Strobogrammatic Number III

Strobogrammatic Number III - Problem

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

Given two strings low and high that represent two integers where low <= high, return the number of strobogrammatic numbers in the range [low, high].

The digits that remain the same when rotated 180 degrees are: 0, 1, 6, 8, 9 where:

0 → 0
1 → 1
6 → 9
8 → 8
9 → 6

Input & Output

Example 1 — Small Range

$ Input: low = "50", high = "100"

› Output: 3

💡 Note: Strobogrammatic numbers in range [50,100] are: 69, 88, 96. Each looks the same when rotated 180°.

Example 2 — Single Digit Range

$ Input: low = "0", high = "9"

› Output: 3

💡 Note: Single digit strobogrammatic numbers are: 0, 1, 8. Numbers 2,3,4,5,6,7,9 don't look the same when rotated.

Example 3 — No Valid Numbers

$ Input: low = "20", high = "30"

› Output: 0

💡 Note: No strobogrammatic numbers exist between 20 and 30.

Constraints

1 ≤ low.length, high.length ≤ 15
low and high consist of only digits
low ≤ high
low and high do not contain any leading zeros except for zero itself

Visualization

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Asked in

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The key insight is that only digits 0,1,6,8,9 can form strobogrammatic numbers, and we can generate them recursively by adding valid pairs to shorter numbers. The optimal approach uses backtracking to generate only valid strobogrammatic numbers rather than checking every number in the range. Time: O(5^L), Space: O(5^L) where L is the maximum length.

Common Approaches

✓ Brute Force

⏱️ Time: O((high-low) × L) Space: O(L)

Iterate through every number from low to high, convert each to string, and check if rotating it 180 degrees gives the same number. This involves checking each digit and its rotated counterpart.

Generate and Count

⏱️ Time: O(5^L) Space: O(5^L)

Use recursion to generate all strobogrammatic numbers of different lengths, then count how many fall within the given range. This avoids checking invalid numbers.

Brute Force — Algorithm Steps

Iterate from low to high
For each number, check if it's strobogrammatic by comparing with its rotated version
Count valid numbers

Visualization

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Step-by-Step Walkthrough

Range Check

Iterate through all numbers from low to high

Rotation Test

For each number, rotate 180° and compare

Count Valid

Increment counter for matching numbers

Code -

solution.c — C

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

char getMapping(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';
    }
}

int isStrobogrammatic(const char* numStr) {
    int len = strlen(numStr);
    char rotated[20] = {0};
    
    for (int i = len - 1; i >= 0; i--) {
        char c = numStr[i];
        char mapped = getMapping(c);
        if (mapped == '\0') {
            return 0;
        }
        rotated[len - 1 - i] = mapped;
    }
    rotated[len] = '\0';
    
    return strcmp(rotated, numStr) == 0;
}

int solution(const char* low, const char* high) {
    int count = 0;
    long long lowNum = atoll(low);
    long long highNum = atoll(high);
    
    for (long long num = lowNum; num <= highNum; num++) {
        char numStr[20];
        sprintf(numStr, "%lld", num);
        if (isStrobogrammatic(numStr)) {
            count++;
        }
    }
    return count;
}

int main() {
    char low[20], high[20];
    fgets(low, sizeof(low), stdin);
    fgets(high, sizeof(high), stdin);
    low[strcspn(low, "\n")] = 0;
    high[strcspn(high, "\n")] = 0;
    
    int result = solution(low, high);
    printf("%d\n", result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O((high-low) × L)

For each number in range, we check L digits where L is average length

✓ Linear Growth

Space Complexity

O(L)

Extra space for string conversion and rotation check

✓ Linear Space

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Medium Frequency

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Input & Output

Constraints

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Related Problems

Common Approaches

Brute Force — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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