Count Odd Numbers in an Interval Range

Count Odd Numbers in an Interval Range - Problem

Math Easy

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

An odd number is any integer that is not divisible by 2. For example, 1, 3, 5, 7, 9 are odd numbers.

Example: If low = 3 and high = 7, the numbers in range are [3, 4, 5, 6, 7]. The odd numbers are 3, 5, 7, so the answer is 3.

Input & Output

Example 1 — Basic Range

$ Input: low = 3, high = 7

› Output: 3

💡 Note: Numbers in range [3,4,5,6,7]. Odd numbers are 3, 5, 7, so count is 3.

Example 2 — Even Range

$ Input: low = 8, high = 10

› Output: 1

💡 Note: Numbers in range [8,9,10]. Only 9 is odd, so count is 1.

Example 3 — Single Number

$ Input: low = 1, high = 1

› Output: 1

💡 Note: Range contains only 1, which is odd, so count is 1.

Constraints

0 ≤ low ≤ high ≤ 10⁹

Visualization

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Asked in

M Microsoft 25 🍎 Apple 20 G Google 15

The key insight is that odd numbers follow a predictable pattern in any range. Instead of iterating through each number, we can use the mathematical formula (high + 1) / 2 - low / 2 to calculate the count instantly. Best approach is the mathematical formula. Time: O(1), Space: O(1)

Common Approaches

✓ Brute Force

⏱️ Time: O(n) Space: O(1)

Loop through every number from low to high and check if each number is odd using modulo operation. Count all odd numbers found.

Mathematical Formula

⏱️ Time: O(1) Space: O(1)

Calculate the count of odd numbers using a mathematical formula based on the range endpoints. Odd numbers in a range follow a predictable pattern.

Brute Force — Algorithm Steps

Step 1: Initialize counter to 0
Step 2: Loop from low to high
Step 3: For each number, check if num % 2 == 1
Step 4: If odd, increment counter

Visualization

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Step-by-Step Walkthrough

Initialize

Start with count = 0

Check Each

Test if each number is odd

Count

Increment counter for odd numbers

Code -

solution.c — C

#include <stdio.h>

int solution(int low, int high) {
    int count = 0;
    for (int num = low; num <= high; num++) {
        if (num % 2 == 1) {
            count++;
        }
    }
    return count;
}

int main() {
    int low, high;
    scanf("%d", &low);
    scanf("%d", &high);
    int result = solution(low, high);
    printf("%d\n", result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n)

Loop iterates through (high - low + 1) numbers

✓ Linear Growth

Space Complexity

O(1)

Only uses a counter variable, no extra data structures

✓ Linear Space

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Input & Output

Constraints

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Related Problems

Common Approaches

Brute Force — Algorithm Steps

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Code -

Time & Space Complexity

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