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246. Strobogrammatic Number

easy

Decide whether a number reads the same after being rotated 180 degrees. A two-pointer scan with a tiny lookup map — clean and elegant.

By Sam K., Founder, InterviewChamp.AI · Last verified

Problem

Given a string num which represents an integer, return true if num is a strobogrammatic number. A strobogrammatic number is a number that looks the same when rotated 180 degrees (looked at upside down).

Constraints

  • 1 <= num.length <= 50
  • num consists of only digits.
  • num does not contain any leading zeros except for zero itself.

Examples

Example 1

Input
num = "69"
Output
true

Example 2

Input
num = "88"
Output
true

Example 3

Input
num = "962"
Output
false

Example 4

Input
num = "1"
Output
true

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Output

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Hints

Progressive — try the first before opening the next.

Hint 1

Only 5 digits are valid when rotated 180 degrees: 0 -> 0, 1 -> 1, 6 -> 9, 8 -> 8, 9 -> 6. Everything else fails.

Hint 2

Walk two pointers from the ends inward. The left digit must rotate to the right digit.

Hint 3

If at any pair num[left] doesn't rotate to num[right], return false.

Hint 4

For odd-length strings, the middle digit rotates to itself — so it must be 0, 1, or 8.

Solution approach

Reveal approach

Lookup map: {'0':'0', '1':'1', '6':'9', '8':'8', '9':'6'}. Two pointers left = 0, right = n - 1. While left <= right: if num[left] not in the map, return false; if map[num[left]] != num[right], return false. Then left++, right--. Return true. The left <= right (rather than <) condition handles odd-length strings: when left == right (middle digit), the digit must rotate to itself — only 0, 1, 8 satisfy that, and the map check enforces it. O(n) time, O(1) space (the map is constant size).

Complexity

Time
O(n)
Space
O(1)

Related patterns

  • math
  • two-pointers
  • hash-map

Related problems

Asked at

Companies reported asking this problem (sourced from public Glassdoor, Blind, and Levels.fyi interview posts).

  • Google
  • Microsoft

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