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.
- Find the sum of missing elements(Y), that is = X*(K + N) - sum(arr).
- If this is less than K or greater than K*M then the array cannot be created. So return an empty array.
- Otherwise, try to equally divide the value Y in K elements, i.e assign all the K elements with value Y/K.
- 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++ 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 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
# 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# 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.
<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