Counting Odd Numbers in a Range

Given two non-negative integers low and high, count how many odd numbers fall between them, inclusive. (LeetCode 1528)

Key observations:

1. Odd numbers satisfy n % 2 == 1.
2. We need to count all odd numbers in [low, high], including the boundaries if they’re odd.
3. Odd numbers occur every other number, so roughly half of any range is odd.

Approach: the number of integers between low and high is high - low + 1. If both boundaries are odd, the count increases by one (since both are included); if only one is odd, the basic pattern holds; if both are even, the count is exactly half the range size.

class Solution {
    public int countOdds(int low, int high) {
        // Calculate the difference between high and low
        int result = high - low;

        // If the result (range) is even and high is odd, we can count (result/2) + 1 odd numbers
        if(result % 2 == 0 && high % 2 == 1){
            return result / 2 + 1;
        }
        // If the result is odd, we also have (result / 2) + 1 odd numbers
        else if(result % 2 == 1){
            return result / 2 + 1;
        }
        // In all other cases, return result / 2
        else{
            return result / 2;
        }
    }
}

For low = 3, high = 7: range [3,4,5,6,7], odd numbers [3,5,7] → output 3. For low = 8, high = 10: range [8,9,10], odd number [9] → output 1.

This solution runs in O(1) since it’s based on simple arithmetic calculations.