What is the time complexity of Next Greater Numerically Balanced Number?

The optimal solution has a time complexity of O(1).

What data structures are used to solve Next Greater Numerically Balanced Number?

This problem uses Hash Table, Math, Backtracking, Counting, Enumeration. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Next Greater Numerically Balanced Number?

Insight 1: Pre-calculation: Since the input range is limited (0 to 10^6), all possible numerically balanced numbers within a reasonable upper bound can be pre-calculated and stored. Insight 2: Generating balanced numbers: Efficiently generating balanced numbers requires exploring combinations of digits and their corresponding counts. We only need to consider digits from 1 to 9.

Which companies ask LeetCode 2048. Next Greater Numerically Balanced Number in interviews?

This problem is frequently asked at Sprinklr.

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Next Greater Numerically Balanced Number - LeetCode 2048 Solution

Next Greater Numerically Balanced Number - Complete Solution Guide

Next Greater Numerically Balanced Number is LeetCode problem 2048, 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

An integer x is numerically balanced if for every digit d in the number x , there are exactly d occurrences of that digit in x . Given an integer n , return the smallest numerically balanced number strictly greater than n . Example 1: Input: n = 1 Output: 22 Explanation: 22 is numerically balanced since: - The digit 2 occurs 2 times. It is also the smallest numerically balanced number strictly greater than 1. Example 2: Input: n = 1000 Output: 1333 Explanation: 1333 is numerically balanced since

Detailed Explanation

The problem defines a 'numerically balanced' number as one where each digit 'd' appears exactly 'd' times in the number. The goal is to find the smallest numerically balanced number that is strictly greater than a given input number 'n'. The input 'n' is constrained to be between 0 and 10^6.

Solution Approach

The solution pre-calculates all possible numerically balanced numbers and then uses binary search to find the smallest one greater than the input 'n'. The pre-calculation involves finding combinations of digits and their corresponding counts, generating permutations of these combinations, and then converting these permutations into integers. The resulting set of balanced numbers is sorted for efficient binary search.

Step-by-Step Algorithm

Step 1: Generate Beautiful Numbers: Define a function `_generate_beautiful_numbers` (or equivalent in each language). This function first generates combinations of digits (1-9) where the sum of the digits' values equals the length of the number. The combinations are used to construct strings of digits repeated according to their value. For example, the combination [1, 2] would become the string "122".
Step 2: Generate Permutations: For each generated string, calculate all unique permutations of the characters within the string. Convert each permutation into an integer.
Step 3: Store and Sort: Store all generated integers in a set to remove duplicates. Convert the set into a sorted list or array. The list is sorted in ascending order to allow for efficient binary search.
Step 4: Binary Search: In the `nextBeautifulNumber` function, perform a binary search on the pre-calculated list of beautiful numbers to find the smallest number that is strictly greater than 'n'. Specifically, use `bisect_right` in python or equivalent methods in other languages that finds the insertion point of the input in the list.
Step 5: Return Result: Return the beautiful number found by the binary search.

Key Insights

Insight 1: Pre-calculation: Since the input range is limited (0 to 10^6), all possible numerically balanced numbers within a reasonable upper bound can be pre-calculated and stored.
Insight 2: Generating balanced numbers: Efficiently generating balanced numbers requires exploring combinations of digits and their corresponding counts. We only need to consider digits from 1 to 9.
Insight 3: Permutations: After forming a candidate balanced number string from digit-count combinations, we need to generate all its distinct permutations and convert each to an integer to explore all possibilities.
Insight 4: Upper Bound: The maximum number of digits in a balanced number will be 7, since 1+2+3+4+5+6+7+8+9 = 45 > 7. Therefore, consider generating number up to 7 digits.
Insight 5: Efficient Search: Binary search is ideal for finding the smallest balanced number greater than n within the pre-calculated and sorted list of balanced numbers.

Complexity Analysis

Time Complexity: O(1)

Space Complexity: O(1)

Topics

This problem involves: Hash Table, Math, Backtracking, Counting, Enumeration.

Companies

Asked at: Sprinklr.

