1523 - Count Odd Numbers in an Interval Range (Easy)

Problem Link

https://leetcode.com/problems/count-odd-numbers-in-an-interval-range/

Problem Statement

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

Approach 1: Brute Force

Iterate from $low$ to $high$ and check if $i$ is odd.

Written by @wingkwong

func countOdds(low int, high int) int {
    ans := 0
    for i := low; i <= high; i += 1 {
        ans += i & 1
    }
    return ans
}

Written by @wingkwong

impl Solution {
    pub fn count_odds(low: i32, high: i32) -> i32 {
        let mut ans = 0;
        for i in low .. high + 1 {
            ans += i & 1;
        }
        return ans;
    }
}

Approach 2: Math

Numbers of odd numbers in $[low, high]$ is same as $[1, high] - [1 , low - 1]$. Therefore, we just find out those two numbers to get the answer. There are $(high + 1) / 2$ odd numbers in $[1, high]$ and $low/2$ odd numbers in $[1, low - 1]$.

Written by @wingkwong

func countOdds(low int, high int) int {
    return (high + 1) / 2 - (low / 2);
}

Written by @wingkwong

impl Solution {
    pub fn count_odds(low: i32, high: i32) -> i32 {
        return (high + 1) / 2 - (low / 2);
    }
}