Generate N sized Array with mean K and minimum difference between min and max

Last Updated : 11 May, 2022

Given two integers N and X, the task is to find an output array arr[] containing distinct integers of length N such that their average is K and the difference between the minimum and the maximum is minimum possible.

Input: N = 4, X = 8
Output :- 6 7 9 10
Explanation: Clearly the mean of 6, 7, 9, 10 is 8 and
The difference between the max and the min is (10 - 6) = 4.

Input: N = 5, X = 15
Output: 13 14 15 16 17

Approach: The problem can be solved based on the following mathematical observation:

For the difference between max and min to be the minimum, the gap between the elements in sorted order must be minimum.
For that every element when in sorted order should have minimum difference from the average of array.
So half of the elements should be in the left of K and half in right and they should have difference = 1 between them
If the value of N is odd, then K can be present in the array.
If the value of N is even, then K cannot be a part, (because then elements in left of K and in right of K will not be the same). The two middle elements will be K-1 and K+1 and all elements in the left part and right part have adjacent difference = 1.

Follow the steps below to solve this problem based on the above idea:

Check the size of output array (N) is even or odd
- If even, then print all the numbers from (M-N/2 to M+N/2) except M itself.
- Otherwise, print all the numbers from (M-N/2 to M+N/2) including M.

Below is the implementation of the above approach :

C++

// C++ code to implement the approach  #include <bits/stdc++.h> using namespace std;  // Function to find the output array // of length N whose mean is equal to X void findArray(int n, int x) {     int p, q, l;      // If array size to be constructed is odd.     if (n % 2 != 0) {         l = n / 2;         p = x - l;         q = l + x;         for (int i = p; i <= q; i++)             cout << i << " ";     }      // If array size to be constructed is even     else {         l = n / 2;         p = x - l;         q = x + l;         for (int i = p; i <= q; i++) {             if (i != x)                 cout << i << " ";         }     } }  // Driver code int main() {     int N = 4, X = 8;      // Function call     findArray(N, X);     return 0; }

Java

// Java code to implement the approach import java.io.*;  class GFG  {    // Function to find the output array   // of length N whose mean is equal to X   public static void findArray(int n, int x)   {     int p = 0, q = 0, l = 0;      // If array size to be constructed is odd.     if (n % 2 != 0) {       l = n / 2;       p = x - l;       q = l + x;       for (int i = p; i <= q; i++)         System.out.print(i + " ");     }      // If array size to be constructed is even     else {       l = n / 2;       p = x - l;       q = x + l;       for (int i = p; i <= q; i++) {         if (i != x)           System.out.print(i + " ");       }     }   }    // Driver Code   public static void main(String[] args)   {     int N = 4, X = 8;      // Function call     findArray(N, X);   } }  // This code is contributed by Rohit Pradhan

Python3

# Python3 code to implement the approach  # Function to find the output array # of length N whose mean is equal to X def findArray(n,x):        # If array size to be constructed is odd.     a=n%2     if (a is not 0):         l = int(n / 2)         p = int(x - l)         q = int(l + x)         for i in range(p,q+1):             print(i,"",end='')                  # If array size to be constructed is even     else:         l = int(n / 2)         p = int(x - l)         q = int(x + l)         for i in range(p,q+1):             if (i is not x):                 print(i,"",end='') # Driver code N = 4 X = 8  # Function call findArray(N,X)  # This code is contributed by ashishsingh13122000.

// C# program to implement // the above approach using System; class GFG {    // Function to find the output array   // of length N whose mean is equal to X   public static void findArray(int n, int x)   {     int p = 0, q = 0, l = 0;      // If array size to be constructed is odd.     if (n % 2 != 0) {       l = n / 2;       p = x - l;       q = l + x;       for (int i = p; i <= q; i++)         Console.Write(i + " ");     }      // If array size to be constructed is even     else {       l = n / 2;       p = x - l;       q = x + l;       for (int i = p; i <= q; i++) {         if (i != x)           Console.Write(i + " ");       }     }   }  // Driver Code public static void Main() {     int N = 4, X = 8;      // Function call     findArray(N, X); } }  // This code is contributed by code_hunt.

JavaScript

<script>         // Javascript code to implement the approach                  // Function to find the output array         // of length N whose mean is equal to X         function findArray(n, x){             let p;             let q;             let l;                      // If array size to be constructed is odd.             if (n % 2 !== 0) {                 l = n / 2;                 p = x - l;                 q = l + x;                 for (let i = p; i <= q; i++)                     document.write(i+" ");             }                      // If array size to be constructed is even             else {                 l = n / 2;                 p = x - l;                 q = x + l;                 for (let i = p; i <= q; i++) {                     if (i !== x)                         document.write(i+" ");                 }             }         }                  // Driver code         let N = 4;         let X = 8;                  // Function call         findArray(N, X);                  // This code is contributed by ashishsingh13122000.         </script>

Output

6 7 9 10

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

Min difference between maximum and minimum element in all Y size subarrays

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Generate N sized Array with mean K and minimum difference between min and max

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