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2544. Alternating Digit Sum

Description

You are given a positive integer n. Each digit of n has a sign according to the following rules:

  • The most significant digit is assigned a positive sign.
  • Each other digit has an opposite sign to its adjacent digits.

Return the sum of all digits with their corresponding sign.

 

Example 1:

Input: n = 521
Output: 4
Explanation: (+5) + (-2) + (+1) = 4.

Example 2:

Input: n = 111
Output: 1
Explanation: (+1) + (-1) + (+1) = 1.

Example 3:

Input: n = 886996
Output: 0
Explanation: (+8) + (-8) + (+6) + (-9) + (+9) + (-6) = 0.

 

Constraints:

  • 1 <= n <= 109

 

Solutions

Solution 1: Simulation

We can directly simulate the process as described in the problem.

We define an initial symbol $sign=1$. Starting from the most significant digit, we take out one digit $x$ each time, multiply it by $sign$, add the result to the answer, then negate $sign$, and continue to process the next digit until all digits are processed.

The time complexity is $O(\log n)$, and the space complexity is $O(\log n)$. Here, $n$ is the given number.

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class Solution:
    def alternateDigitSum(self, n: int) -> int:
        return sum((-1) ** i * int(x) for i, x in enumerate(str(n)))

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class Solution {
    public int alternateDigitSum(int n) {
        int ans = 0, sign = 1;
        for (char c : String.valueOf(n).toCharArray()) {
            int x = c - '0';
            ans += sign * x;
            sign *= -1;
        }
        return ans;
    }
}

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class Solution {
public:
    int alternateDigitSum(int n) {
        int ans = 0, sign = 1;
        for (char c : to_string(n)) {
            int x = c - '0';
            ans += sign * x;
            sign *= -1;
        }
        return ans;
    }
};

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func alternateDigitSum(n int) (ans int) {
    sign := 1
    for _, c := range strconv.Itoa(n) {
        x := int(c - '0')
        ans += sign * x
        sign *= -1
    }
    return
}

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function alternateDigitSum(n: number): number {
    let ans = 0;
    let sign = 1;
    while (n) {
        ans += (n % 10) * sign;
        sign = -sign;
        n = Math.floor(n / 10);
    }
    return ans * -sign;
}

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impl Solution {
    pub fn alternate_digit_sum(mut n: i32) -> i32 {
        let mut ans = 0;
        let mut sign = 1;
        while n != 0 {
            ans += (n % 10) * sign;
            sign = -sign;
            n /= 10;
        }
        ans * -sign
    }
}

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int alternateDigitSum(int n) {
    int ans = 0;
    int sign = 1;
    while (n) {
        ans += (n % 10) * sign;
        sign = -sign;
        n /= 10;
    }
    return ans * -sign;
}

Solution 2

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class Solution:
    def alternateDigitSum(self, n: int) -> int:
        ans, sign = 0, 1
        for c in str(n):
            x = int(c)
            ans += sign * x
            sign *= -1
        return ans

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impl Solution {
    pub fn alternate_digit_sum(n: i32) -> i32 {
        let mut ans = 0;
        let mut sign = 1;

        for c in format!("{}", n).chars() {
            let x = c.to_digit(10).unwrap() as i32;
            ans += x * sign;
            sign *= -1;
        }

        ans
    }
}

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