Python Program to Find Grade from Percentage

Use a Python program with if-elif-else statements to check the percentage ranges and print the corresponding grade, for example: if percentage >= 90: grade = 'A'.

Examples

Input95

OutputGrade: A

Input76

OutputGrade: C

Input45

OutputGrade: F

How to Think About It

To find the grade from a percentage, compare the percentage value against fixed ranges using if and elif conditions. Assign a grade based on which range the percentage falls into, such as 90 and above for 'A', 80 to 89 for 'B', and so on.

Algorithm

1

Get the percentage input from the user.

2

Check if the percentage is 90 or above, assign grade 'A'.

3

Else if percentage is 80 or above, assign grade 'B'.

4

Else if percentage is 70 or above, assign grade 'C'.

5

Else if percentage is 60 or above, assign grade 'D'.

6

Else assign grade 'F'.

7

Print the grade.

Code

python

percentage = float(input('Enter percentage: '))
if percentage >= 90:
    grade = 'A'
elif percentage >= 80:
    grade = 'B'
elif percentage >= 70:
    grade = 'C'
elif percentage >= 60:
    grade = 'D'
else:
    grade = 'F'
print('Grade:', grade)

Output

Enter percentage: 85
Grade: B

Dry Run

Let's trace the input 85 through the code

1

Input percentage

percentage = 85.0

2

Check if percentage >= 90

85.0 >= 90 is False

3

Check if percentage >= 80

85.0 >= 80 is True, so grade = 'B'

4

Print grade

Output: Grade: B

Step	Condition	Result	Grade
1	Input percentage	85.0
2	percentage >= 90	False
3	percentage >= 80	True	B
4	Print grade		B

Why This Works

Step 1: Input percentage

We get the percentage value from the user as a number to check.

Step 2: Check ranges

Using if-elif, we compare the percentage to fixed ranges to find the correct grade.

Step 3: Assign grade

Once the correct range is found, we assign the corresponding grade letter.

Step 4: Output grade

Finally, we print the grade so the user can see the result.

Alternative Approaches

Using dictionary with thresholds

python

percentage = float(input('Enter percentage: '))
grades = {90: 'A', 80: 'B', 70: 'C', 60: 'D', 0: 'F'}
for threshold in sorted(grades.keys(), reverse=True):
    if percentage >= threshold:
        grade = grades[threshold]
        break
print('Grade:', grade)

This method uses a dictionary and a loop to find the grade, making it easier to update thresholds.

Using function with return

python

def find_grade(p):
    if p >= 90:
        return 'A'
    elif p >= 80:
        return 'B'
    elif p >= 70:
        return 'C'
    elif p >= 60:
        return 'D'
    else:
        return 'F'
percentage = float(input('Enter percentage: '))
grade = find_grade(percentage)
print('Grade:', grade)

This approach wraps the logic in a function for reuse and cleaner code.

Complexity: O(1) time, O(1) space

Time Complexity

The program uses a fixed number of comparisons regardless of input size, so it runs in constant time O(1).

Space Complexity

Only a few variables are used to store input and grade, so space complexity is O(1).

Which Approach is Fastest?

All approaches run in constant time; using if-elif is simplest, dictionary method is more flexible but slightly more complex.

Approach	Time	Space	Best For
If-elif-else	O(1)	O(1)	Simple and clear grading logic
Dictionary with loop	O(1)	O(n) where n is number of thresholds	Easily update or add grade thresholds
Function with return	O(1)	O(1)	Reusable and clean code structure