The Grading Logic Hackerran Problem Explained
The Grading Students Hackerrank Solution is a beginner-level coding challenge that asks you to convert numerical test scores into letter grades based on a defined threshold system. It sounds simple because it is simple, but there are enough edge cases in the rounding rule that people still get this wrong on their first attempt. Here is how the grading logic actually works. If a score is less than 38, you leave it as-is. If the score is 38 or higher, you check the difference between the score and the next multiple of 5. If that difference is less than 3, you round up to the nearest multiple of 5. Otherwise, you keep the original score. That last rule exists to prevent someone with a 37 from getting a D instead of an F, which is a specific policy decision by the problem setter.
How to Implement the Grading Students Hackerrank Solution in Python
I will walk through the code directly because that is what most people searching for the Grading Students Hackerrank Solution actually need right now. Start with a function that takes an integer n. Check if n is less than 38. If it is, return n immediately. If not, calculate the remainder of n divided by 5. Subtract that remainder from n to get the current lower multiple, then add 5 to get the next multiple. Compare the difference between n and that next multiple. If it is less than 3, return the next multiple. Otherwise, return n. Here is the full function structure:
def gradingStudents(n): if n < 38: return n
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remainder = n % 5 if 5 - remainder < 3: return n + (5 - remainder)
return n Then you handle the input loop. Read an integer for the number of test cases, loop through each one, apply the function, and print the result. The HackerRank template usually provides a stub where you just need to fill in the function body. Copy-pasting the function above into the provided template is all that is required to pass every hidden test case.
Common Pitfalls People Miss
Most beginners write the rounding logic backward. They subtract from the next multiple instead of adding to the current score. The difference is subtle in code but fatal in results. A score of 73 should stay 73 because the next multiple of 5 is 75 and the gap is 2, which is less than 3, so it rounds up to 75. But a score of 74 also rounds to 75. A score of 76 stays 76 because the gap to 80 is 4, which is not less than 3. I remember spending about twenty minutes debugging a submission once because I had written the condition as n % 5 >= 3 instead of 5 - n % 5 < 3. Both look equivalent at a glance. They are not. When n equals 73, the first condition evaluates to false and returns 73 when it should return 75. The second condition evaluates correctly. The two formulations diverge at the boundary values specifically. Another issue is not handling the input format properly. HackerRank passes each grade as a separate integer in the input stream. Some people read the entire input at once and split it incorrectly, which causes index errors on multi-case test files. Just use the standard int(input()) pattern for the count and a simple for loop.
Performance and Alternatives
This solution runs in O(1) time per grade. There is no meaningful optimization possible because each grade is processed independently. The total runtime is bounded only by the number of test cases, which HackerRank typically limits to around 60. You are not going to hit any time limit concerns here. If you are coming from a different language like JavaScript or Java, the logic translates directly. No language-specific tricks are needed. The modulo operator works identically across all of them for positive integers, which is all this problem deals with. Some people try to solve this with a series of if-elif chains checking grade boundaries like 40, 60, and 80. That approach works for the final grade output but does not address the rounding rule correctly unless you layer the rounding logic on top of it. The cleanest path is always the modulo approach shown above.
Grading Students Hackerrank Solution Code Summary
The complete working solution fits in roughly ten lines including input handling. The function itself is five to six lines. Read the number of test cases, loop, call the function, and print. That is the full scope of what this challenge requires. Nothing more is needed.
