Find K missing numbers from given Array in range [1, M] such that total average is X

Given an array arr[] of integers of size N where each element can be in the range [1, M], and two integers X and K. The task is to find K possible numbers in range [1, M] such that the average of all the (N + K) numbers is equal to X. If there are multiple valid answers, any one is acceptable.

Examples:

Input: arr[] = {3, 2, 4, 3}, M = 6, K = 2, X = 4
Output: 6 6
Explanation: The mean of all elements is (3 + 2 + 4 + 3 + 6 + 6) / 6 = 4.

Input: arr[] = {1}, M = 8, K = 1, X = 4
Output: 7
Explanation: The mean of all elements is (1 + 7) / 2 = 4.

Input: arr[] = {1, 2, 3, 4}, M = 6, K = 4, X = 6
Output: []
Explanation: It is impossible for the mean to be 6 no matter what the 4 missing elements are.

Approach: The approach is based on the following mathematical observation. If the total expected sum after adding M elements is such that the sum of the M elements added needs to be less than M or greater than K*M then no solution is possible. Otherwise, there is always a possible solution.

1. Find the sum of missing elements(Y), that is = X*(K + N) – sum(arr).
2. If this is less than K or greater than K*M then the array cannot be created. So return an empty array.
3. Otherwise, try to equally divide the value Y in K elements, i.e assign all the K elements with value Y/K.
4. if still some value remains to be assigned then after assigning every element equally, the value left will be = (Y%K). So add 1 to (Y%K) elements of the new array.

Below is the implementation of the above approach:

C++

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

// Function to get the missing elements
vector<int> missing(vector<int>& arr,
                    int M, int X, int K)
{
    int N = arr.size(),
        sum = accumulate(arr.begin(),
                         arr.end(), 0),
        newsum = 0;
    newsum = X * (K + N) - sum;
    // If this newsum is less than M
    // or greater than K*M then
    // no array can be created.
    if (newsum < K || newsum > K * M)
        return {};

    int mod = newsum % K;
    vector<int> ans(K, newsum / K);
    for (int i = 0; i < mod; i++)
        ans[i] += 1;
    return ans;
}

// Driver code
int main()
{
    vector<int> arr{ 3, 2, 4, 3 };
    int X = 4;
    int K = 2;
    int M = 6;

    // Vector to store resultant list
    vector<int> ans = missing(arr, M, X, K);

    for (auto i : ans)
        cout << i << " ";
    return 0;
}

Java

// Java code to implement above approach
import java.util.*;
class GFG{

// Function to get the missing elements
static  int []missing(int []arr,
                    int M, int X, int K)
{
    int N = arr.length,
        sum = accumulate(arr,0,N),
        newsum = 0;
    newsum = X * (K + N) - sum;

    // If this newsum is less than M
    // or greater than K*M then
    // no array can be created.
    if (newsum < K || newsum > K * M)
        return new int[]{};

    int mod = newsum % K;
    int []ans = new int[K];
    Arrays.fill(ans, newsum / K);
    for (int i = 0; i < mod; i++)
        ans[i] += 1;
    return ans;
}
static int accumulate(int[] arr, int start, int end){
    int sum=0;
    for(int i= 0; i < arr.length; i++)
        sum+=arr[i];
    return sum;
}

// Driver code
public static void main(String[] args)
{
    int[]arr = { 3, 2, 4, 3 };
    int X = 4;
    int K = 2;
    int M = 6;

    // Vector to store resultant list
    int []ans = missing(arr, M, X, K);

    for (int i : ans)
        System.out.print(i+ " ");
}
}

// This code is contributed by shikhasingrajput

Python3

# Python code for the above approach

# Function to get the missing elements
def missing(arr, M, X, K):
    N = len(arr)
    sum = 0
    for i in range(len(arr)):
        sum += arr[i]
    newsum = 0
    newsum = X * (K + N) - sum

    # If this newsum is less than M
    # or greater than K*M then
    # no array can be created.
    if (newsum < K or newsum > K * M):
        return []

    mod = newsum % K
    ans = [newsum // K] * K

    for i in range(mod):
        ans[i] += 1
    return ans

# Driver code
arr = [3, 2, 4, 3]
X = 4
K = 2
M = 6

# Vector to store resultant list
ans = missing(arr, M, X, K)

for i in ans:
    print(i, end=" ")

# This code is contributed by gfgking

C#

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

class GFG {

    // Function to get the missing elements
    static int[] missing(int[] arr, int M, int X, int K)
    {
        int N = arr.Length, sum = accumulate(arr, 0, N),
            newsum = 0;
        newsum = X * (K + N) - sum;

        // If this newsum is less than M
        // or greater than K*M then
        // no array can be created.
        if (newsum < K || newsum > K * M)
            return new int[] {};

        int mod = newsum % K;
        int[] ans = new int[K];
        Array.Fill(ans, newsum / K);
        for (int i = 0; i < mod; i++)
            ans[i] += 1;
        return ans;
    }
    static int accumulate(int[] arr, int start, int end)
    {
        int sum = 0;
        for (int i = 0; i < arr.Length; i++)
            sum += arr[i];
        return sum;
    }

    // Driver code
    public static void Main(string[] args)
    {
        int[] arr = { 3, 2, 4, 3 };
        int X = 4;
        int K = 2;
        int M = 6;

        // Vector to store resultant list
        int[] ans = missing(arr, M, X, K);

        foreach(int i in ans) Console.Write(i + " ");
    }
}

// This code is contributed by ukasp.

Javascript

  <script>
      // JavaScript code for the above approach

      // Function to get the missing elements
      function missing(arr,
          M, X, K) {
          let N = arr.length;
          let sum = 0;
          for (let i = 0; i < arr.length; i++) {
              sum += arr[i];
          }
          newsum = 0;
          newsum = X * (K + N) - sum;

          // If this newsum is less than M
          // or greater than K*M then
          // no array can be created.
          if (newsum < K || newsum > K * M)
              return [];

          let mod = newsum % K;

          let ans = new Array(K).fill(Math.floor(newsum / K))

          for (let i = 0; i < mod; i++)
              ans[i] += 1;
          return ans;
      }

      // Driver code
      let arr = [3, 2, 4, 3];
      let X = 4;
      let K = 2;
      let M = 6;

      // Vector to store resultant list
      let ans = missing(arr, M, X, K);

      for (let i of ans)
          document.write(i + " ");

// This code is contributed by Potta Lokesh
  </script>

Output

6 6 

Time Complexity: O(N)
Auxiliary Space: O(1) When space of the resultant list is not considered