What is the time complexity of Lexicographically Smallest Generated String?

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

What data structures are used to solve Lexicographically Smallest Generated String?

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

What are the key insights for solving Lexicographically Smallest Generated String?

Insight 1: The 'T' conditions are restrictive; they directly specify portions of the generated string 'word'. We should fill these in first. Insight 2: The 'F' conditions are constraints that prevent certain substrings. After satisfying the 'T' constraints, we must ensure the 'F' constraints are met while minimizing 'word' lexicographically.

Which companies ask LeetCode 3474. Lexicographically Smallest Generated String in interviews?

This problem is frequently asked at Barclays.

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Lexicographically Smallest Generated String - LeetCode 3474 Solution

Q: How to solve Lexicographically Smallest Generated String?

The solution uses a greedy approach. First, it initializes an array `word` of the correct length (n + m - 1) to hold the characters of the generated string. It iterates through `str1`, and whenever it encounters a 'T', it fills in the corresponding substring of `word` with `str2`. During this process, it checks for inconsistencies. If a cell in word has already been filled and is different from str2[j] at the current index, it immediately returns an empty string because a contradiction is present. After addressing all 'T' conditions, it proceeds to address the 'F' conditions, by iterating through `word`. To satisfy an 'F' constraint at index `i`, it looks for a valid character to fill in `word`. To minimize lexicographical order, 'a' is tried first. If using 'a' causes `word` to contain `str2` at the index, then 'b' is used instead.

Q: What is the space complexity of Lexicographically Smallest Generated String?

The space complexity is O(n+m).

Lexicographically Smallest Generated String - Complete Solution Guide

Lexicographically Smallest Generated String is LeetCode problem 3474, 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 two strings, str1 and str2 , of lengths n and m , respectively. A string word of length n + m - 1 is defined to be generated by str1 and str2 if it satisfies the following conditions for each index 0 <= i <= n - 1 : If str1[i] == 'T' , the substring of word with size m starting at index i is equal to str2 , i.e., word[i..(i + m - 1)] == str2 . If str1[i] == 'F' , the substring of word with size m starting at index i is not equal to str2 , i.e., word[i..(i + m - 1)] != str2 . Return

Detailed Explanation

The problem asks us to generate the lexicographically smallest string of length n + m - 1 from two input strings, str1 of length n and str2 of length m. The generated string, 'word', must satisfy certain conditions based on the characters in str1. If str1[i] is 'T', then the substring of 'word' of length m starting at index i must be equal to str2. If str1[i] is 'F', then the substring of 'word' of length m starting at index i must *not* be equal to str2. If no such 'word' can be generated, we should return an empty string. Constraints specify lowercase English characters in str2 and only 'T' or 'F' in str1. The goal is to find the smallest possible string that adheres to all the 'T' and 'F' rules imposed by str1 with respect to str2.

Solution Approach

The solution uses a greedy approach. First, it initializes an array `word` of the correct length (n + m - 1) to hold the characters of the generated string. It iterates through `str1`, and whenever it encounters a 'T', it fills in the corresponding substring of `word` with `str2`. During this process, it checks for inconsistencies. If a cell in word has already been filled and is different from str2[j] at the current index, it immediately returns an empty string because a contradiction is present. After addressing all 'T' conditions, it proceeds to address the 'F' conditions, by iterating through `word`. To satisfy an 'F' constraint at index `i`, it looks for a valid character to fill in `word`. To minimize lexicographical order, 'a' is tried first. If using 'a' causes `word` to contain `str2` at the index, then 'b' is used instead.

