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.

• 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;
}
}

• 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;
}
};

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


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

• 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;
}


• 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
}
}