# Maximize count of distinct profits possible by N transactions

• Difficulty Level : Hard
• Last Updated : 07 Jun, 2021

Given three integers N, X, and Y, representing the total number of transactions, minimum and maximum profit respectively, the task is to find the count of distinct total profits that can be earned in N ( N > 1) transactions using X and Y at least once.

Examples:

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Input: N = 3, X = 13, Y = 15
Output: 3
Explanation:
The different possible transactions satisfying the conditions are as follows:

• In first two transactions, the profit earned is 13. In the last transaction, the profit earned is 15. Therefore, the total profit earned = 13 + 13 + 15 = 41.
• In first two transactions, the profit earned is 13 and 14 respectively. In the last transaction, the profit earned is 15. Therefore, the total profit earned = 13 + 14 + 15 = 42.
• In first transaction, profit earned is 13. In the last two transactions, profit earned is 15. Therefore, the total profit = 13 + 15 + 15 = 43.

Therefore, the total distinct profits earned is 3.

Input: N = 2, X = 10, Y = 17
Output: 1

Approach: The given problem can be solved based on the following observations:

• The minimum total profit that can be earned is:
• S1 = (N-1)*X +Y.
• The maximum total profit that can be earned is:
• S2 = (N-1)*Y + X.
• Now it can be observed that all the total profit will lie within the range [S1, S2].

Follow the steps below to solve the problem:

• Print the value (S2-S1+1).

Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach` `#include ``using` `namespace` `std;` `// Function to count distinct``// profits possible``int` `numberOfWays(``int` `N, ``int` `X, ``int` `Y)``{``    ``// Stores the minimum total profit``    ``int` `S1 = (N - 1) * X + Y;` `    ``// Stores the maximum total profit``    ``int` `S2 = (N - 1) * Y + X;` `    ``// Return count of distinct profits``    ``return` `(S2 - S1 + 1);``}` `// Driver code``int` `main()``{``    ``// Input``    ``int` `N = 3;``    ``int` `X = 13;``    ``int` `Y = 15;` `    ``// Function call``    ``cout << numberOfWays(N, X, Y);``    ``return` `0;``}`

## Java

 `// Java program for the above approach` `import` `java.util.Arrays;` `class` `GFG``{` `// Function to count distinct``// profits possible``static` `int` `numberOfWays(``int` `N, ``int` `X, ``int` `Y)``{``    ``// Stores the minimum total profit``    ``int` `S1 = (N - ``1``) * X + Y;` `    ``// Stores the maximum total profit``    ``int` `S2 = (N - ``1``) * Y + X;` `    ``// Return count of distinct profits``    ``return` `(S2 - S1 + ``1``);``}`  `// Driver code``public` `static` `void` `main(String[] args)``{``    ``// Input``    ``int` `N = ``3``;``    ``int` `X = ``13``;``    ``int` `Y = ``15``;` `    ``// Function call``    ``System.out.println(numberOfWays(N, X, Y));``    ``}``}` `// This code is contributed by jana_sayantan.`

## Python3

 `# Python3 program for the above approach` `# Function to count distinct``# profits possible``def` `numberOfWays(N, X, Y):``    ` `    ``# Stores the minimum total profit``    ``S1 ``=` `(N ``-` `1``) ``*` `X ``+` `Y` `    ``# Stores the maximum total profit``    ``S2 ``=` `(N ``-` `1``) ``*` `Y ``+` `X` `    ``# Return count of distinct profits``    ``return` `(S2 ``-` `S1 ``+` `1``)` `# Driver code``if` `__name__ ``=``=` `'__main__'``:``    ` `    ``# Input``    ``N ``=` `3``    ``X ``=` `13``    ``Y ``=` `15` `    ``# Function call``    ``print``(numberOfWays(N, X, Y))``    ` `# This code is contributed by SURENDRA_GANGWAR`

## C#

 `// C# program for the above approach``using` `System;``        ` `class` `GFG``{` `// Function to count distinct``// profits possible``static` `int` `numberOfWays(``int` `N, ``int` `X, ``int` `Y)``{``    ``// Stores the minimum total profit``    ``int` `S1 = (N - 1) * X + Y;` `    ``// Stores the maximum total profit``    ``int` `S2 = (N - 1) * Y + X;` `    ``// Return count of distinct profits``    ``return` `(S2 - S1 + 1);``}``    ` `// Driver Code``public` `static` `void` `Main()``{``    ``// Input``    ``int` `N = 3;``    ``int` `X = 13;``    ``int` `Y = 15;` `    ``// Function call``    ``Console.WriteLine(numberOfWays(N, X, Y));``    ` `}``}` `// This code is contributed by code_hunt.`

## Javascript

 ``
Output:
`3`

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

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