What is the time complexity of Minimum Addition to Make Integer Beautiful?

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

What data structures are used to solve Minimum Addition to Make Integer Beautiful?

This problem uses Math, Greedy. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Minimum Addition to Make Integer Beautiful?

Insight 1: The core idea is to progressively round 'n' up to the nearest power of 10, starting from the least significant digit. We repeatedly add to 'n' until its digit sum is less than or equal to the target. Insight 2: The `digit_sum` function efficiently calculates the sum of the digits of an integer.

Which companies ask LeetCode 2457. Minimum Addition to Make Integer Beautiful in interviews?

This problem is frequently asked at Infosys.

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Minimum Addition to Make Integer Beautiful - LeetCode 2457 Solution

Minimum Addition to Make Integer Beautiful - Complete Solution Guide

Minimum Addition to Make Integer Beautiful is LeetCode problem 2457, 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

You are given two positive integers n and target . An integer is considered beautiful if the sum of its digits is less than or equal to target . Return the minimum non-negative integer x such that n + x is beautiful . The input will be generated such that it is always possible to make n beautiful. Example 1: Input: n = 16, target = 6 Output: 4 Explanation: Initially n is 16 and its digit sum is 1 + 6 = 7. After adding 4, n becomes 20 and digit sum becomes 2 + 0 = 2. It can be shown that we can n

Detailed Explanation

The problem requires finding the smallest non-negative integer 'x' that, when added to a given integer 'n', results in a 'beautiful' integer. An integer is considered 'beautiful' if the sum of its digits is less than or equal to a given 'target' value. The input 'n' is a positive integer, and 'target' is a positive integer. The constraints guarantee that a solution always exists.

Solution Approach

The solution uses a greedy approach. It starts by checking if the initial number 'n' is already beautiful. If not, it iteratively adds a value 'x' to 'n' to make the last digits zero, effectively rounding 'n' up to the nearest power of 10. This process continues until the resulting number's digit sum is less than or equal to 'target'. The algorithm finds the smallest 'x' by incrementing the power of 10 being considered. The loop stops when `new_n` becomes 'beautiful'.

Step-by-Step Algorithm

Step 1: Define a helper function `digit_sum(num)` that calculates the sum of the digits of a given number 'num'.
Step 2: Check if the initial number 'n' is already beautiful (i.e., `digit_sum(n) <= target`). If it is, return 0.
Step 3: Initialize a variable 'p' to 10. This represents the current power of 10 we are considering.
Step 4: Enter a `while` loop that continues until a beautiful number is found.
Step 5: Calculate the remainder of 'n' when divided by 'p'. This represents the part of 'n' that needs to be rounded up.
Step 6: Calculate 'x' as `(p - remainder) % p`. This ensures 'x' is the smallest non-negative integer that, when added to 'n', makes the last digits (up to the current power of 10) zero.
Step 7: Calculate the new number `new_n = n + x`.
Step 8: Check if `new_n` is beautiful (i.e., `digit_sum(new_n) <= target`). If it is, return 'x'.
Step 9: If `new_n` is not beautiful, multiply 'p' by 10 to consider the next power of 10 and continue the loop.

Key Insights

Insight 1: The core idea is to progressively round 'n' up to the nearest power of 10, starting from the least significant digit. We repeatedly add to 'n' until its digit sum is less than or equal to the target.
Insight 2: The `digit_sum` function efficiently calculates the sum of the digits of an integer.
Insight 3: The modulo operator (%) is used effectively to isolate the rightmost digits of 'n' and calculate the value 'x' needed to round it up.

Complexity Analysis

Time Complexity: O(log(n))

Space Complexity: O(1)

Topics

This problem involves: Math, Greedy.

Companies

Asked at: Infosys.

Minimum Addition to Make Integer Beautiful | LeetCode 2457

Medium

Math Greedy

Time: O(log(n))Space: O(1)

Solution Code

Available in:

class Solution {
    public long makeIntegerBeautiful(long n, int target) {
        
        
        if (digit_sum(n) <= target) {
            return 0;
        }
        
        long p = 10;
        while (true) {
            long remainder = n % p;
            long x = (p - remainder) % p;
            
            long new_n = n + x;
            
            if (digit_sum(new_n) <= target) {
                return x;
            }
            
            p *= 10;
        }
    }

    private int digit_sum(long num) {
        int s = 0;
        while (num > 0) {
            s += num % 10;
            num /= 10;
        }
        return s;
    }
}

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Problem Info

Description

Problem 2457: Minimum Addition to Make Integer Beautiful - Detailed explanation and solution approaches.

