What is the time complexity of Number of Beautiful Integers in the Range?

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

What is the space complexity of Number of Beautiful Integers in the Range?

The space complexity is O(n*k*n^2).

What data structures are used to solve Number of Beautiful Integers in the Range?

This problem uses Math, Dynamic Programming. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Number of Beautiful Integers in the Range?

Insight 1: The problem can be solved efficiently using dynamic programming (specifically, digit DP) due to the overlapping subproblems arising from considering digits individually and the constraints on 'k'. Insight 2: We can decompose the problem into counting beautiful numbers up to 'high' and then subtracting the count of beautiful numbers up to 'low - 1'. This simplifies the range counting problem to a prefix counting problem.

Which companies ask LeetCode 2827. Number of Beautiful Integers in the Range in interviews?

This problem is frequently asked at Infosys.

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Number of Beautiful Integers in the Range - LeetCode 2827 Solution

Q: How to solve Number of Beautiful Integers in the Range?

The provided solution employs a digit DP approach to efficiently count beautiful integers. It uses recursion with memoization (lru_cache in Python, Integer[][][][][] in Java, vector > dp in C++) to store and reuse results of overlapping subproblems. The core idea is to build numbers digit by digit, keeping track of the current position, whether the digits chosen so far match the given number (tight constraint), the count of even and odd digits seen so far, and the remainder when divided by 'k'. By exploring all possible digits for each position, we can determine the number of beautiful integers that can be formed.

Number of Beautiful Integers in the Range - Complete Solution Guide

Number of Beautiful Integers in the Range is LeetCode problem 2827, a Hard level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

You are given positive integers low , high , and k . A number is beautiful if it meets both of the following conditions: The count of even digits in the number is equal to the count of odd digits. The number is divisible by k . Return the number of beautiful integers in the range [low, high] . Example 1: Input: low = 10, high = 20, k = 3 Output: 2 Explanation: There are 2 beautiful integers in the given range: [12,18]. - 12 is beautiful because it contains 1 odd digit and 1 even digit, and is di

Detailed Explanation

The problem asks us to count the number of 'beautiful' integers within a given range [low, high]. A number is considered beautiful if it satisfies two conditions: 1) the number of even digits equals the number of odd digits in the integer, and 2) the integer is divisible by a given number 'k'. The inputs are low, high, and k, and the output is the count of beautiful numbers in the range [low, high]. Constraints include 0 < low <= high <= 10^9 and 0 < k <= 20.

Solution Approach

The provided solution employs a digit DP approach to efficiently count beautiful integers. It uses recursion with memoization (lru_cache in Python, Integer[][][][][] in Java, vector<vector<...>> dp in C++) to store and reuse results of overlapping subproblems. The core idea is to build numbers digit by digit, keeping track of the current position, whether the digits chosen so far match the given number (tight constraint), the count of even and odd digits seen so far, and the remainder when divided by 'k'. By exploring all possible digits for each position, we can determine the number of beautiful integers that can be formed.

Step-by-Step Algorithm

Step 1: Define a recursive function `dp(pos, is_tight, even_count, odd_count, rem, is_leading_zero)` that represents the number of beautiful integers formed from position `pos` onwards, given the current state (is_tight, even/odd digit counts, remainder, and leading zero status).
Step 2: The base case for the recursion is when `pos == n` (we have considered all digits). If `even_count == odd_count` and `rem == 0` and `is_leading_zero` is false, it means we have found a beautiful number, so return 1; otherwise, return 0.
Step 3: In the recursive step, iterate through all possible digits (0 to 9) or up to the digit at the current position of input number if `is_tight` is true.
Step 4: Update the state variables for each digit: `is_tight` becomes true if the current digit equals the digit in input number at the current position. Increment `even_count` or `odd_count` based on whether the digit is even or odd. Update the remainder `rem` by multiplying the previous remainder by 10 and adding the current digit, modulo `k`.
Step 5: Recursively call the `dp` function with the updated state and add the result to the `res` variable. The case of leading zeros requires special handling to prevent incorrect counts.
Step 6: Memoize the result of the `dp` function to avoid redundant calculations.
Step 7: The `solve` function converts the input integer to a string and calls the `dp` function starting from the leftmost digit (position 0).
Step 8: Finally, to find the number of beautiful integers in the range [low, high], compute `solve(str(high)) - solve(str(low - 1))`, where `str` represents the conversion of an integer to a string.

