Problem Description

A confusing number is a number that becomes a different number when rotated 180 degrees, with all digits remaining valid after rotation.

When we rotate individual digits by 180 degrees:

• 0 stays 0
• 1 stays 1
• 6 becomes 9
• 8 stays 8
• 9 becomes 6
• 2, 3, 4, 5, and 7 become invalid (cannot be rotated)

The rotation process works by rotating each digit individually and reading the result from right to left. For example, if we have 69:

• The 6 rotates to 9
• The 9 rotates to 6
• Reading from right to left gives us 96

Since 96 ≠ 69, this is a confusing number.

Important notes:

• If any digit in the number cannot be rotated (like 2, 3, 4, 5, or 7), the entire number is not a confusing number
• Leading zeros after rotation are ignored (e.g., 8000 rotated becomes 0008, which is just 8)
• A confusing number must be different from the original number after rotation

Given an integer n, determine if it's a confusing number by returning true if it is, or false otherwise.

The solution uses a mapping array d where:

• Valid rotations are stored at their index positions (d[0]=0, d[1]=1, d[6]=9, d[8]=8, d[9]=6)
• Invalid digits are marked with -1

The algorithm extracts each digit from right to left, checks if it's rotatable, builds the rotated number, and finally compares it with the original.

Intuition

To determine if a number is confusing, we need to simulate the rotation process and check two conditions: all digits must be rotatable, and the final result must differ from the original.

The key insight is that we can process the number digit by digit from right to left, simultaneously building the rotated number from left to right. This naturally handles the rotation's "flip" effect where the rightmost digit becomes the leftmost after rotation.

Consider how rotation works visually: when you rotate 69 by 180 degrees, the 6 (on the left) becomes 9 and moves to the right position, while the 9 (on the right) becomes 6 and moves to the left position. This is equivalent to:

1. Taking each digit from right to left
2. Rotating it individually
3. Appending it to build a new number from left to right

We can use a lookup table to make rotation efficient. Since we only have 10 possible digits (0-9), we can pre-define what each digit becomes after rotation:

• Rotatable digits: 0→0, 1→1, 6→9, 8→8, 9→6
• Non-rotatable digits: 2, 3, 4, 5, 7 → invalid (we can mark these as -1)

The algorithm naturally emerges:

1. Extract the last digit using divmod(n, 10)
2. Check if it's rotatable using our lookup table
3. If not rotatable, immediately return false
4. If rotatable, append its rotated value to build our result
5. Continue until all digits are processed
6. Compare the rotated number with the original

This approach elegantly handles leading zeros since we build the number mathematically (y = y * 10 + d[v]), which automatically ignores leading zeros.

Learn more about Math patterns.

Solution Approach

The implementation uses a digit-by-digit extraction and reconstruction approach with a lookup table for efficient rotation mapping.

Data Structure:

• Lookup array d = [0, 1, -1, -1, -1, -1, 9, -1, 8, 6] where the index represents the original digit and the value represents its rotated form. Invalid rotations are marked as -1.

Algorithm Steps:

1. Initialize variables:

   • x = n: Copy of the original number for processing
   • y = 0: Will store the rotated number
   • d: The rotation mapping array

2. Process each digit from right to left:

   while x:
       x, v = divmod(x, 10)

   • divmod(x, 10) gives us the quotient and remainder
   • v is the current rightmost digit
   • x becomes the remaining number after removing the rightmost digit

3. Check validity and rotate:

   if d[v] < 0:
       return False

   • If d[v] < 0, the digit cannot be rotated (it's one of 2, 3, 4, 5, 7)
   • Immediately return False as the number cannot be a confusing number

4. Build the rotated number:

   y = y * 10 + d[v]

   • Multiply current y by 10 to shift digits left
   • Add the rotated digit d[v] to the rightmost position
   • This builds the rotated number from left to right as we extract digits from right to left

5. Final comparison:

   return y != n

   • A confusing number must be different from the original after rotation
   • Return True if they differ, False if they're the same

Example walkthrough with n = 69:

• First iteration: x=69, v=9, d[9]=6, y=0*10+6=6, x=6
• Second iteration: x=6, v=6, d[6]=9, y=6*10+9=69, x=0
• Loop ends, compare: y=96 != n=69, return True

Example with invalid digit n = 25:

• First iteration: x=25, v=5, d[5]=-1 < 0, return False immediately

This approach has O(log n) time complexity (number of digits) and O(1) space complexity.

