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2180. Count Integers With Even Digit Sum

Description

Given a positive integer num, return the number of positive integers less than or equal to num whose digit sums are even.

The digit sum of a positive integer is the sum of all its digits.

Β 

Example 1:

Input: num = 4
Output: 2
Explanation:
The only integers less than or equal to 4 whose digit sums are even are 2 and 4.

Example 2:

Input: num = 30
Output: 14
Explanation:
The 14 integers less than or equal to 30 whose digit sums are even are
2, 4, 6, 8, 11, 13, 15, 17, 19, 20, 22, 24, 26, and 28.

Β 

Constraints:

  • 1 <= num <= 1000

Solutions

Solution 1

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class Solution:
    def countEven(self, num: int) -> int:
        ans = 0
        for x in range(1, num + 1):
            s = 0
            while x:
                s += x % 10
                x //= 10
            ans += s % 2 == 0
        return ans

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class Solution {
    public int countEven(int num) {
        int ans = 0;
        for (int i = 1; i <= num; ++i) {
            int s = 0;
            for (int x = i; x > 0; x /= 10) {
                s += x % 10;
            }
            if (s % 2 == 0) {
                ++ans;
            }
        }
        return ans;
    }
}

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class Solution {
public:
    int countEven(int num) {
        int ans = 0;
        for (int i = 1; i <= num; ++i) {
            int s = 0;
            for (int x = i; x; x /= 10) {
                s += x % 10;
            }
            ans += s % 2 == 0;
        }
        return ans;
    }
};

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func countEven(num int) (ans int) {
    for i := 1; i <= num; i++ {
        s := 0
        for x := i; x > 0; x /= 10 {
            s += x % 10
        }
        if s%2 == 0 {
            ans++
        }
    }
    return
}

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function countEven(num: number): number {
    let ans = 0;
    for (let i = 1; i <= num; ++i) {
        let s = 0;
        for (let x = i; x; x = Math.floor(x / 10)) {
            s += x % 10;
        }
        if (s % 2 == 0) {
            ++ans;
        }
    }
    return ans;
}

Solution 2

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class Solution:
    def countEven(self, num: int) -> int:
        ans = num // 10 * 5 - 1
        x, s = num // 10, 0
        while x:
            s += x % 10
            x //= 10
        ans += (num % 10 + 2 - (s & 1)) >> 1
        return ans

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class Solution {
    public int countEven(int num) {
        int ans = num / 10 * 5 - 1;
        int s = 0;
        for (int x = num / 10; x > 0; x /= 10) {
            s += x % 10;
        }
        ans += (num % 10 + 2 - (s & 1)) >> 1;
        return ans;
    }
}

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class Solution {
public:
    int countEven(int num) {
        int ans = num / 10 * 5 - 1;
        int s = 0;
        for (int x = num / 10; x > 0; x /= 10) {
            s += x % 10;
        }
        ans += (num % 10 + 2 - (s & 1)) >> 1;
        return ans;
    }
};

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func countEven(num int) (ans int) {
    ans = num/10*5 - 1
    s := 0
    for x := num / 10; x > 0; x /= 10 {
        s += x % 10
    }
    ans += (num%10 + 2 - (s & 1)) >> 1
    return
}

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function countEven(num: number): number {
    let ans = Math.floor(num / 10) * 5 - 1;
    let s = 0;
    for (let x = Math.floor(num / 10); x; x = Math.floor(x / 10)) {
        s += x % 10;
    }
    ans += ((num % 10) + 2 - (s & 1)) >> 1;
    return ans;
}

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