How do you solve Count Odd Numbers in an Interval Range on LeetCode?

Count Odd Numbers in an Interval Range (LeetCode #1523) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(1) time complexity and O(1) space complexity. Key insight: If the range has an even number of integers, the count of odd and even numbers will be the same.

What is the brute force solution for Count Odd Numbers in an Interval Range?

The brute force approach for Count Odd Numbers in an Interval Range is Brute Force, with O(n) time complexity and O(1) space complexity. The brute force approach involves iterating through each number in the range from low to high and counting how many of them are odd. This is straightforward but inefficient for large ranges. The optimal Optimal Solution approach improves this to O(1).

What algorithm or data structure is used to solve Count Odd Numbers in an Interval Range?

Count Odd Numbers in an Interval Range uses the following concepts: Math. The recommended patterns to study are: Mathematical Counting, Parity Check.

What are the interview tips for Count Odd Numbers in an Interval Range?

Always start by clarifying the problem and constraints with the interviewer. Discuss your thought process and approach before jumping into coding. Consider edge cases, such as when low equals high or when both are even.

What are common mistakes when solving Count Odd Numbers in an Interval Range?

Not considering the case when low and high are the same. Confusing the parity of low and high when calculating the count.

#1523

Count Odd Numbers in an Interval Range

Easy

MathMathematical CountingParity Check

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n)	O(1)
Space	O(1)	O(1)

💡

Intuition

Time O(1)Space O(1)

Instead of iterating through all numbers, we can use a mathematical approach to directly calculate the number of odd numbers in the range. This is much faster and efficient, especially for large ranges.

⚙️

Algorithm

3 steps

1Step 1: Calculate the total number of integers in the range: total = high - low + 1.
2Step 2: If total is even, return total / 2 (half will be odd).
3Step 3: If total is odd, check the parity of low. If low is odd, return (total / 2) + 1; otherwise, return total / 2.

solution.py8 lines

1# Full working Python code
2low, high = 3, 7
3total = high - low + 1
4if total % 2 == 0:
5    count = total // 2
6else:
7    count = (total // 2) + (1 if low % 2 != 0 else 0)
8print(count)

ℹ

Complexity note: The time complexity is O(1) because we are performing a constant number of operations regardless of the size of the range. The space complexity is also O(1) as we are using a constant amount of space.

1If the range has an even number of integers, the count of odd and even numbers will be the same.
2If the range has an odd number of integers, the count of odd numbers will depend on whether the starting number (low) is odd or even.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.