Skip to content

1523. Count Odd Numbers in an Interval Range

Description

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

Solutions

Solution 1: Prefix Sum Concept

Thinking

Count the odds in the closed interval \([low,high]\). The span can reach \(10^9\), so we cannot test each integer.

The range \([0,x]\) contains \(\lfloor(x+1)/2\rfloor\) odds. Subtracting the count for \([0,low-1]\) from that for \([0,high]\) gives \(\lfloor(high+1)/2\rfloor-\lfloor low/2\rfloor\), which a shift computes in constant time.

We know that the count of odd numbers in the range \([0, x]\) is \(\lfloor\frac{x+1}{2}\rfloor\). Therefore, the count of odd numbers in the range \([low, high]\) is \(\lfloor\frac{high+1}{2}\rfloor - \lfloor\frac{low}{2}\rfloor\).

The time complexity is \(O(1)\), and the space complexity is \(O(1)\).

1
2
3
class Solution:
    def countOdds(self, low: int, high: int) -> int:
        return ((high + 1) >> 1) - (low >> 1)

1
2
3
4
5
class Solution {
    public int countOdds(int low, int high) {
        return ((high + 1) >> 1) - (low >> 1);
    }
}

1
2
3
4
5
6
class Solution {
public:
    int countOdds(int low, int high) {
        return (high + 1 >> 1) - (low >> 1);
    }
};

1
2
3
func countOdds(low int, high int) int {
    return ((high + 1) >> 1) - (low >> 1)
}

1
2
3
function countOdds(low: number, high: number): number {
    return ((high + 1) >> 1) - (low >> 1);
}

1
2
3
4
5
impl Solution {
    pub fn count_odds(low: i32, high: i32) -> i32 {
        ((high + 1) >> 1) - (low >> 1)
    }
}

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
class Solution {
    /**
     * @param Integer $low
     * @param Integer $high
     * @return Integer
     */
    function countOdds($low, $high) {
        return (($high + 1) >> 1) - ($low >> 1);
    }
}

1
2
3
int countOdds(int low, int high) {
    return ((high + 1) >> 1) - (low >> 1);
}

Comments