What is the time complexity of Numbers With Same Consecutive Differences?

The optimal solution has a time complexity of O(2^n).

What is the space complexity of Numbers With Same Consecutive Differences?

The space complexity is O(2^n).

What data structures are used to solve Numbers With Same Consecutive Differences?

This problem uses Backtracking, Breadth-First Search. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Numbers With Same Consecutive Differences?

Insight 1: Backtracking is well-suited for generating all possible numbers meeting the given criteria because we explore different digits at each position. Insight 2: Using a set (or a boolean array in C) avoids duplicate computations when k=0, preventing infinite recursion or redundant results.

Which companies ask LeetCode 967. Numbers With Same Consecutive Differences in interviews?

This problem is frequently asked at Flipkart.

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Numbers With Same Consecutive Differences - LeetCode 967 Solution

Numbers With Same Consecutive Differences - Complete Solution Guide

Numbers With Same Consecutive Differences is LeetCode problem 967, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

Given two integers n and k, return an array of all the integers of length n where the difference between every two consecutive digits is k . You may return the answer in any order . Note that the integers should not have leading zeros. Integers as 02 and 043 are not allowed. Example 1: Input: n = 3, k = 7 Output: [181,292,707,818,929] Explanation: Note that 070 is not a valid number, because it has leading zeroes. Example 2: Input: n = 2, k = 1 Output: [10,12,21,23,32,34,43,45,54,56,65,67,76,78,

Detailed Explanation

The problem asks us to generate all integers of a given length 'n' such that the absolute difference between any two consecutive digits in each integer is equal to a given value 'k'. The output should be an array of these integers. Leading zeros are not allowed, meaning integers like '02' or '043' are invalid. The inputs are the length 'n' and the difference 'k', and the output is a list of integers that satisfy the given conditions. The constraints specify that 'n' is between 2 and 9, and 'k' is between 0 and 9.

Solution Approach

The solution uses a Depth-First Search (DFS) algorithm to explore all possible numbers that satisfy the given conditions. It starts with the first digit ranging from 1 to 9 and recursively appends digits such that the absolute difference between the consecutive digits equals 'k'. The DFS function maintains the current length of the number being generated and the current number itself. Once the length reaches 'n', the number is added to the result list.

Step-by-Step Algorithm

Step 1: Initialize an empty list `result` to store valid numbers.
Step 2: Define a recursive DFS function `dfs(length, num)` where `length` is the current number of digits and `num` is the current number being formed.
Step 3: Base case for the DFS: If `length` equals `n`, add `num` to `result` and return.
Step 4: Get the last digit of `num` using the modulo operator (`num % 10`).
Step 5: Calculate potential next digits by adding and subtracting `k` from the last digit.
Step 6: Use a set (or boolean array in C) to store the possible next digits to avoid duplicates when `k` is 0.
Step 7: Iterate through the potential next digits in the set.
Step 8: For each valid next digit (between 0 and 9), construct a new number by appending the digit to the current number (`new_num = num * 10 + next_digit`).
Step 9: Recursively call `dfs(length + 1, new_num)` to explore further digits.
Step 10: Iterate through the digits 1 to 9 and call the DFS function for each as the starting digit `dfs(1, i)`.
Step 11: Return the `result` list.

Key Insights

Insight 1: Backtracking is well-suited for generating all possible numbers meeting the given criteria because we explore different digits at each position.
Insight 2: Using a set (or a boolean array in C) avoids duplicate computations when k=0, preventing infinite recursion or redundant results.
Insight 3: The first digit cannot be 0, so the search space starts from 1 to 9.

Complexity Analysis

Time Complexity: O(2^n)

Space Complexity: O(2^n)

Topics

This problem involves: Backtracking, Breadth-First Search.

Companies

Asked at: Flipkart.

Numbers With Same Consecutive Differences | LeetCode 967

Medium

Backtracking Breadth-First Search

Time: O(2^n)Space: O(2^n)

Solution Code

Available in:

import java.util.ArrayList;
import java.util.HashSet;
import java.util.List;
import java.util.Set;

class Solution {
    public int[] numsSameConsecDiff(int n, int k) {
        List<Integer> result = new ArrayList<>();

        function dfs(int length, int num, List<Integer> result, int n, int k) {
            if (length == n) {
                result.add(num);
                return;
            }

            int lastDigit = num % 10;

            // The next digit can be last_digit + k or last_digit - k.
            // Using a set handles the case where k=0, so we don't process the same next digit twice.
            Set<Integer> nextDigits = new HashSet<>();
            nextDigits.add(lastDigit + k);
            nextDigits.add(lastDigit - k);

            for (int nextD : nextDigits) {
                if (0 <= nextD && nextD <= 9) {
                    int newNum = num * 10 + nextD;
                    dfs(length + 1, newNum, result, n, k);
                }
            }
        }

