Given two integers N, K, and an array arr[] consisting of positive integers. The task is to find any possible array of size N such that the mean of arr[] and the array to be found together is K.
Examples:
Input: arr[] = {1, 5, 6}, N = 4, K = 3
Output: {1, 2, 3, 3}
Explanation: Mean of {1, 5, 6} and {1, 2, 3, 3} together is the following:
(1 + 5 + 6 + 1 + 2 + 3 + 3) / (3+4) = 3.
Therefore {1, 2, 3, 3} is possible array to make the mean = 3.Input: arr[] = {7, 8, 6}, N = 2, K = 3
Output: Not Possible
Approach: This problem can be solved by using simple Number System Average concepts. Let’s say sumArr be the sum of array new_arr[] and M to be the size of new_arr[]. Then according to average concept the equation can be formed as:
(sumArr + X) / (N+M) = K, where X is sum of required array.
X = K * (N + M) – sumArr.
Therefore, sum of the required array should be X to make the mean equal to K. If X is less than N, then there is no way to form the array.
The simplest way to form the required array is
1, 1, 1, …(M-1 times), X-(M-1)
.
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Find the required array such that mean of both// the arrays is equal to Kvector<int> findArray(vector<int> arr, int N, int K){ // To store sum of elements in arr[] int sumArr = 0; int M = int(arr.size()); // Iterate to find sum for (int i = 0; i < M; i++) { sumArr += arr[i]; } // According to the formula derived above int X = K * (N + M) - sumArr; // If requiredSum if less than N if (X < N) { cout << "Not Possible"; return {}; } // Otherwise create an array to store answer vector<int> res(N); // Putting all 1s till N-1 for (int i = 0; i < N - 1; i++) { res[i] = 1; } res[N - 1] = X - (N - 1); // Return res as the final result return res;}// Driver Codeint main(){ vector<int> arr = { 1, 5, 6 }; int N = 4, K = 3; vector<int> ans = findArray(arr, N, K); for (int i = 0; i < ans.size(); i++) cout << ans[i] << " ";} |
Java
// Java program for the above approachpublic class GFG { // Find the required array such that mean of both // the arrays is equal to K static void findArray(int []arr, int N, int K) { // To store sum of elements in arr[] int sumArr = 0; int M = arr.length; // Iterate to find sum for (int i = 0; i < M; i++) { sumArr += arr[i]; } // According to the formula derived above int X = K * (N + M) - sumArr; // If requiredSum if less than N if (X < N) { System.out.println("Not Possible"); } // Otherwise create an array to store answer int []res = new int[N]; // Putting all 1s till N-1 for (int i = 0; i < N - 1; i++) { res[i] = 1; } res[N - 1] = X - (N - 1); // Return res as the final result for (int i = 0; i < res.length; i++) System.out.print(res[i] + " "); } // Driver Code public static void main (String[] args) { int []arr = { 1, 5, 6 }; int N = 4, K = 3; findArray(arr, N, K); }}// This code is contributed by AnkThon |
Python3
# python program for the above approach# Find the required array such that mean of both# the arrays is equal to Kdef findArray(arr, N, K): # To store sum of elements in arr[] sumArr = 0 M = len(arr) # Iterate to find sum for i in range(0, M): sumArr += arr[i] # According to the formula derived above X = K * (N + M) - sumArr # If requiredSum if less than N if (X < N): print("Not Possible") return [] # Otherwise create an array to store answer res = [0 for _ in range(N)] # Putting all 1s till N-1 for i in range(0, N-1): res[i] = 1 res[N - 1] = X - (N - 1) # Return res as the final result return res# Driver Codeif __name__ == "__main__": arr = [1, 5, 6] N = 4 K = 3 ans = findArray(arr, N, K) for i in range(0, len(ans)): print(ans[i], end=" ") # This code is contributed by rakeshsahni |
C#
// C# program for the above approachusing System;public class GFG { // Find the required array such that mean of both // the arrays is equal to K static void findArray(int []arr, int N, int K) { // To store sum of elements in arr[] int sumArr = 0; int M = arr.Length; // Iterate to find sum for (int i = 0; i < M; i++) { sumArr += arr[i]; } // According to the formula derived above int X = K * (N + M) - sumArr; // If requiredSum if less than N if (X < N) { Console.WriteLine("Not Possible"); } // Otherwise create an array to store answer int []res = new int[N]; // Putting all 1s till N-1 for (int i = 0; i < N - 1; i++) { res[i] = 1; } res[N - 1] = X - (N - 1); // Return res as the final result for (int i = 0; i < res.Length; i++) Console.Write(res[i] + " "); } // Driver Code public static void Main (string[] args) { int []arr = { 1, 5, 6 }; int N = 4, K = 3; findArray(arr, N, K); }}// This code is contributed by AnkThon |
Javascript
<script>// Javascript program for the above approach// Find the required array such that mean of both// the arrays is equal to Kfunction findArray(arr, N, K) { // To store sum of elements in arr[] let sumArr = 0; let M = (arr.length); // Iterate to find sum for (let i = 0; i < M; i++) { sumArr += arr[i]; } // According to the formula derived above let X = K * (N + M) - sumArr; // If requiredSum if less than N if (X < N) { document.write("Not Possible"); return []; } // Otherwise create an array to store answer let res = new Array(N); // Putting all 1s till N-1 for (let i = 0; i < N - 1; i++) { res[i] = 1; } res[N - 1] = X - (N - 1); // Return res as the final result return res;}// Driver Codelet arr = [1, 5, 6];let N = 4, K = 3;let ans = findArray(arr, N, K);for (let i = 0; i < ans.length; i++) document.write(ans[i] + " "); // This code is contributed by gfgking.</script> |
Output
1 1 1 6
Time Complexity: O(N)
Auxiliary Space: O(N)
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