Example Walkthrough

Let's walk through the solution with n = 89:

Step 1: Initialize

• x = 89 (working copy)
• y = 0 (will build rotated number)
• d = [0, 1, -1, -1, -1, -1, 9, -1, 8, 6] (rotation mapping)

Step 2: Extract rightmost digit (9)

• x, v = divmod(89, 10) → x = 8, v = 9
• Check: d[9] = 6 (valid, not -1)
• Build rotated number: y = 0 * 10 + 6 = 6

Step 3: Extract next digit (8)

• x, v = divmod(8, 10) → x = 0, v = 8
• Check: d[8] = 8 (valid, not -1)
• Build rotated number: y = 6 * 10 + 8 = 68

Step 4: No more digits

• x = 0, exit loop

Step 5: Compare results

• Original: n = 89
• Rotated: y = 68
• Since 68 ≠ 89, return True

Visual representation:

Original:  8 9
           ↓ ↓ (rotate each digit)
           8 6
           ← ← (read right to left)
Rotated:   6 8

The number 89 becomes 68 after rotation, making it a confusing number.

Solution Implementation

Time and Space Complexity

Time Complexity: O(log n)

The algorithm iterates through each digit of the number n. Since a number n has ⌊log₁₀(n)⌋ + 1 digits, the while loop executes O(log n) times. In each iteration, we perform constant time operations:

• divmod(x, 10): O(1)
• Array lookup d[v]: O(1)
• Multiplication and addition y * 10 + d[v]: O(1)

Therefore, the overall time complexity is O(log n).

Space Complexity: O(1)

The algorithm uses a fixed amount of extra space:

• Variable x to store a copy of n: O(1)
• Variable y to build the rotated number: O(1)
• Variable v to store the current digit: O(1)
• Array d with 10 elements (constant size): O(1)

The space usage doesn't grow with the input size, making the space complexity O(1).

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Modifying the Original Number Variable

A critical mistake is using the original number variable n for digit extraction, which destroys the original value needed for the final comparison.

Incorrect approach:

def confusingNumber(self, n: int) -> bool:
    rotation_map = [0, 1, -1, -1, -1, -1, 9, -1, 8, 6]
    rotated = 0
  
    # ERROR: Modifying n directly loses the original value
    while n > 0:
        n, digit = divmod(n, 10)  # n is being modified!
        if rotation_map[digit] < 0:
            return False
        rotated = rotated * 10 + rotation_map[digit]
  
    return rotated != n  # This will always compare with 0!

Solution: Create a copy of n for processing:

original_number = n  # Keep n intact
while original_number > 0:
    original_number, digit = divmod(original_number, 10)
    # ... rest of the logic
return rotated != n  # Now n still has the original value

2. Incorrect Rotation Building Order

Another common mistake is building the rotated number in the wrong order, either by prepending digits instead of appending or by not understanding the rotation reversal.

Incorrect approach:

# Wrong: Trying to prepend digits (this doesn't work as intended)
rotated = d[v] * (10 ** digit_position) + rotated

Correct approach: The digits should be added from left to right as we extract from right to left:

rotated = rotated * 10 + rotation_map[digit]

3. Forgetting to Handle Single Valid Digits

Numbers like 0, 1, and 8 rotate to themselves. Beginners might incorrectly assume these should return True because they're "valid for rotation."

Key insight: A confusing number must be different after rotation. So:

• 8 → 8: Not confusing (returns False)
• 6 → 9: Confusing (returns True)
• 88 → 88: Not confusing (returns False)
• 68 → 89: Confusing (returns True)

4. Index Out of Bounds for Large Digits

If the rotation map array has exactly 10 elements but you don't validate the digit range, you could get an index error.

Prevention: While digits from divmod are guaranteed to be 0-9, if you're getting digits differently, always validate:

if current_digit < 0 or current_digit > 9:
    return False

5. Not Handling Zero Correctly

The number 0 is a special case that rotates to itself (0 → 0), so it should return False. Make sure your logic handles this correctly and doesn't have any division by zero or infinite loop issues.