Step-by-Step Algorithm

Step 1: Initialize an array `word` of characters with length `n + m - 1`. In the C example, this is intialized to all null characters, while in the Java and Python examples it is intialized to None/null values.
Step 2: Iterate through `str1` from index 0 to `n - 1`. If `str1[i]` is 'T', then iterate through `str2` from index 0 to `m - 1`.
Step 3: For each `str1[i] == 'T'` condition, check if `word[i + j]` is already set. If it's set to a different character than `str2[j]`, then the problem has no solution, return "". Otherwise, set `word[i + j] = str2[j]`.
Step 4: After addressing all 'T' conditions, iterate through `word` from index 0 to `n + m - 2`.
Step 5: For each index `k` in `word`, check if it's already assigned a value. If not, determine whether the string `word` contains `str2` at index `i = k - m + 1`. If `str1[i]` is `'F'` and word contains `str2` at that point, assign `word[k]` either `'b'` or `'a'` to ensure the 'F' condition. If `word` does *not* contain `str2` at `i`, simply assign `word[k] = 'a'` to satisfy that spot, as the goal is to return the *lexicographically smallest* possible generated string.
Step 6: If at any time we cannot satisfy an F condition without violating an existing assignment return "".
Step 7: After processing all positions, convert `word` to string and return it.

Key Insights

Insight 1: The 'T' conditions are restrictive; they directly specify portions of the generated string 'word'. We should fill these in first.
Insight 2: The 'F' conditions are constraints that prevent certain substrings. After satisfying the 'T' constraints, we must ensure the 'F' constraints are met while minimizing 'word' lexicographically.
Insight 3: The problem requires careful handling of overlapping conditions imposed by 'T' and 'F' characters. If a contradiction arises (e.g., a position must be both 'a' and 'b'), no solution exists.

Complexity Analysis

Time Complexity: O(n*m)

Space Complexity: O(n+m)

Topics

This problem involves: String, Greedy, String Matching.

Companies

Asked at: Barclays.

Lexicographically Smallest Generated String | LeetCode 3474

Hard

String Greedy String Matching

Time: O(n*m)Space: O(n+m)

Solution Code

Available in:

class Solution {
    public String generateString(String str1, String str2) {
        int n = str1.length();
        int m = str2.length();
        int L = n + m - 1;
        
        Character[] word = new Character[L];
        
        for (int i = 0; i < n; i++) {
            if (str1.charAt(i) == 'T') {
                for (int j = 0; j < m; j++) {
                    int k = i + j;
                    if (word[k] != null && word[k] != str2.charAt(j)) {
                        return "";
                    }
                    word[k] = str2.charAt(j);
                }
            }
        }

        for (int k = 0; k < L; k++) {
            int i = k - m + 1;
            boolean is_prefix_match = false;
            if (i >= 0 && str1.charAt(i) == 'F') {
                boolean match = true;
                for (int j = 0; j < m - 1; j++) {
                    if (word[i + j] == null || word[i + j] != str2.charAt(j)) {
                        match = false;
                        break;
                    }
                }
                if (match) {
                    is_prefix_match = true;
                }
            }

            if (word[k] != null) {
                if (is_prefix_match && word[k] == str2.charAt(m - 1)) {
                    return "";
                }
            } else {
                if (is_prefix_match) {
                    char forbidden_char = str2.charAt(m - 1);
                    if (forbidden_char == 'a') {
                        word[k] = 'b';
                    } else {
                        word[k] = 'a';
                    }
                } else {
                    word[k] = 'a';
                }
            }
        }
        
        StringBuilder sb = new StringBuilder();
        for (Character c : word) {
            sb.append(c);
        }
        return sb.toString();
    }
}

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Problem Info

Description

Problem 3474: Lexicographically Smallest Generated String - Detailed explanation and solution approaches.

Topics

String Greedy String Matching

Companies

Barclays

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:3474

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 array `word` of characters with length `n + m - 1`. In the C example, this is intialized to all null characters, while in the Java and Python examples it is intialized to None/null values.
2
Step 2: Iterate through `str1` from index 0 to `n - 1`. If `str1[i]` is 'T', then iterate through `str2` from index 0 to `m - 1`.
3
Step 3: For each `str1[i] == 'T'` condition, check if `word[i + j]` is already set. If it's set to a different character than `str2[j]`, then the problem has no solution, return "". Otherwise, set `word[i + j] = str2[j]`.
4
Step 4: After addressing all 'T' conditions, iterate through `word` from index 0 to `n + m - 2`.
5
Step 5: For each index `k` in `word`, check if it's already assigned a value. If not, determine whether the string `word` contains `str2` at index `i = k - m + 1`. If `str1[i]` is `'F'` and word contains `str2` at that point, assign `word[k]` either `'b'` or `'a'` to ensure the 'F' condition. If `word` does *not* contain `str2` at `i`, simply assign `word[k] = 'a'` to satisfy that spot, as the goal is to return the *lexicographically smallest* possible generated string.
6
Step 6: If at any time we cannot satisfy an F condition without violating an existing assignment return "".
7
Step 7: After processing all positions, convert `word` to string and return it.

