Check if Array with mean X can be made using N elements of given Array

Given an array arr[] and two integers N and X, the task is to find if it is possible to create an array using N distinct elements from arr[] such that the mean of the newly formed array is X.

Examples:

Input: N = 5,  X = 8, arr[] = {1, 10, 3, 2, 6, 7, 4, 5}
Output: YES
Explanation: Many arrays using 5 distinct elements from the array are possible like {10, 6, 7, 4, 5, 10, 10, 10, 10}

Input: N = 3, X = 4, arr[] = {9, 7, 5}
Output: NO
Explanation: There is no possible array by a given finite set. So the mean of the array becomes exact to X.Therefore, the answer is NO.

Approach: Implement the idea below to solve the problem:

It is always possible to make mean X from a given arr[] of the integer if it lies in between the minimum and maximum integer in an arr[]. Formally, if ( X >= MinValue &&  X <= MaxValue ) then, the answer will be YES only for those cases otherwise NO.

Steps were taken to solve the problem:

1. Create an integer variable min and store the minimum value in this variable by iterating on arr[].
2. Create an integer variable max and store the maximum value in this variable by iterating on arr[].
3. if X lies in the range of min and max value of arr[], Then print YES else print NO.

Below is the implementation of the above approach.

// C++ code to implement the approach
#include <bits/stdc++.h>
using namespace std;

// Function to check if it is possible
// to create the array satisfying the conditions
string isPossible(int N, int X, int arr[])
{
  
    // Min variable to store minimum
    // value from input arr[]
    int sz = sizeof(arr) / sizeof(arr[0]);
    int min = INT_MAX;

    // Max variable to store maximum
    // value from input arr[]
    int max = INT_MIN;

    // Loop for iterating on arr[]
    for (int i = 0; i < sz; i++) {

        // Updating value of min variable
        min = arr[i] < min ? arr[i] : min;

        // Updating value of max variable
        max = arr[i] > max ? arr[i] : max;
    }

    // Printing answer as YES/NO by
    // comparing mean with min and max
    // values
    return X >= min && X <= max ? "YES" : "NO";
}

// Driver Code
int main()
{
    int N = 5, X = 8;
    int arr[] = { 1, 10, 3, 2, 6, 7, 4, 5 };

    // Function call
    cout << isPossible(N, X, arr);
    return 0;
}

// This code is contributed by rohit768.

// Java code to implement the approach

import java.io.*;

// Driver Class
class GFG {

    // Driver code
    public static void main(String[] args)
    {
        int N = 5, X = 8;
        int[] arr = { 1, 10, 3, 2, 6, 7, 4, 5 };

        // Function call
        System.out.println(isPossible(N, X, arr));
    }

    // Function to check if it is possible
    // to create the array satisfying the conditions
    static String isPossible(int N, int X, int arr[])
    {
        // Min variable to store minimum
        // value from input arr[]
        int min = Integer.MAX_VALUE;

        // Max variable to store maximum
        // value from input arr[]
        int max = Integer.MIN_VALUE;

        // Loop for iterating on arr[]
        for (int i = 0; i < arr.length; i++) {

            // Updating value of min variable
            min = arr[i] < min ? arr[i] : min;

            // Updating value of max variable
            max = arr[i] > max ? arr[i] : max;
        }

        // Printing answer as YES/NO by
        // comparing mean with min and max
        // values
        return X >= min && X <= max ? "YES" : "NO";
    }
}

# python code to implement the approach

# Function to check if it is possible
# to create the array satisfying the conditions
def isPossible(N, X, arr):
    # Min variable to store minimum
    # value from input arr[]
    sz = len(arr)
    mini = 9223372036854775807

    # Max variable to store maximum
    # value from input arr[]
    maxi = -9223372036854775808

    # Loop for iterating on arr[]
    for i in range(0, sz):
        # Updating value of min variable
        if(arr[i] < mini):
            mini = arr[i]

        # Updating value of max variable
        if(arr[i] > maxi):
            maxi = arr[i]
    # Print answer as YES/NO by
    # comparing mean with min and max
    # values
    if(X >= mini and X <= maxi):
        return "YES"
    else:
        return "NO"


# Driver Code
N = 5
X = 8
arr = [1, 10, 3, 2, 6, 7, 4, 5]
# Function call
print(isPossible(N, X, arr))

# This code is contributed by ksam24000

// C# code to implement the approach
using System;

// Driver Class
class GFG {

  // Driver code
  public static void Main()
  {
    int N = 5, X = 8;
    int[] arr = { 1, 10, 3, 2, 6, 7, 4, 5 };

    // Function call
    Console.WriteLine(isPossible(N, X, arr));
  }

  // Function to check if it is possible
  // to create the array satisfying the conditions
  static string isPossible(int N, int X, int[] arr)
  {
    // Min variable to store minimum
    // value from input arr[]
    int min = Int32.MaxValue;

    // Max variable to store maximum
    // value from input arr[]
    int max = Int32.MinValue;

    // Loop for iterating on arr[]
    for (int i = 0; i < arr.Length; i++) {

      // Updating value of min variable
      min = arr[i] < min ? arr[i] : min;

      // Updating value of max variable
      max = arr[i] > max ? arr[i] : max;
    }

    // Printing answer as YES/NO by
    // comparing mean with min and max
    // values
    return X >= min && X <= max ? "YES" : "NO";
  }
}

// This code is contributed by Samim Hossain Mondal.

// JS code to implement the approach

// Function to check if it is possible
// to create the array satisfying the conditions
function isPossible(N,X,arr)
{
  
    // Min variable to store minimum
    // value from input arr[]
    let sz = arr.length;
    let mn = Number.MAX_VALUE;

    // Max variable to store maximum
    // value from input arr[]
    let mx = Number.MIN_VALUE;

    // Loop for iterating on arr[]
    for (let i = 0; i < sz; i++) {

        // Updating value of min variable
        mn = arr[i] < mn ? arr[i] : mn;

        // Updating value of max variable
        mx = arr[i] > mx ? arr[i] : mx;
    }
    // Printing answer as YES/NO by
    // comparing mean with min and max
    // values
    return X >= mn && X <= mx ? "YES" : "NO";
}

// Driver Code

    let N = 5, X = 8;
    let arr = [ 1, 10, 3, 2, 6, 7, 4, 5 ];

    // Function call
    console.log(isPossible(N, X, arr));
    

// This code is contributed by ksam24000.

YES

Time Complexity: O(N)
Auxiliary Space: O(1)