Alternating Digit Sum - Problem
You are given a positive integer n. Imagine each digit as having a sign (positive or negative) based on its position:
- The most significant digit (leftmost) is always positive
- Each subsequent digit has the opposite sign of the digit before it
Your task is to calculate the alternating sum of all digits with their corresponding signs.
Example: For n = 521:
• Digit 5 (position 0): +5
• Digit 2 (position 1): -2
• Digit 1 (position 2): +1
• Result: 5 - 2 + 1 = 4
Input & Output
example_1.py — Basic Case
$
Input:
n = 521
›
Output:
4
💡 Note:
The digits are 5, 2, 1 with signs +, -, + respectively. So the sum is +5 - 2 + 1 = 4.
example_2.py — Even Number of Digits
$
Input:
n = 111
›
Output:
1
💡 Note:
The digits are 1, 1, 1 with signs +, -, + respectively. So the sum is +1 - 1 + 1 = 1.
example_3.py — Single Digit
$
Input:
n = 886996
›
Output:
0
💡 Note:
The digits are 8,8,6,9,9,6 with signs +,-,+,-,+,- respectively. So the sum is +8-8+6-9+9-6 = 0.
Visualization
Tap to expand
Understanding the Visualization
1
Identify Positions
Number the digit positions from left starting at 0
2
Apply Sign Rule
Even positions (0,2,4...) are positive, odd positions (1,3,5...) are negative
3
Calculate Sum
Sum all digits with their corresponding signs
Key Takeaway
🎯 Key Insight: The sign alternates based on position - we can determine this mathematically without needing to track state!
Time & Space Complexity
Time Complexity
O(d)
Where d is the number of digits. We process each digit exactly once.
✓ Linear Growth
Space Complexity
O(1)
Only using a few variables regardless of input size
✓ Linear Space
Constraints
- 1 ≤ n ≤ 109
- n is a positive integer
- The number will have at most 10 digits
💡
Explanation
AI Ready
💡 Suggestion
Tab
to accept
Esc
to dismiss
// Output will appear here after running code