Sequential Digits

Sequential Digits - Problem

Enumeration Medium

An integer has sequential digits if and only if each digit in the number is one more than the previous digit.

Return a sorted list of all the integers in the range [low, high] inclusive that have sequential digits.

Example: The number 234 has sequential digits because 2→3→4 where each digit is exactly one more than the previous. The number 235 does NOT have sequential digits because 3→5 skips 4.

Input & Output

Example 1 — Basic Range

$ Input: low = 100, high = 300

› Output: [123,234]

💡 Note: 123 has sequential digits (1→2→3), 234 has sequential digits (2→3→4). No other numbers in range [100,300] have sequential digits.

Example 2 — Wider Range

$ Input: low = 1000, high = 13000

› Output: [1234,2345,3456,4567,5678,6789,12345]

💡 Note: All 4-digit sequential numbers (1234-6789) and the 5-digit number 12345 fall within this range.

Example 3 — Single Digits

$ Input: low = 1, high = 10

› Output: [1,2,3,4,5,6,7,8,9]

💡 Note: All single digits from 1-9 have sequential digits by definition (only one digit).

Constraints

10 ≤ low ≤ high ≤ 10⁹

Visualization

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Asked in

G Google 15 a Amazon 12 M Microsoft 8

The key insight is that there are only 36 possible sequential digit numbers total (1-9, 12-89, 123-789, etc.). Instead of checking every number in the range, we can generate all sequential numbers using the base pattern '123456789' and filter by the given range. Best approach generates all possibilities in O(1) time since the count is fixed, then filters by range.

Common Approaches

✓ Brute Force Check

⏱️ Time: O(n × d) Space: O(k)

Iterate through all numbers from low to high. For each number, convert to string and check if each digit is exactly one more than the previous digit.

Generate Sequential Numbers

⏱️ Time: O(1) Space: O(k)

Since there are only 36 possible sequential digit numbers (1-9, 12-89, 123-789, etc.), generate them all systematically using the base string '123456789' and sliding windows.

Brute Force Check — Algorithm Steps

Step 1: Loop through all numbers from low to high
Step 2: For each number, check if digits are sequential
Step 3: Add valid numbers to result list

Visualization

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Step-by-Step Walkthrough

Check 100

Digits: 1,0,0 - not sequential (0 ≠ 1+1)

Check 123

Digits: 1,2,3 - sequential! (2=1+1, 3=2+1)

Result

Add 123 to result list

Code -

solution.c — C

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

int hasSequentialDigits(int num) {
    char s[20];
    sprintf(s, "%d", num);
    int len = strlen(s);
    if (len == 1) return 1;
    for (int i = 1; i < len; i++) {
        if (s[i] - s[i-1] != 1) {
            return 0;
        }
    }
    return 1;
}

int main() {
    int low, high;
    scanf("%d", &low);
    scanf("%d", &high);
    
    int result[1000];
    int count = 0;
    
    for (int num = low; num <= high; num++) {
        if (hasSequentialDigits(num)) {
            result[count++] = num;
        }
    }
    
    printf("[");
    for (int i = 0; i < count; i++) {
        printf("%d", result[i]);
        if (i < count - 1) printf(",");
    }
    printf("]\n");
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n × d)

n numbers in range, d digits per number to check

✓ Linear Growth

Space Complexity

O(k)

k sequential numbers found in the result list

✓ Linear Space

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Input & Output

Constraints

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Related Problems

Common Approaches

Brute Force Check — Algorithm Steps

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Code -

Time & Space Complexity

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