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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 highΒ (inclusive).

Β 

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].

Β 

Constraints:

  • 0 <= low <= highΒ <= 10^9

Solutions

Solution 1: Prefix Sum Concept

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 \(O(1)\).

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

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

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

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

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

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

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

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

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