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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 <bits/stdc++.h>
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




<script>
 
// JavaScript program for the above approach
 
// Function to count distinct
// profits possible
function numberOfWays( N, X, Y)
{
    // Stores the minimum total profit
    let S1 = (N - 1) * X + Y;
 
    // Stores the maximum total profit
    let S2 = (N - 1) * Y + X;
 
    // Return count of distinct profits
    return (S2 - S1 + 1);
}
 
// Driver code
    // Input
    let N = 3;
    let X = 13;
    let Y = 15;
 
    // Function call
    document.write(numberOfWays(N, X, Y));
     
// This code is contributed by shivanisinghss2110
 
</script>
Output: 
3

 

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

 




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