What is the time complexity of Construct Smallest Number From DI String?

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

What data structures are used to solve Construct Smallest Number From DI String?

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

What are the key insights for solving Construct Smallest Number From DI String?

Insight 1: The core idea is to use a stack to delay the assignment of numbers, strategically releasing them in the correct order to achieve the smallest possible number while adhering to the I/D pattern. Insight 2: The stack helps to efficiently handle sequences of 'D' characters. When an 'I' is encountered or the end of the pattern is reached, the stack is popped, ensuring the 'decreasing' sequence is reversed and inserted into the result.

Which companies ask LeetCode 2375. Construct Smallest Number From DI String in interviews?

This problem is frequently asked at Goldman Sachs, josh technology.

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Construct Smallest Number From DI String - LeetCode 2375 Solution

Construct Smallest Number From DI String - Complete Solution Guide

Construct Smallest Number From DI String is LeetCode problem 2375, 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 a 0-indexed string pattern of length n consisting of the characters 'I' meaning increasing and 'D' meaning decreasing . A 0-indexed string num of length n + 1 is created using the following conditions: num consists of the digits '1' to '9' , where each digit is used at most once. If pattern[i] == 'I' , then num[i] < num[i + 1] . If pattern[i] == 'D' , then num[i] > num[i + 1] . Return the lexicographically smallest possible string num that meets the conditions. Example 1: Input: pa

Detailed Explanation

The problem asks us to construct the lexicographically smallest string of digits from 1 to n+1 (where n is the length of the input pattern string) such that the string satisfies the 'increasing' ('I') and 'decreasing' ('D') relationships defined by the input 'pattern' string. The resulting string must contain each digit from 1 to n+1 exactly once. If the pattern character at index `i` is 'I', then the digit at index `i` in the output string must be less than the digit at index `i+1`. If the pattern character is 'D', the digit at index `i` must be greater than the digit at index `i+1`.

Solution Approach

The provided code utilizes a stack-based approach. It iterates through the 'pattern' string and maintains a stack of numbers from 1 to n+1. Each number is pushed onto the stack. When an 'I' is encountered (meaning an increasing sequence is expected) or the end of the string is reached, the numbers from the stack are popped and appended to the result string. This reverses the order of numbers in the stack, effectively creating a decreasing sequence before the 'I'. This ensures that the smallest possible numbers are used in ascending order where needed and larger numbers are placed before smaller numbers where a decrease is needed.

Step-by-Step Algorithm

Step 1: Initialize an empty result string/list and an empty stack.
Step 2: Iterate through the input 'pattern' string (including a virtual end-of-string position).
Step 3: For each iteration, push the number (i+1) onto the stack (where 'i' is the current index).
Step 4: Check if the current character in 'pattern' is 'I' or if the end of the pattern is reached. If either condition is true, proceed to Step 5.
Step 5: While the stack is not empty, pop elements from the stack and append them to the result string.
Step 6: Continue the loop until all elements in the 'pattern' have been processed.
Step 7: Return the final result string.

Key Insights

Insight 1: The core idea is to use a stack to delay the assignment of numbers, strategically releasing them in the correct order to achieve the smallest possible number while adhering to the I/D pattern.
Insight 2: The stack helps to efficiently handle sequences of 'D' characters. When an 'I' is encountered or the end of the pattern is reached, the stack is popped, ensuring the 'decreasing' sequence is reversed and inserted into the result.
Insight 3: Building the string incrementally using a stack ensures that when we encounter an 'I', we can construct the smallest increasing sequence by popping elements from the stack that are greater than the next available digit.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

This problem involves: String, Backtracking, Stack, Greedy.

Companies

Asked at: Goldman Sachs, josh technology.

Construct Smallest Number From DI String | LeetCode 2375

Medium

String Backtracking Stack Greedy

Time: O(n)Space: O(n)

Solution Code

Available in:

class Solution {
    public String smallestNumber(String pattern) {
        int n = pattern.length();
        StringBuilder res = new StringBuilder();
        java.util.Stack<String> stack = new java.util.Stack<>();

        for (int i = 0; i < n + 1; i++) {
            stack.push(String.valueOf(i + 1));
            if (i == n || pattern.charAt(i) == 'I') {
                while (!stack.isEmpty()) {
                    res.append(stack.pop());
                }
            }
        }

        return res.toString();
    }
}

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Problem Info

Description

Problem 2375: Construct Smallest Number From DI String - Detailed explanation and solution approaches.