Next Greater Numerically Balanced Number | LeetCode 2048

Medium

Hash Table Math Backtracking Counting Enumeration

Time: O(1)Space: O(1)

Solution Code

Available in:

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

class Solution {

    private static List<Integer> _beautiful_numbers = generateBeautifulNumbers();

    private static List<Integer> generateBeautifulNumbers() {
        Set<Integer> res = new HashSet<>();

        function findCombinations = (combination, start) -> {
            int length = combination.stream().mapToInt(Integer::intValue).sum();

            if (length > 7) {
                return;
            }

            if (!combination.isEmpty()) {
                StringBuilder s = new StringBuilder();
                for (int d : combination) {
                    s.append(String.valueOf(d).repeat(d));
                }
                Set<String> permutations = permutations(s.toString());
                for (String p : permutations) {
                    try {
                        res.add(Integer.parseInt(p));
                    } catch (NumberFormatException e) {
                        //ignore very large numbers
                    }
                }
            }

            for (int i = start; i < 10; i++) {
                if (length + i > 7) {
                    break;
                }
                combination.add(i);
                findCombinations.apply(combination, i + 1);
                combination.remove(combination.size() - 1);
            }
        };

        findCombinations.apply(new ArrayList<>(), 1);
        List<Integer> sortedList = new ArrayList<>(res);
        Collections.sort(sortedList);
        return sortedList;
    }

    interface function {
        void apply(List<Integer> combination, int start);
    }


    private static Set<String> permutations(String str) {
        Set<String> permutations = new HashSet<>();
        permute(str, 0, str.length() - 1, permutations);
        return permutations;
    }

    private static void permute(String str, int l, int r, Set<String> permutations) {
        if (l == r) {
            permutations.add(str);
        } else {
            for (int i = l; i <= r; i++) {
                str = swap(str, l, i);
                permute(str, l + 1, r, permutations);
                str = swap(str, l, i); // backtrack
            }
        }
    }

    private static String swap(String a, int i, int j) {
        char temp;
        char[] charArray = a.toCharArray();
        temp = charArray[i];
        charArray[i] = charArray[j];
        charArray[j] = temp;
        return String.valueOf(charArray);
    }
    
    public int nextBeautifulNumber(int n) {
        int idx = binarySearchRight(_beautiful_numbers, n);
        return _beautiful_numbers.get(idx);
    }

    private int binarySearchRight(List<Integer> sortedList, int target) {
        int low = 0;
        int high = sortedList.size() - 1;

        while (low <= high) {
            int mid = low + (high - low) / 2;
            if (sortedList.get(mid) <= target) {
                low = mid + 1;
            } else {
                high = mid - 1;
            }
        }
        return low;
    }
}

View Full Solution on GitHubGitHub Solution View on LeetCodeLeetCode

Problem Info

Description

Problem 2048: Next Greater Numerically Balanced Number - Detailed explanation and solution approaches.

Topics

Hash Table Math Backtracking Counting Enumeration

Companies

Sprinklr

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:2048

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

5 steps

1
Step 1: Generate Beautiful Numbers: Define a function `_generate_beautiful_numbers` (or equivalent in each language). This function first generates combinations of digits (1-9) where the sum of the digits' values equals the length of the number. The combinations are used to construct strings of digits repeated according to their value. For example, the combination [1, 2] would become the string "122".
2
Step 2: Generate Permutations: For each generated string, calculate all unique permutations of the characters within the string. Convert each permutation into an integer.
3
Step 3: Store and Sort: Store all generated integers in a set to remove duplicates. Convert the set into a sorted list or array. The list is sorted in ascending order to allow for efficient binary search.
4
Step 4: Binary Search: In the `nextBeautifulNumber` function, perform a binary search on the pre-calculated list of beautiful numbers to find the smallest number that is strictly greater than 'n'. Specifically, use `bisect_right` in python or equivalent methods in other languages that finds the insertion point of the input in the list.
5
Step 5: Return Result: Return the beautiful number found by the binary search.

💡

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(1)

⚡ Constant time - Super fast! Same speed regardless of input size.

