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Formatted question description: https://leetcode.ca/all/1523.html

# 1523. Count Odd Numbers in an Interval Range (Easy)

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

Related Topics:
Math

## Solution 1.

There are high - low + 1 numbers in the range, so there are at least (high - low + 1) / 2 odd numbers. If low is odd, then there will be (high - low + 1 + (low % 2)) / 2 odd numbers.

• class Solution {
public int countOdds(int low, int high) {
if (low % 2 == 0)
low++;
if (high % 2 == 0)
high--;
return (high - low) / 2 + 1;
}
}

############

class Solution {
public int countOdds(int low, int high) {
return ((high + 1) >> 1) - (low >> 1);
}
}

• // OJ: https://leetcode.com/problems/count-odd-numbers-in-an-interval-range/
// Time: O(1)
// Space: O(1)
class Solution {
public:
int countOdds(int low, int high) {
return (high - low + 1 + low % 2) / 2;
}
};

• class Solution:
def countOdds(self, low: int, high: int) -> int:
return ((high + 1) >> 1) - (low >> 1)


• func countOdds(low int, high int) int {
return ((high + 1) >> 1) - (low >> 1)
}

• function countOdds(low: number, high: number): number {
return ((high + 1) >> 1) - (low >> 1);
}


• class Solution {

/**
* @param Integer $low * @param Integer$high
* @return Integer
*/
function countOdds($low,$high) {
return (($high + 1) >> 1) - ($low >> 1);
}
}

• impl Solution {
pub fn count_odds(low: i32, high: i32) -> i32 {
((high + 1) >> 1) - (low >> 1)
}
}