i++)
        {
            if (arr[i] == max_value)
                max_count++;
        }
        return (power(2, max_count) - 1 + mod) % mod;
    }
     
    let arr = [ 1, 2, 1, 2 ];
    let n = arr.length;
    document.write(numberOfWays(arr, n));
 
// This code is contributed by divyeshrabadiya07.
</script>
Output: 
3

 

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