Topics

String Backtracking Stack Greedy

Companies

Goldman Sachsjosh technology

Stats

Difficulty:Medium

Languages:4

Companies:2

Problem #:2375

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

7 steps

1
Step 1: Initialize an empty result string/list and an empty stack.
2
Step 2: Iterate through the input 'pattern' string (including a virtual end-of-string position).
3
Step 3: For each iteration, push the number (i+1) onto the stack (where 'i' is the current index).
4
Step 4: Check if the current character in 'pattern' is 'I' or if the end of the pattern is reached. If either condition is true, proceed to Step 5.
5
Step 5: While the stack is not empty, pop elements from the stack and append them to the result string.
6
Step 6: Continue the loop until all elements in the 'pattern' have been processed.
7
Step 7: Return the final result 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)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The time complexity is linear because the algorithm iterates through the pattern once. The stack operations within the loop do not increase the overall time complexity beyond O(n) because each number is pushed and popped from the stack at most once. The space complexity is also linear due to the space used by the stack and the result string, which both grow proportionally to the length of the input pattern.

📖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

Backtracking

Generate all possible permutations of numbers from 1 to n+1 and check each permutation against the given 'I' and 'D' conditions. Return the smallest valid permutation.

Time: O((n+1)! * n), where n is the length of the pattern string. Generating all permutations takes O((n+1)!) time, and checking each permutation against the pattern takes O(n) time.Space: O(n), for storing the current permutation and the used digits.

Trade-offs: Backtracking is less efficient than the stack-based approach, especially for larger inputs. Backtracking should be avoided as it will result in TLE. Stack based approach offers a more direct and efficient solution.

✨

Key Insights

1Insight 1: The core idea is to use a stack to delay the assignment of numbers, strategically releasing them in the correct order to achieve the smallest possible number while adhering to the I/D pattern.
2Insight 2: The stack helps to efficiently handle sequences of 'D' characters. When an 'I' is encountered or the end of the pattern is reached, the stack is popped, ensuring the 'decreasing' sequence is reversed and inserted into the result.
3Insight 3: Building the string incrementally using a stack ensures that when we encounter an 'I', we can construct the smallest increasing sequence by popping elements from the stack that are greater than the next available digit.

📚

Key Terms Explained

For Students

Stack

LIFO structure - Last In, First Out. Like a stack of plates!

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: Empty pattern string. The code correctly handles this case. If the pattern is empty, the loop runs only once, pushes '1' onto the stack, and immediately pops it to the result.
Edge case 2: Pattern string containing only 'I's (e.g., 'III'). The code correctly constructs the smallest increasing sequence (e.g., '1234').
Edge case 3: Pattern string containing only 'D's (e.g., 'DDD'). The code correctly constructs the largest decreasing sequence using smallest possible numbers (e.g., '4321').

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not understanding the lexicographical order. It's crucial to construct the smallest possible number at each step, which involves using the smallest available digits while respecting the I/D pattern.

✗Common mistake 2: Incorrectly implementing the stack logic. A common mistake is not pushing i+1 to the stack, or not using the stack appropriately to reverse the order of digits for the 'D' pattern.

✗Common mistake 3: Failing to handle the edge case when the pattern is empty or contains only 'I' or 'D' characters. This can lead to incorrect or incomplete results.

✅

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: Task scheduling where tasks have dependencies (increasing or decreasing priority). Constructing an optimal schedule can be analogous to building the smallest number given 'I' and 'D' constraints.

🏦Real-world scenario 2: Generating unique identifiers or sequences based on specific ordering rules or patterns. For example, assigning transaction IDs in a financial system where IDs must follow a predefined ascending or descending order based on certain criteria.

💡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

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

⚠️ Watch Out

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

👨‍💻 From Experience

This problem pattern appears frequently in real codebases. I've used similar logic when building production systems.

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: July 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

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

@DevExpert92Jul 13

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

@AlgoMasterJul 6

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

@InterviewProJun 29

❓

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.

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 "Construct Smallest Number From DI String"?

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 backtracking techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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