Topics

Math Greedy

Companies

Infosys

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:2457

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

9 steps

1
Step 1: Define a helper function `digit_sum(num)` that calculates the sum of the digits of a given number 'num'.
2
Step 2: Check if the initial number 'n' is already beautiful (i.e., `digit_sum(n) <= target`). If it is, return 0.
3
Step 3: Initialize a variable 'p' to 10. This represents the current power of 10 we are considering.
4
Step 4: Enter a `while` loop that continues until a beautiful number is found.
5
Step 5: Calculate the remainder of 'n' when divided by 'p'. This represents the part of 'n' that needs to be rounded up.
6
Step 6: Calculate 'x' as `(p - remainder) % p`. This ensures 'x' is the smallest non-negative integer that, when added to 'n', makes the last digits (up to the current power of 10) zero.
7
Step 7: Calculate the new number `new_n = n + x`.
8
Step 8: Check if `new_n` is beautiful (i.e., `digit_sum(new_n) <= target`). If it is, return 'x'.
9
Step 9: If `new_n` is not beautiful, multiply 'p' by 10 to consider the next power of 10 and continue the loop.

💡

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(log(n))

🚀 Logarithmic - Very efficient! Halves the problem each step.

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The `digit_sum` function takes O(log(n)) time because it iterates through the digits of 'n'. The `while` loop in `makeIntegerBeautiful` also iterates a number of times proportional to the number of digits in 'n', also O(log(n)). The space complexity is O(1) because we use only a constant amount of extra space, regardless of the size of the input 'n'.

📖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 possible values of 'x' starting from 0, adding 'x' to 'n' and checking if the resulting number is beautiful. Stop when a beautiful number is found.

Time: O(K * log(n+K)), where K is the value of x that results in beautiful numberSpace: O(1)

Trade-offs: This approach is simple to understand but can be very slow if the required 'x' is large. The primary solution is much more efficient.

✨

Key Insights

1Insight 1: The core idea is to progressively round 'n' up to the nearest power of 10, starting from the least significant digit. We repeatedly add to 'n' until its digit sum is less than or equal to the target.
2Insight 2: The `digit_sum` function efficiently calculates the sum of the digits of an integer.
3Insight 3: The modulo operator (%) is used effectively to isolate the rightmost digits of 'n' and calculate the value 'x' needed to round it up.

📚

Key Terms Explained

For Students

Greedy Algorithm

Making the best choice at each step without looking ahead.

Edge Cases

Edge case 1: When the input 'n' is already beautiful, the function correctly returns 0.
Edge case 2: Large values of 'n' (close to 10^12) are handled correctly because the algorithm iteratively rounds up to higher powers of 10 until a beautiful number is found.
Edge case 3: Target = 1. Requires the algorithm to aggressively round to higher powers of 10 until the leading digit is 1 or 0.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not handling the case where the initial number 'n' is already beautiful (i.e., forgetting to return 0 in this case).

✗Common mistake 2: Incorrectly calculating the value of 'x' needed to round 'n' up to the nearest power of 10.

✗Common mistake 3: Integer overflow when calculating `new_n` for very large `n`. (Although the prompt states the inputs are generated so this won't happen, good to be aware of.)

✅

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: Financial calculations where rounding to the nearest multiple of a certain value is required to simplify transactions or reporting.

🏦Real-world scenario 2: Optimizing resource allocation by rounding up resource requirements to the nearest available unit, while minimizing the total amount of resources used.

💡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

Medium problems are the bread and butter of technical interviews. Practice explaining your approach before coding.

⚠️ 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: September 26, 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

What are common greedy problem patterns?

Activity selection (sort by end time), interval scheduling (sort by start/end), Huffman coding (always combine smallest), fractional knapsack (sort by value/weight ratio), and minimum spanning tree. Look for problems where local decisions don't affect future options.

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 "Minimum Addition to Make Integer Beautiful"?

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

🤖 Need help with this problem?