Key Insights

Insight 1: The problem can be solved efficiently using dynamic programming (specifically, digit DP) due to the overlapping subproblems arising from considering digits individually and the constraints on 'k'.
Insight 2: We can decompose the problem into counting beautiful numbers up to 'high' and then subtracting the count of beautiful numbers up to 'low - 1'. This simplifies the range counting problem to a prefix counting problem.
Insight 3: Handling the 'tight' constraint (ensuring the generated numbers don't exceed the upper bound) and leading zeros are crucial aspects of digit DP, as they affect the state transitions and valid counts.

Complexity Analysis

Time Complexity: O(n*k*n^2)

Space Complexity: O(n*k*n^2)

Topics

This problem involves: Math, Dynamic Programming.

Companies

Asked at: Infosys.

Number of Beautiful Integers in the Range | LeetCode 2827

Hard

Math Dynamic Programming

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

Solution Code

Available in:

import java.util.Arrays;

class Solution {
    private int k;

    public int numberOfBeautifulIntegers(int low, int high, int k) {
        this.k = k;
        return solve(String.valueOf(high)) - solve(String.valueOf(low - 1));
    }

    private int solve(String s) {
        int n = s.length();
        Integer[][][][][] dp = new Integer[n][2][n + 1][n + 1][k];
        return dp(0, 1, 0, 0, 0, 1, s, dp);
    }

    private int dp(int pos, int isTight, int evenCount, int oddCount, int rem, int isLeadingZero, String s, Integer[][][][][] dp) {
        int n = s.length();
        if (pos == n) {
            return (isLeadingZero == 0 && evenCount == oddCount && rem == 0) ? 1 : 0;
        }

        if (dp[pos][isTight][evenCount][oddCount][rem] != null) {
            return dp[pos][isTight][evenCount][oddCount][rem];
        }

        int res = 0;
        int upperBound = (isTight == 1) ? (s.charAt(pos) - '0') : 9;

        for (int digit = 0; digit <= upperBound; digit++) {
            int newTight = (isTight == 1 && digit == upperBound) ? 1 : 0;

            if (isLeadingZero == 1 && digit == 0) {
                res += dp(pos + 1, newTight, evenCount, oddCount, rem, 1, s, dp);
            } else {
                int newEvenCount = evenCount + ((digit % 2 == 0) ? 1 : 0);
                int newOddCount = oddCount + ((digit % 2 != 0) ? 1 : 0);
                int newRem = (rem * 10 + digit) % k;
                res += dp(pos + 1, newTight, newEvenCount, newOddCount, newRem, 0, s, dp);
            }
        }

        return dp[pos][isTight][evenCount][oddCount][rem] = res;
    }
}

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Problem Info

Description

Problem 2827: Number of Beautiful Integers in the Range - Detailed explanation and solution approaches.

Topics

Math Dynamic Programming

Companies

Infosys

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:2827

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

8 steps

1
Step 1: Define a recursive function `dp(pos, is_tight, even_count, odd_count, rem, is_leading_zero)` that represents the number of beautiful integers formed from position `pos` onwards, given the current state (is_tight, even/odd digit counts, remainder, and leading zero status).
2
Step 2: The base case for the recursion is when `pos == n` (we have considered all digits). If `even_count == odd_count` and `rem == 0` and `is_leading_zero` is false, it means we have found a beautiful number, so return 1; otherwise, return 0.
3
Step 3: In the recursive step, iterate through all possible digits (0 to 9) or up to the digit at the current position of input number if `is_tight` is true.
4
Step 4: Update the state variables for each digit: `is_tight` becomes true if the current digit equals the digit in input number at the current position. Increment `even_count` or `odd_count` based on whether the digit is even or odd. Update the remainder `rem` by multiplying the previous remainder by 10 and adding the current digit, modulo `k`.
5
Step 5: Recursively call the `dp` function with the updated state and add the result to the `res` variable. The case of leading zeros requires special handling to prevent incorrect counts.
6
Step 6: Memoize the result of the `dp` function to avoid redundant calculations.
7
Step 7: The `solve` function converts the input integer to a string and calls the `dp` function starting from the leftmost digit (position 0).
8
Step 8: Finally, to find the number of beautiful integers in the range [low, high], compute `solve(str(high)) - solve(str(low - 1))`, where `str` represents the conversion of an integer to a string.