        // The first digit cannot be 0, so we start our DFS from 1 to 9.
        for (int i = 1; i <= 9; i++) {
            dfs(1, i, result, n, k);
        }

        int[] resultArray = new int[result.size()];
        for (int i = 0; i < result.size(); i++) {
            resultArray[i] = result.get(i);
        }
        return resultArray;
    }
    
    private void dfs(int length, int num, List<Integer> result, int n, int k) {
        if (length == n) {
            result.add(num);
            return;
        }

        int lastDigit = num % 10;
        
        // The next digit can be last_digit + k or last_digit - k.
        // Using a set handles the case where k=0, so we don't process the same next digit twice.
        Set<Integer> nextDigits = new HashSet<>();
        nextDigits.add(lastDigit + k);
        nextDigits.add(lastDigit - k);
        
        for (int nextD : nextDigits) {
            if (0 <= nextD && nextD <= 9) {
                int newNum = num * 10 + nextD;
                dfs(length + 1, newNum, result, n, k);
            }
        }
    }
}

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Problem Info

Description

Problem 967: Numbers With Same Consecutive Differences - Detailed explanation and solution approaches.

Topics

Backtracking Breadth-First Search

Companies

Flipkart

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:967

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

11 steps

1
Step 1: Initialize an empty list `result` to store valid numbers.
2
Step 2: Define a recursive DFS function `dfs(length, num)` where `length` is the current number of digits and `num` is the current number being formed.
3
Step 3: Base case for the DFS: If `length` equals `n`, add `num` to `result` and return.
4
Step 4: Get the last digit of `num` using the modulo operator (`num % 10`).
5
Step 5: Calculate potential next digits by adding and subtracting `k` from the last digit.
6
Step 6: Use a set (or boolean array in C) to store the possible next digits to avoid duplicates when `k` is 0.
7
Step 7: Iterate through the potential next digits in the set.
8
Step 8: For each valid next digit (between 0 and 9), construct a new number by appending the digit to the current number (`new_num = num * 10 + next_digit`).
9
Step 9: Recursively call `dfs(length + 1, new_num)` to explore further digits.
10
Step 10: Iterate through the digits 1 to 9 and call the DFS function for each as the starting digit `dfs(1, i)`.
11
Step 11: Return the `result` list.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(2^n)

🐌 Exponential - Very slow! Often needs optimization.

💾

Space Complexity

O(2^n)

📝

Why?

The time complexity is exponential because, at each digit position, we have up to two choices for the next digit (last_digit + k or last_digit - k). The DFS explores all possible combinations. The space complexity accounts for the storage of the resulting numbers (which can grow exponentially) and the call stack during the recursion.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Breadth-First Search (BFS)

Instead of using DFS, you could use BFS. Starting with the initial digits (1-9), you would add them to a queue. Then, you would iteratively process each number in the queue by generating possible next digits and adding new numbers to the queue until you reach the desired length 'n'.

Time: O(2^n)Space: O(2^n)

Trade-offs: BFS and DFS have similar time and space complexity for this problem. The choice between them might depend on personal preference. BFS could be argued to be slightly better if the number of possible numbers is expected to be lower, due to its level-by-level exploration.

✨

Key Insights

1Insight 1: Backtracking is well-suited for generating all possible numbers meeting the given criteria because we explore different digits at each position.
2Insight 2: Using a set (or a boolean array in C) avoids duplicate computations when k=0, preventing infinite recursion or redundant results.
3Insight 3: The first digit cannot be 0, so the search space starts from 1 to 9.

📚

Key Terms Explained

For Students

Edge Cases

Edge case 1: k = 0. The solution correctly handles this by using a set (or a boolean array in C) to avoid adding the same digit twice, which would lead to infinite recursion.
Edge case 2: n = 1. While the problem states n >= 2, handling the case n=1 would only require returning digits 0-9. This is not handled in the solution but is a trivial extension.
Edge case 3: The result can contain duplicate if `k == 0`. The approach addresses this by using a set to keep only unique next digits. Without the set it is possible to have duplicate numbers if a number can be created from two different starting points.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not handling the case when `k=0` correctly, leading to infinite recursion.

✗Common mistake 2: Forgetting to check if the next digit is within the valid range (0 to 9).

✗Common mistake 3: Starting the DFS from 0, resulting in numbers with leading zeros.

✗Common mistake 4: Allocating insufficiently large array in C.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Generating valid ID numbers or serial numbers with specific digit difference constraints.

🏦Real-world scenario 2: Creating test cases for financial simulations where specific patterns in numbers are required.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

Backtracking = DFS + pruning. The key is identifying when to backtrack. Always restore state after recursion.

🎯 Interview Tip

If stuck, try working through a small example by hand. The pattern often reveals itself.

⚠️ Watch Out

Always test your solution with edge cases: empty input, single element, and boundary values.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: August 7, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 14

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 7

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJun 30

❓

Frequently Asked Questions

How do I avoid duplicate solutions in backtracking?

Sort the input first, then skip consecutive duplicates at the same recursion level: if (i > start && nums[i] == nums[i-1]) continue. For permutations, use a visited array or swap-based approach. Always consider if order matters (permutations vs combinations).

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Numbers With Same Consecutive Differences"?

For this medium-level Backtracking problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether breadth-first search techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?