Leetcode 2544: Alternating Digit Sum
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.
Problem
Approach
Steps
Complexity
Input: You are given a positive integer n.
Example: n = 324
Constraints:
• 1 <= n <= 10^9
Output: Return the sum of all digits with their corresponding signs.
Example: 1
Constraints:
Goal: Calculate the sum of the digits of the integer n, where each digit has a sign based on its position.
Steps:
• 1. Extract each digit from the number n, starting from the least significant to the most significant.
• 2. Alternate the sign for each digit based on its position.
• 3. Compute the sum of the digits, considering their corresponding signs.
Goal: The solution must handle numbers up to 10^9 efficiently.
Steps:
• 1 <= n <= 10^9
Assumptions:
• The input number n is a positive integer.
• The number has at least one digit.
• Input: n = 324
• Explanation: For n = 324, the sum is (+3) + (-2) + (+4) = 1.
• Input: n = 101
• Explanation: For n = 101, the sum is (+1) + (-0) + (+1) = 2.
Approach: The approach is to alternate the signs of the digits in the number and compute their sum.
Observations:
• The problem asks us to sum digits with alternating signs based on their position.
• This is a straightforward problem of alternating signs and summing values.
• We can extract digits one by one and alternate their signs using a simple loop.
Steps:
• 1. Initialize a sum variable to 0.
• 2. Start from the least significant digit and alternate the sign for each digit.
• 3. Update the sum as you traverse through the digits.
Empty Inputs:
• The input is always a positive integer, so no need to handle empty inputs.
Large Inputs:
• Handle inputs where n is close to the upper constraint of 10^9.
Special Values:
• Handle cases where all digits are the same, such as n = 111.
Constraints:
• The algorithm should run efficiently even for large numbers up to 10^9.
int alternateDigitSum(int n) {
int sum = 0;
int sgn = 1;
while(n) {
sum += sgn * (n % 10);
n /= 10;
sgn *= -1;
}
return sgn == -1? sum : -sum;
}
1 : Function Definition
int alternateDigitSum(int n) {
The function 'alternateDigitSum' is defined to compute the alternating sum of digits of an integer 'n'.
2 : Variable Initialization
int sum = 0;
The variable 'sum' is initialized to zero. This will accumulate the alternating sum of the digits.
3 : Variable Initialization
int sgn = 1;
The variable 'sgn' is initialized to 1. This will help alternate the sign (positive or negative) for each digit in the number.
4 : Loop Setup
while(n) {
A while loop is initiated to iterate over the digits of the integer 'n'. The loop will continue until all digits are processed.
5 : Digit Extraction and Summing
sum += sgn * (n % 10);
The last digit of 'n' (n % 10) is multiplied by 'sgn' (which alternates between 1 and -1) and added to 'sum'.
6 : Digit Removal
n /= 10;
The last digit of 'n' is removed by dividing it by 10, effectively shifting to the next digit.
7 : Sign Alternation
sgn *= -1;
The sign alternator 'sgn' is multiplied by -1 to switch between adding and subtracting for the next digit.
8 : Return Statement
return sgn == -1? sum : -sum;
Once the loop is complete, the function returns the sum. If 'sgn' is -1, the sum is returned as is; otherwise, it is negated to account for the final sign.
Best Case: O(d), where d is the number of digits in n.
Average Case: O(d), where d is the number of digits in n.
Worst Case: O(d), where d is the number of digits in n.
Description: The time complexity is O(d), where d is the number of digits in n (which is log(n) with respect to the input number).
Best Case: O(1)
Worst Case: O(1)
Description: The space complexity is constant, O(1), as we only use a fixed amount of space for the sum and a few variables.
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