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Maximum possible number in a Sequence of distinct elements with given Average

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  • Last Updated : 21 Feb, 2022

Given integers N and K, the task is to find the maximum possible number an N length sequence of distinct positive integers can have if the average of the elements of the sequence is K.

Examples:

Input: N = 4, K = 3
Output: 6
Explanation: The desired sequence would be – {1, 2, 3, 6}. 
There could be other sequences as well satisfying the given values of N and K, such as {1, 2, 4, 5}.
But , since the maximum possible number in the sequence is needed, 6 would be the answer.

Input: N = 5, K = 4
Output: 10
Explanation: The desired sequence would be – {1, 2, 3, 4, 10}.

Input: N = 5, K = 2
Output: -1
Explanation: Forming a sequence having 5 distinct positive integer and average as 2 is not possible.

 

Approach: The solution to the problem is based on the following observation:

  • From the number of integers in the sequence and the average of the sequence, the total sum of all numbers of the sequence can be easily calculated by using the following formula: 
    sum of all numbers in sequence = (number of terms) x (average of all terms)
  • Now , suppose the maximum term (or number) is M and sum of the sequence is S. Then, to maximize M, (S-M) must be minimized.
  • Since , the terms must be distinct  and positive, so the series with minimum possible sum would be 1, 2, 3 . . . (N-1).
    Sum of such a sequence of natural numbers can be easily calculated for (N-1) terms , and would be: (N * (N – 1)) /  2.
  • So , the maximum possible integer would be: M = sum of the given sequence – sum of first (N-1) natural numbers.

Follow the steps mentioned below:

  • Calculate the sum of the given sequence.
  • Find the sum of first (N-1) natural numbers.
  • Calculate the maximum possible number from the above observation.
  • If the maximum value is less than N then no such sequence is possible.
  • Otherwise, the maximum value is the required answer.

Below is the implementation of the above approach: 

C++




// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to calculate the
// maximum possible number in a sequence
// with given average and number of terms.
int maxNumber(int N, int K)
{
    // Sum of the sequence
    int sum = N * K;
 
    // Minimum possible sum of a sequence
    // having N-1 distinct positive integers
    int minSum = N * (N - 1) / 2;
 
    // Maximum number of the given sequence
    int maxNum = sum - minSum;
 
    // If such sequence is not possible
    if (maxNum < N)
        return -1;
   
    return maxNum;
}
 
// Driver Code
int main()
{
    int N = 5, K = 4;
    int maximum_number = maxNumber(N, K);
    cout << maximum_number;
    return 0;
}


Java




// Java program for the above approach
import java.util.*;
public class GFG
{
     
// Function to calculate the
// maximum possible number in a sequence
// with given average and number of terms.
static int maxNumber(int N, int K)
{
   
    // Sum of the sequence
    int sum = N * K;
 
    // Minimum possible sum of a sequence
    // having N-1 distinct positive integers
    int minSum = N * (N - 1) / 2;
 
    // Maximum number of the given sequence
    int maxNum = sum - minSum;
 
    // If such sequence is not possible
    if (maxNum < N)
        return -1;
   
    return maxNum;
}
 
// Driver Code
public static void main(String args[])
{
    int N = 5, K = 4;
    int maximum_number = maxNumber(N, K);
    System.out.println(maximum_number);
}
}
 
// This code is contributed by Samim Hossain Mondal.


Python3




# Python code for the above approach
# Function to calculate the
# maximum possible number in a sequence
# with given average and number of terms.
def maxNumber(N, K):
 
    # Sum of the sequence
    sum = N * K;
 
    # Minimum possible sum of a sequence
    # having N-1 distinct positive integers
    minSum = (N * (N - 1) // 2);
 
    # Maximum number of the given sequence
    maxNum = sum - minSum;
 
    # If such sequence is not possible
    if (maxNum < N):
        return -1;
 
    return maxNum;
 
# Driver Code
N = 5
K = 4;
maximum_number = maxNumber(N, K);
print(maximum_number);
 
# This code is contributed by gfgking


C#




using System;
 
public class GFG{
 
  // Function to calculate the
  // maximum possible number in a sequence
  // with given average and number of terms.
  static int maxNumber(int N, int K)
  {
    // Sum of the sequence
    int sum = N * K;
 
    // Minimum possible sum of a sequence
    // having N-1 distinct positive integers
    int minSum = N * (N - 1) / 2;
 
    // Maximum number of the given sequence
    int maxNum = sum - minSum;
 
    // If such sequence is not possible
    if (maxNum < N)
      return -1;
 
    return maxNum;
  }
 
  // Driver Code
  static public void Main (){
 
    int N = 5, K = 4;
    int maximum_number = maxNumber(N, K);
    Console.Write(maximum_number);
  }
}
 
// This code is contributed by hrithikgarg03188


Javascript




<script>
       // JavaScript code for the above approach
       // Function to calculate the
       // maximum possible number in a sequence
       // with given average and number of terms.
       function maxNumber(N, K)
       {
        
           // Sum of the sequence
           let sum = N * K;
 
           // Minimum possible sum of a sequence
           // having N-1 distinct positive integers
           let minSum = Math.floor(N * (N - 1) / 2);
 
           // Maximum number of the given sequence
           let maxNum = sum - minSum;
 
           // If such sequence is not possible
           if (maxNum < N)
               return -1;
 
           return maxNum;
       }
 
       // Driver Code
       let N = 5, K = 4;
       let maximum_number = maxNumber(N, K);
       document.write(maximum_number);
 
      // This code is contributed by Potta Lokesh
   </script>


 
 

Output

10

 

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

 


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