💡

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*m)

💾

Space Complexity

O(n+m)

📝

Why?

The time complexity is O(n*m) because the code iterates through str1 (length n) and potentially iterates through str2 (length m) within the loop when str1[i] is 'T'. The nested loop contributes the n*m factor. The second loop iterates through the resulting `word` which is O(n+m) but does constant work or another linear scan in a window of `m`. Thus, the overall time complexity is still dominated by O(n*m). The space complexity is O(n+m) due to the `word` array that stores the generated string, which is of length n+m-1.

📖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

A backtracking approach could be used to explore all possible combinations of characters in 'word'. This would involve trying 'a' and 'b' (or other characters) at each position and recursively checking if the current assignment satisfies the 'T' and 'F' constraints. However, this approach is much less efficient.

Time: O(2^(n+m)) in worst case.Space: O(n+m)

Trade-offs: Backtracking guarantees finding a solution if one exists, but its exponential time complexity makes it unsuitable for larger inputs. The greedy approach is much faster but might not work for all variations of the problem.

✨

Key Insights

1Insight 1: The 'T' conditions are restrictive; they directly specify portions of the generated string 'word'. We should fill these in first.
2Insight 2: The 'F' conditions are constraints that prevent certain substrings. After satisfying the 'T' constraints, we must ensure the 'F' constraints are met while minimizing 'word' lexicographically.
3Insight 3: The problem requires careful handling of overlapping conditions imposed by 'T' and 'F' characters. If a contradiction arises (e.g., a position must be both 'a' and 'b'), no solution exists.

📚

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: `str1` contains only 'T's, and the implied constraints on 'word' contradict each other. The code correctly identifies the contradiction and returns "".
Edge case 2: `str1` contains only 'F's. The algorithm will fill the entire string with 'a's, provided this doesn't contradict any 'F' condition. If putting 'a' in a certain index makes a substring equal to str2 when an F condition says it can't, the code will change it to 'b'. If it still contradicts it, it will return "".

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Failing to handle overlapping 'T' conditions correctly, leading to incorrect or invalid generated strings.

✗Common mistake 2: Incorrectly handling the 'F' conditions. Beginners may assume they can simply skip substrings where 'F' is specified without trying 'b' or other characters. Failing to properly consider all 'F' constraints.

✗Common mistake 3: Forgetting to initialize the `word` array/string correctly. The elements must be in a state indicating that they are not yet determined.

✅

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: Constraint satisfaction problems in planning and scheduling where certain events must occur at specific times (like the 'T' constraints) while others must not ('F' constraints).

🏦Real-world scenario 2: DNA sequence assembly, where some sequences are known with certainty ('T'), and others are known to be absent ('F'), and the goal is to assemble the shortest possible valid sequence.

💡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

Hard problems often combine multiple techniques. Break the problem into sub-problems.

⚠️ 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: July 25, 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

❓

Frequently Asked Questions

How do I handle string manipulation efficiently?

In languages with immutable strings (Java, Python, C#), use StringBuilder/list to avoid O(n) concatenation cost. Consider using character arrays for in-place modifications. For pattern matching, consider KMP or Rabin-Karp for O(n+m) complexity.

How do I verify if a greedy approach is correct?

Prove the greedy choice property: the locally optimal choice leads to globally optimal solution. Use exchange argument: show that swapping any element with the greedy choice doesn't improve the solution. If you can't prove it, test extensively or consider DP.

How do I approach hard algorithm problems?

Hard problems often require: 1) Breaking down into subproblems, 2) Combining multiple advanced techniques, 3) Careful optimization of brute force. Start with a working solution (even O(n²) or worse), then optimize. Draw examples, identify patterns, and think about what makes this problem unique.

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 "Lexicographically Smallest Generated String"?

For this hard-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.

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