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Problem 2544 Alternating Digit Sum

Problem Statement

Link - Problem 2544

Question

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

Solution

class Solution {
public:
    int alternateDigitSum(int n) {
        string num = to_string(n);
        int var = 0, sign = 1;
        for(int i = 0; i< num.size(); i++){
            var += (num[i]-'0')*sign;
            sign*=-1;
        }
        return var;
    }
};

Complexity Analysis

| Algorithm         | Time Complexity | Space Complexity |
| ----------------- | --------------- | ---------------- |
| String conversion | O(log(n))       | O(1)             |
| Computation       | O(log(n))       | O(1)             |

Explanation

1. Intuition

• To handle the MSB condition, convert to string and then process the string.

2. Implementation

- Initialize the variable `var` to `0` and `sign` to `1`.
- Convert the integer to string.
- Iterate over the string and add the digit to `var` with the sign.
- Change the sign after each digit.
- Return the `var`.