Strobogrammatic Number

Strobogrammatic Number - Problem

A strobogrammatic number is a number that looks the same when rotated 180 degrees (looked at upside down).

Given a string num which represents an integer, return true if num is a strobogrammatic number.

When rotated 180 degrees, the digits transform as follows:

0 → 0
1 → 1
6 → 9
8 → 8
9 → 6
All other digits (2, 3, 4, 5, 7) are invalid when rotated

Input & Output

Example 1 — Valid Strobogrammatic

$ Input: num = "69"

› Output: true

💡 Note: When rotated 180°: '6' becomes '9' and '9' becomes '6', giving us '96'. Reversed: '69' equals original.

Example 2 — Invalid Digit

$ Input: num = "88"

› Output: true

💡 Note: Both '8' digits remain '8' when rotated 180°, so '88' rotated is still '88'.

Example 3 — Contains Invalid Digit

$ Input: num = "962"

› Output: false

💡 Note: The digit '2' cannot be rotated 180° to form a valid digit, so this is not strobogrammatic.

Constraints

1 ≤ num.length ≤ 50
num consists of only digits.
num does not contain any leading zeros except for the zero itself.

Visualization

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Asked in

G Google 45 f Facebook 38 M Microsoft 25

The key insight is that only digits 0,1,6,8,9 are valid when rotated 180°, and we can use two pointers to validate pairs efficiently. Best approach uses two pointers for O(1) space. Time: O(n), Space: O(1)

Common Approaches

✓ Build Rotated String

⏱️ Time: O(n) Space: O(n)

Build the complete rotated string by mapping each digit, then reverse it and compare with the original string.

Two Pointers Validation

⏱️ Time: O(n) Space: O(1)

Use two pointers starting from both ends. For each pair, verify they form a valid strobogrammatic pair when rotated.

Build Rotated String — Algorithm Steps

Create mapping for valid rotations (0→0, 1→1, 6→9, 8→8, 9→6)
Iterate through each digit and build rotated string
Reverse the rotated string and compare with original

Visualization

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Step-by-Step Walkthrough

Map Each Digit

Transform each digit using rotation mapping

Reverse Result

Reverse the mapped string to simulate 180° rotation

Compare

Check if rotated string equals original

Code -

solution.c — C

#include <stdio.h>
#include <string.h>
#include <stdbool.h>

char getRotation(char c) {
    switch (c) {
        case '0': return '0';
        case '1': return '1';
        case '6': return '9';
        case '8': return '8';
        case '9': return '6';
        default: return '\0';
    }
}

bool solution(char* num) {
    int len = strlen(num);
    char rotated[len + 1];
    
    for (int i = 0; i < len; i++) {
        char rotation = getRotation(num[i]);
        if (rotation == '\0') {
            return false;
        }
        rotated[len - 1 - i] = rotation;
    }
    rotated[len] = '\0';
    
    return strcmp(rotated, num) == 0;
}

int main() {
    char num[1000];
    fgets(num, sizeof(num), stdin);
    num[strcspn(num, "\n")] = 0;
    
    bool result = solution(num);
    printf(result ? "true\n" : "false\n");
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n)

Single pass to build rotated string plus O(n) for comparison

✓ Linear Growth

Space Complexity

O(n)

Extra string to store the rotated version

✓ Linear Space

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Smart Actions

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Input & Output

Constraints

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Related Problems

Common Approaches

Build Rotated String — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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