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Find the remaining balance after the transaction

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Given the initial balance as bal and the amount X to be debited, where X must be a multiple of 10 and rupees 1.50 is deducted as the debit charge for each successful debit. The task is to find the remaining balance left after the transaction, which can be successful, or unsuccessful. The balances are in 2 floating-point precision. 

Examples: 

Input: X = 50, bal = 100.50
Output: 49.00
Transaction successful

Input: X = 55, bal = 99.00
Output: 99.00
Transaction unsuccessful

Approach: Find out if the transaction can be successful or not. 

  • The transaction can be successful if: 
    • X is a multiple of 10, and
    • The person has at least (X+1.50) rupees, that is the money to be withdrawn plus the charges, in the account.
  • In any other case, the transaction will be unsuccessful.
  • If the transaction is successful, then deduct the (X + 1.50) amount from the balance and return it
  • Else just return the balance.

Below is the implementation of the above approach: 

C++




// C++ program to find the remaining balance
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the balance
void findBalance(int x, float bal)
{
 
    // Check if the transaction
    // can be successful or not
    if (x % 10 == 0
        && ((float)x + 1.50) <= bal) {
 
        // Transaction is successful
        cout << fixed << setprecision(2)
             << (bal - x - 1.50) << endl;
    }
    else {
 
        // Transaction is unsuccessful
        cout << fixed << setprecision(2)
             << (bal) << endl;
    }
}
 
int main()
{
 
    int x = 50;
    float bal = 100.50;
 
    findBalance(x, bal);
 
    return 0;
}


Java




// Java program to find the remaining balance
import java.util.*;
 
class GFG
{
 
// Function to find the balance
static void findBalance(int x, float bal)
{
 
    // Check if the transaction
    // can be successful or not
    if (x % 10 == 0 && ((float)x + 1.50) <= bal)
    {
 
        // Transaction is successful
        System.out.printf("%.2f\n", bal - x - 1.50);
    }
    else
    {
 
        // Transaction is unsuccessful
        System.out.printf("%.2f\n", bal);
    }
}
 
// Driver Code
public static void main(String[] args)
{
    int x = 50;
    float bal = (float) 100.50;
 
    findBalance(x, bal);
}
}
 
// This code is contributed by Princi Singh


Python3




# Python3 program to find the remaining balance
 
# Function to find the balance
def findBalance(x,bal):
 
    # Check if the transaction
    # can be successful or not
    if (x % 10 == 0 and (x + 1.50) <= bal):
 
        #Transaction is successful
        print(round(bal - x - 1.50, 2))
    else:
 
        # Transaction is unsuccessful
        print(round(bal, 2))
 
# Driver Code
x = 50
bal = 100.50
 
findBalance(x, bal)
 
# This code is contributed by Mohit Kumar


C#




// C# program to find the remaining balance
using System;
 
class GFG
{
 
// Function to find the balance
static void findBalance(int x, float bal)
{
 
    // Check if the transaction
    // can be successful or not
    if (x % 10 == 0 && ((float)x + 1.50) <= bal)
    {
 
        // Transaction is successful
        Console.Write("{0:F2}\n", bal - x - 1.50);
    }
    else
    {
 
        // Transaction is unsuccessful
        Console.Write("{0:F2}\n", bal);
    }
}
 
// Driver Code
public static void Main(String[] args)
{
    int x = 50;
    float bal = (float) 100.50;
 
    findBalance(x, bal);
}
}
 
// This code is contributed by PrinciRaj1992


Javascript




<script>
 
 
// JavaScript program to find the remaining balance
 
// Function to find the balance
function findBalance(x, bal)
{
 
    // Check if the transaction
    // can be successful or not
    if (x % 10 == 0
        && (x + 1.50) <= bal) {
 
        // Transaction is successful
        document.write( (bal - x - 1.50).toFixed(2));
    }
    else {
 
        // Transaction is unsuccessful
        document.write( (bal).toFixed(2));
    }
}
 
var x = 50;
var bal = 100.50;
findBalance(x, bal);
 
 
</script>


Output: 

49.00

 

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



Last Updated : 03 Jan, 2023
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