# Find permutation with maximum remainder Sum

Given an integer N, the task is to find a permutation of the integers from 1 to N such that is maximum.

Examples:

Input: N = 3
Output: 3 1 2
Sum of the remainder values is (0 + 1 + 2) = 3
which is the maximum possible.

Input: N = 5
Output: 5 1 2 3 4

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: As it is known that the maximum value of a number X after doing the mod with Y is Y-1. The permutation that will yield the maximum sum of the mosulus values will be {N, 1, 2, 3, …., N – 1}.

After evaluating the expression on the above array the output array will be {0, 1, 2, 3, …., N – 1} and this is the maximum value that can be obtained.

Below is the implementation of the above approach:

## C++

 // C++ implementation of the approach  #include  using namespace std;     // Function to find the permutation  vector<int> Findpermutation(int n)  {      vector<int> a(n + 1);         // Put n at the first index 1      a = n;         // Put all the numbers from      // 2 to n sequentially      for (int i = 2; i <= n; i++)          a[i] = i - 1;         return a;  }     // Driver code  int main()  {      int n = 8;         vector<int> v = Findpermutation(n);         // Display the permutation      for (int i = 1; i <= n; i++)          cout << v[i] << ' ';         return 0;  }

## Java

 // Java implementation of the approach  import java.util.*;  class GFG   {     // Function to find the permutation  static int[] Findpermutation(int n)  {      int [] a = new int[n + 1];         // Put n at the first index 1      a = n;         // Put all the numbers from      // 2 to n sequentially      for (int i = 2; i <= n; i++)          a[i] = i - 1;         return a;  }     // Driver code  public static void main(String[] args)   {      int n = 8;         int []v = Findpermutation(n);         // Display the permutation      for (int i = 1; i <= n; i++)          System.out.print(v[i] + " ");  }  }      // This code is contributed by 29AjayKumar

## Python3

 # Python3 implementation of the approach      # Function to find the permutation   def Findpermutation(n) :         a =  * (n + 1);          # Put n at the first index 1       a = n;          # Put all the numbers from       # 2 to n sequentially       for i in range(2, n + 1) :          a[i] = i - 1;          return a;      # Driver code   if __name__ == "__main__" :          n = 8;          v = Findpermutation(n);          # Display the permutation       for i in range(1, n + 1) :          print(v[i], end = ' ');      # This code is contributed by AnkitRai01

## C#

 // C# implementation of the approach  using System;     class GFG   {     // Function to find the permutation  static int[] Findpermutation(int n)  {      int [] a = new int[n + 1];         // Put n at the first index 1      a = n;         // Put all the numbers from      // 2 to n sequentially      for (int i = 2; i <= n; i++)          a[i] = i - 1;         return a;  }     // Driver code  public static void Main(String[] args)   {      int n = 8;         int []v = Findpermutation(n);         // Display the permutation      for (int i = 1; i <= n; i++)          Console.Write(v[i] + " ");  }  }     // This code is contributed by 29AjayKumar

Output:

8 1 2 3 4 5 6 7


Time Complexity: O(N)

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