1141EasyMath

Odd Numbers Inside a Range

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Trains the technique from

LeetCode 1523Count Odd Numbers in an Interval Range

This is an original problem, written from a brief that listed the technique, the difficulty, the topics, the function shape and the input bounds — none of that problem's wording, examples, hints or editorials. The link is there so you can map your practice onto the standard set.

Same function shape, different story and different numbers.

Return how many odd numbers lie in the range from low to high, counting both ends.

Examples

Example 1

Input: low = 1, high = 10
Output: 5

The odd numbers from 1 to 10 are 1, 3, 5, 7 and 9.

Example 2

Input: low = 2, high = 2
Output: 0

The range holds a single number and it is even.

Example 3

Input: low = 5, high = 5
Output: 1

The range holds a single number and it is odd.

Constraints

0 <= low <= high <= 10^9

The signature

The editor is preloaded with this. It matches the parent problem's shape, so a solution that works here transfers to a judge unchanged.

Python

def count_odds(low: int, high: int) -> int:

Java

public int countOdds(int low, int high)

The intended approach

Time O(1).Space O(1).

The range reaches a billion, so it cannot be walked. Count the odd numbers up to a value instead: from `0` to `m` there are `(m + 1) / 2` of them once the division rounds down, since odd numbers take every other place starting from one. Then the odd numbers inside the range are those up to `high` less those below `low`. Those below `low` are the odd numbers up to `low - 1`, which is `low / 2` rounded down, and writing it that way avoids handling `low` of zero as a special case. Both ends belong to the range, which is exactly what taking `high` inclusive and `low - 1` exclusive gives.

Pythonentry point: count_odds

def count_odds(low, high):
    return (high + 1) // 2 - low // 2

The obvious version

Walk every number from `low` to `high` and count the ones leaving a remainder against two. Costs the whole width of the range.

This is not filler. It was written independently of the fast solution and the two were run against thousands of random inputs to confirm they agree — which is how the answer above is known to be right. Read it when the intended approach feels like a rabbit pulled out of a hat.

Python

def bf_count_odds(low, high):
    total = 0
    for value in range(low, high + 1):
        if value % 2:
            total += 1
    return total

How this problem was checked

Reference solution executed against 15 test cases — 15 passed
3 worked examples confirmed by running the code
0 eight-word sequences shared with the parent problem's text
Agrees with an independent brute-force implementation on random inputs
Test set proven to catch deliberately broken solutions
Reviewed by a separate quality pass
Function arity matches the parent problem's

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