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The time complexity is O(1) because the beautiful numbers are generated once and stored. The subsequent call to `nextBeautifulNumber` then just requires a single binary search in the small pre-calculated list. The space complexity is O(1) since the pre-calculated list is fixed in size and independent of input size.

📖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

Brute Force

Starting from n+1, check each subsequent number for numerically balanced property until a balanced number is found. Check each number by counting the occurrences of each digit and comparing it to the digit value. This continues until the upper limit is reached.

Time: O(N * log N), where N is the difference between n and the next beautiful number.Space: O(log N) for storing digit counts.

Trade-offs: This is less efficient due to the need to check each subsequent number. Only suitable if the problem constraints significantly reduce the search space such that N is very small.

✨

Key Insights

1Insight 1: Pre-calculation: Since the input range is limited (0 to 10^6), all possible numerically balanced numbers within a reasonable upper bound can be pre-calculated and stored.
2Insight 2: Generating balanced numbers: Efficiently generating balanced numbers requires exploring combinations of digits and their corresponding counts. We only need to consider digits from 1 to 9.
3Insight 3: Permutations: After forming a candidate balanced number string from digit-count combinations, we need to generate all its distinct permutations and convert each to an integer to explore all possibilities.
4Insight 4: Upper Bound: The maximum number of digits in a balanced number will be 7, since 1+2+3+4+5+6+7+8+9 = 45 > 7. Therefore, consider generating number up to 7 digits.
5Insight 5: Efficient Search: Binary search is ideal for finding the smallest balanced number greater than n within the pre-calculated and sorted list of balanced numbers.

📚

Key Terms Explained

For Students

Hash Table

A data structure that maps keys to values for quick lookup. Like a dictionary!

Edge Cases

Edge case 1: n = 0: The code correctly finds the next beautiful number, which is 1.
Edge case 2: n is already a beautiful number: The code correctly finds the next *strictly greater* beautiful number.
Edge case 3: n is close to the upper bound: The pre-calculation ensures that the next beautiful number exists.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not generating *all* permutations of the digit combinations, resulting in missing some potential balanced numbers.

✗Common mistake 2: Failing to remove duplicate balanced numbers. Using a set avoids duplicates.

✗Common mistake 3: Incorrect implementation of the binary search or not handling the edge case where n is already a beautiful number.

✗Common mistake 4: Integer Overflow: Failing to handle potential integer overflow issues during permutation generation or number conversion, especially when dealing with larger numbers (e.g., in Java).

✅

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: Number plate generation: In certain regions, number plates might need to adhere to a specific pattern or rule (though numerically balanced is unlikely!). The idea of generating numbers meeting certain criteria is relevant.

🏦Real-world scenario 2: Data generation: When generating test datasets, one may want to include data that is balanced in some way, ensuring that certain criteria are met (e.g., balancing the occurrence of values).

🎮Real-world scenario 3: Cryptography, where number patterns and properties can be critical for security, although specifically balanced numbers are probably not of direct use.

💡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

Hash tables trade space for time. Before using one, consider if the problem constraints allow the extra memory. In interviews, always mention this trade-off.

🎯 Interview Tip

Consider multiple approaches before coding. Mentioning alternatives shows breadth of knowledge.

⚠️ Watch Out

Don't forget to handle key collisions and null/undefined keys. Consider what happens with duplicate 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 8, 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

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

@CodeNinjaJul 14

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

@TechLeadJul 7

❓

Frequently Asked Questions

What are the trade-offs of using a hash table?

Hash tables provide O(1) average-case lookup, insertion, and deletion, making them ideal for frequency counting and duplicate detection. However, they use O(n) extra space and may degrade to O(n) time in worst-case scenarios with hash collisions. Always consider if the space trade-off is acceptable.

What techniques are commonly tested in medium-difficulty problems?

Medium problems often require combining multiple techniques: two-pointer with hash maps, BFS/DFS with state tracking, dynamic programming with optimization, or divide-and-conquer approaches. They test your ability to recognize patterns and choose the right combination of techniques.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Next Greater Numerically Balanced Number"?

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

🤖 Need help with this problem?