Count of ways to distribute N items among 3 people with one person receiving maximum

Given an integer N, the task is to find the total number of ways to distribute N among 3 people such that:

  • Exactly one person gets the maximum number of items among all the 3 people.
  • Each person gets at least 1 item.

Examples:

Input: N = 5
Output: 3
Explanation:
3 way distribute the item among 3 people are {1, 1, 3}, {1, 3, 1} and {3, 1, 1}.
Distributions like {1, 2, 2} or {2, 1, 2} are not valid as two persons are getting the maximum.

Input: N = 10
Output: 33
Explanation:
For the Input N = 10 there are 33 ways of distribution.

Approach:
To solve the problem mentioned above, we have to observe that if N < 4, then such a distribution is not possible.
For all values of N ≥ 4, follow the steps to solve the problem:



  • Total no of ways to distribute N items among 3 people is given by (N – 1) * (N – 2) / 2.
  • Initialize a variable s = 0 which stores the count of ways the distribution is not possible.
  • Iterate two nested loops, where i ranges between [2, N – 3] and j ranging upto i:
    • For each iteration, check if N = 2 * i + j, that is 2 persons can receive the maximum number of elements
    • If so, then increment s by 1. If N is divisible by 3, update s by 3 * s + 1. Otherwise, update to 3 * s.
  • Finally, return ans – s as the total number of ways to distribute N items among three people.

Below is the implementation of the above approach:

C++

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// C++ program to find the number
// of ways to distribute N item
// among three people such
// that one person always gets
// the maximum value
  
#include <bits/stdc++.h>
using namespace std;
  
// Function to find the number
// of ways to distribute N
// items among 3 people
int countWays(int N)
{
    // No distribution
    // possible
    if (N < 4)
        return 0;
  
    // Total number of ways to
    // distribute N items
    // among 3 people
    int ans = ((N - 1) * (N
                          - 2))
              / 2;
  
    // Store the number of
    // distributions which
    // are not possible
    int s = 0;
  
    for (int i = 2; i <= N - 3;
         i++) {
        for (int j = 1; j < i;
             j++) {
  
            // Count possiblities
            // of two persons
            // receiving the
            // maximum
            if (N == 2 * i + j)
                s++;
        }
    }
  
    // If N is divisible by 3
    if (N % 3 == 0)
        s = 3 * s + 1;
  
    else
        s = 3 * s;
  
    // Return the final
    // count of ways
    // to distribute
    return ans - s;
}
  
// Driver Code
int main()
{
    int N = 10;
    cout << countWays(N);
    return 0;
}

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Java

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// Java program to find the number
// of ways to distribute N item
// among three people such
// that one person always gets
// the maximum value
class GFG{
  
// Function to find the number
// of ways to distribute N
// items among 3 people
static int countWays(int N)
{
    // No distribution
    // possible
    if (N < 4)
        return 0;
  
    // Total number of ways to
    // distribute N items
    // among 3 people
    int ans = ((N - 1) * (N - 2)) / 2;
  
    // Store the number of
    // distributions which
    // are not possible
    int s = 0;
  
    for (int i = 2; i <= N - 3; i++) 
    {
        for (int j = 1; j < i; j++)
        {
  
            // Count possiblities
            // of two persons
            // receiving the
            // maximum
            if (N == 2 * i + j)
                s++;
        }
    }
  
    // If N is divisible by 3
    if (N % 3 == 0)
        s = 3 * s + 1;
    else
        s = 3 * s;
  
    // Return the final
    // count of ways
    // to distribute
    return ans - s;
}
  
// Driver Code
public static void main(String[] args)
{
    int N = 10;
    System.out.println(countWays(N));
}
}
  
// This code is contributed by rock_cool

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Python3

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# Python3 program to find the number
# of ways to distribute N item
# among three people such
# that one person always gets
# the maximum value
  
# Function to find the number
# of ways to distribute N
# items among 3 people
def countWays(N):
      
    # No distribution
    # possible
    if (N < 4):
        return 0
  
    # Total number of ways to
    # distribute N items
    # among 3 people
    ans = ((N - 1) * (N - 2)) // 2
  
    # Store the number of
    # distributions which
    # are not possible
    s = 0
  
    for i in range( 2, N - 2, 1):
        for j in range( 1, i, 1):
              
            # Count possiblities
            # of two persons
            # receiving the
            # maximum
            if (N == 2 * i + j):
                s += 1
  
    # If N is divisible by 3
    if (N % 3 == 0):
        s = 3 * s + 1
  
    else:
        s = 3 * s
  
    # Return the final
    # count of ways
    # to distribute
    return ans - s
  
# Driver Code
N = 10
  
print (countWays(N))
      
# This code is contributed by sanjoy_62

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C#

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// C# program to find the number
// of ways to distribute N item
// among three people such
// that one person always gets
// the maximum value
using System;
class GFG{
  
// Function to find the number
// of ways to distribute N
// items among 3 people
static int countWays(int N)
{
    // No distribution
    // possible
    if (N < 4)
        return 0;
  
    // Total number of ways to
    // distribute N items
    // among 3 people
    int ans = ((N - 1) * (N - 2)) / 2;
  
    // Store the number of
    // distributions which
    // are not possible
    int s = 0;
  
    for (int i = 2; i <= N - 3; i++) 
    {
        for (int j = 1; j < i; j++)
        {
  
            // Count possiblities
            // of two persons
            // receiving the
            // maximum
            if (N == 2 * i + j)
                s++;
        }
    }
  
    // If N is divisible by 3
    if (N % 3 == 0)
        s = 3 * s + 1;
    else
        s = 3 * s;
  
    // Return the final
    // count of ways
    // to distribute
    return ans - s;
}
  
// Driver Code
public static void Main()
{
    int N = 10;
    Console.Write(countWays(N));
}
}
  
// This code is contributed by Code_Mech

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Output:

33

Time complexity: O(N2)
Auxiliary Space: O(1)

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Improved By : rock_cool, Code_Mech, sanjoy_62