LeetCode 2544. Alternating Digit Sum Solution in Java, C++, Python & More | Explanation + Code

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.

PythonJavaC++GoTypeScriptRustC

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

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; } }(code-box)

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; } };(code-box)

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

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; }(code-box)

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 } }(code-box)

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

Solution 2

PythonRust

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(code-box)

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 } }(code-box)