LeetCode 1291: Sequential Digits Solution

Master LeetCode problem 1291 (Sequential Digits), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

1291. Sequential Digits

Medium

Enumeration

Problem Explanation

Explanation

To solve this problem, we can generate all possible sequential digits starting from 1 to 9, and then iterate through these numbers to find those within the range [low, high]. We can achieve this by using a queue to store the sequential digits and keep generating new numbers by appending the next digit if it is less than or equal to 9.

Algorithmic Idea

Create a queue and add all single-digit sequential numbers from 1 to 9.
While the queue is not empty, dequeue a number, check if it is within the range [low, high], and add it to the result list if it satisfies the condition.
If the current number's last digit is less than 9, append a new digit to create the next sequential number and enqueue it.
Continue this process until the queue is empty.

Time Complexity

The time complexity of this solution is O(1) since the maximum number of sequential digits to be generated is fixed.

Space Complexity

The space complexity is O(1) as we are only storing a fixed number of sequential digits at any given time.

Solution Code

class Solution {
    public List<Integer> sequentialDigits(int low, int high) {
        List<Integer> result = new ArrayList<>();
        Queue<Integer> queue = new LinkedList<>();
        
        for (int i = 1; i <= 9; i++) {
            queue.offer(i);
        }
        
        while (!queue.isEmpty()) {
            int num = queue.poll();
            if (num >= low && num <= high) {
                result.add(num);
            }
            int lastDigit = num % 10;
            if (lastDigit < 9) {
                queue.offer(num * 10 + lastDigit + 1);
            }
        }
        
        return result;
    }
}

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