What is the time complexity of Partitioning Into Minimum Number Of Deci-Binary Numbers?

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

What data structures are used to solve Partitioning Into Minimum Number Of Deci-Binary Numbers?

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

What are the key insights for solving Partitioning Into Minimum Number Of Deci-Binary Numbers?

Insight 1: The minimum number of deci-binary numbers needed is equal to the largest digit in the given number `n`. For example, if `n` is "82734", we need at least 8 deci-binary numbers because we need to form the digit '8'. Insight 2: We can create the target number by summing deci-binary numbers. Consider the largest digit. To represent that digit, we need that many '1's in that place value (plus the necessary zeros).

Which companies ask LeetCode 1689. Partitioning Into Minimum Number Of Deci-Binary Numbers in interviews?

This problem is frequently asked at Nutanix.

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Partitioning Into Minimum Number Of Deci-Binary Numbers - LeetCode 1689 Solution

Partitioning Into Minimum Number Of Deci-Binary Numbers - Complete Solution Guide

Partitioning Into Minimum Number Of Deci-Binary Numbers is LeetCode problem 1689, 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

A decimal number is called deci-binary if each of its digits is either 0 or 1 without any leading zeros. For example, 101 and 1100 are deci-binary , while 112 and 3001 are not. Given a string n that represents a positive decimal integer, return the minimum number of positive deci-binary numbers needed so that they sum up to n . Example 1: Input: n = "32" Output: 3 Explanation: 10 + 11 + 11 = 32 Example 2: Input: n = "82734" Output: 8 Example 3: Input: n = "27346209830709182346" Output: 9 Constra

Detailed Explanation

The problem asks us to find the minimum number of 'deci-binary' numbers that sum up to a given positive decimal integer represented as a string `n`. A deci-binary number is defined as a number whose digits are all either 0 or 1 (e.g., 101, 1100). Essentially, we need to decompose the given number `n` into the smallest possible set of deci-binary numbers.

Solution Approach

The solution approach leverages the key insight that the minimum number of deci-binary numbers needed is equal to the largest digit in the input string `n`. We iterate through the string, find the maximum digit, and return its integer value. This works because each digit in `n` needs to be formed by the sum of deci-binary numbers. The largest digit will be the limiting factor.

Step-by-Step Algorithm

Step 1: Initialize a variable `maxDigit` to 0.
Step 2: Iterate through the input string `n` character by character.
Step 3: For each character, convert it to its integer value.
Step 4: Compare the current digit's integer value with `maxDigit`. If the current digit is greater than `maxDigit`, update `maxDigit`.
Step 5: After iterating through all digits in `n`, return the value of `maxDigit`.

Key Insights

Insight 1: The minimum number of deci-binary numbers needed is equal to the largest digit in the given number `n`. For example, if `n` is "82734", we need at least 8 deci-binary numbers because we need to form the digit '8'.
Insight 2: We can create the target number by summing deci-binary numbers. Consider the largest digit. To represent that digit, we need that many '1's in that place value (plus the necessary zeros).
Insight 3: Since we are looking for the *minimum* number, focusing on the largest digit is optimal. Lower digits can always be satisfied within the decomposition needed to satisfy the largest digit.

Complexity Analysis

Time Complexity: O(N)

Space Complexity: O(1)

Topics

This problem involves: String, Greedy.

Companies

Asked at: Nutanix.

Partitioning Into Minimum Number Of Deci-Binary Numbers | LeetCode 1689

Medium

String Greedy

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

Solution Code

Available in:

class Solution {
    public int minPartitions(String n) {
        char maxDigit = '0';
        for (int i = 0; i < n.length(); i++) {
            if (n.charAt(i) > maxDigit) {
                maxDigit = n.charAt(i);
            }
        }
        return Character.getNumericValue(maxDigit);
    }
}

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Problem Info

Description

Problem 1689: Partitioning Into Minimum Number Of Deci-Binary Numbers - Detailed explanation and solution approaches.

Topics

String Greedy

Companies

Nutanix

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:1689

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: Initialize a variable `maxDigit` to 0.
2
Step 2: Iterate through the input string `n` character by character.
3
Step 3: For each character, convert it to its integer value.
4
Step 4: Compare the current digit's integer value with `maxDigit`. If the current digit is greater than `maxDigit`, update `maxDigit`.
5
Step 5: After iterating through all digits in `n`, return the value of `maxDigit`.

💡

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)

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The algorithm iterates through the input string `n` once, performing a constant-time operation (comparison) at each step. Therefore, the time complexity is directly proportional to the length of the string, resulting in O(N) time complexity. The space complexity is constant because we use a fixed number of variables regardless of the input size, resulting in O(1) space complexity.

📖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

Using built-in max function (Python)

Python's built-in `max()` function can directly find the largest character in the string. No explicit loop is needed.

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

Trade-offs: More concise in Python, but conceptually the same and doesn't offer a significant performance advantage.

✨

Key Insights

1Insight 1: The minimum number of deci-binary numbers needed is equal to the largest digit in the given number `n`. For example, if `n` is "82734", we need at least 8 deci-binary numbers because we need to form the digit '8'.
2Insight 2: We can create the target number by summing deci-binary numbers. Consider the largest digit. To represent that digit, we need that many '1's in that place value (plus the necessary zeros).
3Insight 3: Since we are looking for the *minimum* number, focusing on the largest digit is optimal. Lower digits can always be satisfied within the decomposition needed to satisfy the largest digit.

📚

Key Terms Explained

For Students

Greedy Algorithm

Making the best choice at each step without looking ahead.

String

A sequence of characters. Like a word or sentence in programming!

Edge Cases

Edge case 1: Input string `n` consists of only '0' and '1'. The algorithm correctly returns 1 if there is at least one '1', and 0 if the string is all zeros. Note: the problem statement specifies that n represents a *positive* decimal integer. The solution handles cases with digits only 0 and 1 correctly.
Edge case 2: Input string `n` contains a single digit (e.g., "5"). The algorithm correctly returns that digit (5).

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Trying to actually construct the deci-binary numbers. The problem only asks for the *minimum number* of them, not the numbers themselves. This unnecessary computation leads to a much more complex (and incorrect) solution.

✗Common mistake 2: Not converting the characters to their integer values before comparing them. Comparing characters directly would give incorrect results because characters are compared based on their ASCII values, not their numerical values.

✅

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: Resource allocation. Imagine allocating different types of resources (represented by digits) where each unit of resource can only be allocated in whole units (deci-binary). Finding the largest resource requirement helps determine the minimum number of independent allocation cycles needed.

🏦Real-world scenario 2: Decomposing a number into simpler components. Although the 'deci-binary' constraint is specific, the general idea of decomposing a number into the sum of simpler, restricted components has applications in cryptography, data compression, and number theory.

💡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

String problems often involve sliding windows or two pointers. Consider if the character set is limited (lowercase letters = only 26 possibilities).

🎯 Interview Tip

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

⚠️ Watch Out

Mutating strings directly in languages where strings are immutable. Use StringBuilder or arrays.

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 9, 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

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

@TechLeadJul 14

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 7

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJun 30

❓

Frequently Asked Questions

What is the sliding window technique for string problems?

Sliding window maintains a "window" of characters with two pointers. Expand the right pointer to include characters, shrink from left when window violates constraints. Useful for substring problems, anagrams, and longest/shortest substring with properties.

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 "Partitioning Into Minimum Number Of Deci-Binary Numbers"?

For this medium-level String 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?