💡

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(n*k*n^2)

💾

Space Complexity

O(n*k*n^2)

📝

Why?

The time complexity stems from the nested loops within the recursive DP function and the size of the DP table. The `dp` function is called for each possible state (pos, is_tight, even_count, odd_count, rem, is_leading_zero), and the complexity of calculating each state is O(1) after memoization. Therefore, the overall time complexity is proportional to the number of states. The space complexity is directly related to the amount of memory used to store the DP table, which has the same dimensions as the number of states.

📖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

Iterate through all numbers in the range [low, high], check if they are beautiful (count even/odd digits and check divisibility by k).

Time: O((high - low) * log(high)), where log(high) is the time to count even and odd digits.Space: O(1)

Trade-offs: This is much simpler to implement, but it is highly inefficient for larger ranges because it doesn't reuse computation or exploit overlapping subproblems. This can work when 'high' is small.

✨

Key Insights

1Insight 1: The problem can be solved efficiently using dynamic programming (specifically, digit DP) due to the overlapping subproblems arising from considering digits individually and the constraints on 'k'.
2Insight 2: We can decompose the problem into counting beautiful numbers up to 'high' and then subtracting the count of beautiful numbers up to 'low - 1'. This simplifies the range counting problem to a prefix counting problem.
3Insight 3: Handling the 'tight' constraint (ensuring the generated numbers don't exceed the upper bound) and leading zeros are crucial aspects of digit DP, as they affect the state transitions and valid counts.

📚

Key Terms Explained

For Students

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

Edge Cases

Edge case 1: low = 0. The code correctly calculates the number of beautiful integers up to high and subtracts the count up to low-1 = -1 (handled implicitly correctly as if there are 0 beautiful numbers before 1)
Edge case 2: k = 1. All numbers are divisible by 1, so the code correctly counts the numbers with an equal number of even and odd digits.
Edge case 3: Single digit numbers. The code handles numbers with one digit because `even_count` and `odd_count` can be equal to 0 when leading zeros are present. If there are no leading zeros and even_count == odd_count, and the number is divisible by k, then it will correctly increment the count.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not handling the 'tight' constraint correctly, which can lead to generating numbers that are larger than the input number.

✗Common mistake 2: Forgetting to memoize the results of the `dp` function, which can significantly increase the runtime due to redundant calculations. Without memoization, the solution may lead to TLE (Time Limit Exceeded).

✗Common mistake 3: Not correctly handling the leading zero case. When leading zeros are present, `even_count` and `odd_count` should not be updated. Also, the divisibility by k condition should not be checked until all digits are processed.

✗Common mistake 4: Integer overflow when calculating remainders, especially in languages like C/C++. Using modulo operator at each step avoids this.

✅

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: This type of digit DP can be applied to problems involving counting valid numbers within a given range that satisfy specific digit-based criteria or divisibility constraints, such as generating specific types of identifiers or codes.

🏦Real-world scenario 2: In cryptography, some algorithms require generating numbers with certain properties. Digit DP could be useful in counting numbers fitting those requirements efficiently.

💡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

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

It's okay to not finish hard problems in interviews. Show your problem-solving process.

⚠️ 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: October 6, 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

Should I use top-down (memoization) or bottom-up (tabulation)?

Start with top-down (recursion + memoization) as it directly maps to the recursive thinking and handles only needed subproblems. Convert to bottom-up for better space efficiency and to avoid stack overflow. Bottom-up is often faster due to no recursion overhead.

Is it okay to not fully solve hard problems in interviews?

Yes! Interviewers care more about your problem-solving process than the final answer. Demonstrate: clear communication, systematic approach, ability to recognize when to pivot strategies, and graceful handling of hints. Partial progress with good reasoning is valued.

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 "Number of Beautiful Integers in the Range"?

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

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