Is Count Odd Numbers in an Interval Range easy or hard?

Count Odd Numbers in an Interval Range is classified as an Easy problem on LeetCode. The challenge is recognizing the mathematical pattern that allows counting odd numbers without iterating through the entire interval.

Count Odd Numbers in an Interval Range Python/Java solution

Most implementations use the constant-time formula (high + 1) // 2 - low // 2. This logic translates directly across Python, Java, C++, JavaScript, and C#. The implementation typically takes only one or two lines after reading the input values.

How to solve Count Odd Numbers in an Interval Range in O(1)?

Use integer division to count odd numbers up to each boundary. Compute (high + 1) // 2 to get the number of odds up to high and subtract low // 2 to remove odds below the interval. The difference gives the exact number of odd integers in the inclusive range.

What is the best approach for Count Odd Numbers in an Interval Range?

The optimal approach uses a mathematical pattern instead of iteration. The formula (high + 1) // 2 - low // 2 directly computes how many odd numbers exist between two integers. This runs in O(1) time and O(1) space, making it far more efficient than scanning the entire range.

Is Count Odd Numbers in an Interval Range asked at Google/Amazon/Meta?

Problems based on counting patterns and parity appear frequently in interviews at large tech companies. While this exact problem may not always appear, the underlying concept of recognizing mathematical patterns instead of brute-force iteration is commonly tested.

What data structure is used in Count Odd Numbers in an Interval Range?

No special data structure is required. The problem relies purely on arithmetic properties of integers and parity checks using modulo or integer division. It falls under the math category rather than array or hash-based problems.

What is the time complexity of Count Odd Numbers in an Interval Range?

The optimal mathematical solution runs in O(1) time because it performs only a few arithmetic operations. A brute-force iterative solution that checks each number in the interval runs in O(n) time where n = high - low + 1. Both approaches use O(1) extra space.

Description Hints 2 Solution Video 5:40

1523. Count Odd Numbers in an Interval Range

Easy54.4% acceptance5 companies

Given two non-negative integers low and high. Return the count of odd numbers between low and high (inclusive).

Example 1:

Input: low = 3, high = 7
Output: 3
Explanation: The odd numbers between 3 and 7 are [3,5,7].

Example 2:

Input: low = 8, high = 10
Output: 1
Explanation: The odd numbers between 8 and 10 are [9].

Constraints:

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

Asked at 5 companies

TCS Microsoft

Google Amazon Meta

Topics

Math

Hints open one at a time

Hint 1

If the range (high - low + 1) is even, the number of even and odd numbers in this range will be the same.

Hint 2

If the range (high - low + 1) is odd, the solution will depend on the parity of high and low.

5:40

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