Python: Find the number of notes against a specified amount

Python Basic - 1: Exercise-21 with Solution

Notes Count from Amount

Write a Python program to find the number of notes (Samples of notes: 10, 20, 50, 100, 200, 500) against an amount.

Range - Number of notes(n) : n (1 ≤ n ≤ 1000000).

Visual Presentation:

Sample Solution:

Python Code:

# Define a function 'no_notes' that calculates the minimum number of notes required
# to represent a given amount 'a' using denominations of 500, 200, 100, 50, 20, and 10.
def no_notes(a):
    # List of denominations in descending order.
    Q = [500, 200, 100, 50, 20, 10]
    
    # Initialize a variable 'x' to 0.
    x = 0
    
    # Iterate over each denomination in the list.
    for i in range(6):
        q = Q[i]  # Get the current denomination.
        x += int(a / q)  # Add the count of notes for the current denomination to 'x'.
        a = int(a % q)  # Update the remaining amount after using the current denomination.
    
    # Check if there is any remaining amount not covered by the available denominations.
    if a > 0:
        x = -1  # If there is a remainder, set 'x' to -1.
    
    return x  # Return the minimum number of notes required.

# Test the 'no_notes' function with different amounts and print the results.
print(no_notes(880))
print(no_notes(1000))

Sample Output:

6
2

Explanation:

The above Python code defines a function "no_notes()" that calculates the minimum number of banknotes required to represent a given amount 'a' using denominations of 500, 200, 100, 50, 20, and 10. Here's a brief explanation:

• Function Definition:
• def no_notes(a):: Define a function named "no_notes" that takes an integer 'a' as input.
• Denomination List:
• Q = [500, 200, 100, 50, 20, 10]: Initialize a list 'Q' with denominations in descending order.
• Initialization:
• x = 0: Initialize a variable x to 0 to count the number of notes required.
• Iteration over Denominations:
• Iterate over each denomination in the list using a for loop.
• for i in range(6):: Loop through denominations from the largest to the smallest.
• q = Q[i]: Get the current denomination.
• x += int(a / q): Calculate the count of notes for the current denomination and add it to 'x'.
• a = int(a % q): Update the remaining amount after using the current denomination.
• Check Remaining Amount:
• if a > 0:: Check if there is any remaining amount not covered by the available denominations.
• x = -1: If there is a remainder, set 'x' to -1, indicating the amount cannot be represented with the available denominations.
• Return Result:
• return x: Return the minimum number of notes required to represent the amount 'a'.
• Function Testing:
• Test the no_notes function with different amounts (880 and 1000).
• Print the results of the function calls.

Flowchart:

Python Code Editor:

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