What is the time complexity of Lexicographically Smallest String After Applying Operations?

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

What is the space complexity of Lexicographically Smallest String After Applying Operations?

The space complexity is O(n^2).

What data structures are used to solve Lexicographically Smallest String After Applying Operations?

This problem uses String, Depth-First Search, Breadth-First Search, Enumeration. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Lexicographically Smallest String After Applying Operations?

Insight 1: The problem can be modeled as a graph traversal, where each string is a node, and the operations represent edges to other nodes (strings). Insight 2: Breadth-First Search (BFS) is a suitable algorithm to explore all reachable states (strings) because it systematically explores the graph level by level, guaranteeing that the lexicographically smallest string is found first.

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Lexicographically Smallest String After Applying Operations - LeetCode 1625 Solution

Q: Which companies ask LeetCode 1625. Lexicographically Smallest String After Applying Operations in interviews?

This problem is frequently asked at J.P. Morgan.

Lexicographically Smallest String After Applying Operations - Complete Solution Guide

Lexicographically Smallest String After Applying Operations is LeetCode problem 1625, 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 string s of even length consisting of digits from 0 to 9 , and two integers a and b . You can apply either of the following two operations any number of times and in any order on s : Add a to all odd indices of s (0-indexed) . Digits post 9 are cycled back to 0 . For example, if s = "3456" and a = 5 , s becomes "3951" . Rotate s to the right by b positions. For example, if s = "3456" and b = 1 , s becomes "6345" . Return the lexicographically smallest string you can obtain by app

Detailed Explanation

The problem requires finding the lexicographically smallest string obtainable from a given string `s` by repeatedly applying two operations: adding a value `a` to digits at odd indices (with wraparound after 9), and rotating the string to the right by `b` positions. The goal is to explore all possible strings reachable through these operations and identify the one that comes earliest in lexicographical order. The string `s` has an even length and contains only digits. The operations can be applied in any order and any number of times.

Solution Approach

The provided code uses Breadth-First Search (BFS) to explore all possible strings that can be obtained by applying the two operations. The algorithm starts with the initial string `s` and iteratively applies the 'add' and 'rotate' operations, adding the resulting strings to a queue for further exploration. A `visited` set keeps track of strings that have already been processed to avoid cycles. The algorithm maintains a `min_s` variable to store the lexicographically smallest string encountered so far, updating it whenever a smaller string is found. The BFS ensures that the search expands uniformly, finding the lexicographically smallest reachable string.

Step-by-Step Algorithm

Step 1: Initialize a queue `q` with the starting string `s` and a set `visited` to keep track of visited strings. Add `s` to `visited`.
Step 2: Initialize `min_s` with the starting string `s` as the current lexicographically smallest string.
Step 3: While the queue `q` is not empty:
Step 4: Dequeue a string `current_s` from the queue.
Step 5: Compare `current_s` with `min_s`. If `current_s` is lexicographically smaller than `min_s`, update `min_s`.
Step 6: Apply the 'add' operation to `current_s` to obtain `add_res`. If `add_res` has not been visited:
Step 7: Add `add_res` to `visited` and enqueue it into `q`.
Step 8: Apply the 'rotate' operation to `current_s` to obtain `rotate_res`. If `rotate_res` has not been visited:
Step 9: Add `rotate_res` to `visited` and enqueue it into `q`.
Step 10: After the queue is empty, return `min_s`.

Key Insights

Insight 1: The problem can be modeled as a graph traversal, where each string is a node, and the operations represent edges to other nodes (strings).
Insight 2: Breadth-First Search (BFS) is a suitable algorithm to explore all reachable states (strings) because it systematically explores the graph level by level, guaranteeing that the lexicographically smallest string is found first.
Insight 3: Using a `visited` set is crucial to prevent cycles and redundant computations, avoiding infinite loops and improving efficiency.

Complexity Analysis

Time Complexity: O(n^2)

Space Complexity: O(n^2)

Topics

This problem involves: String, Depth-First Search, Breadth-First Search, Enumeration.

Companies

Asked at: J.P. Morgan.

Lexicographically Smallest String After Applying Operations | LeetCode 1625

Medium

String Depth-First Search Breadth-First Search Enumeration

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

Solution Code

Available in:

import java.util.ArrayDeque;
import java.util.HashSet;
import java.util.Queue;
import java.util.Set;

class Solution {
    public String findLexSmallestString(String s, int a, int b) {
        /**
         * Solves the Lexicographically Smallest String After Applying Operations problem.
         *
         * This solution treats the problem as finding the minimum node in a state graph,
         * where each string is a node and operations are edges. It uses Breadth-First
         * Search (BFS) to explore all reachable strings from the initial string `s`.
         *
         * A queue is used to manage the strings to be processed, and a `visited` set
         * is used to avoid cycles and redundant computations. The lexicographically
         * smallest string encountered during the traversal is tracked and returned
         * as the result.
         */
        int n = s.length();
        Queue<String> q = new ArrayDeque<>();
        Set<String> visited = new HashSet<>();
        
        q.offer(s);
        visited.add(s);
        
        String min_s = s;

        while (!q.isEmpty()) {
            String current_s = q.poll();
            
            if (current_s.compareTo(min_s) < 0) {
                min_s = current_s;
            }
            
            // Operation 1: Add 'a' to all odd-indexed digits.
            char[] s_list = current_s.toCharArray();
            for (int i = 1; i < n; i += 2) {
                s_list[i] = (char) (((s_list[i] - '0') + a) % 10 + '0');
            }
            String add_res = new String(s_list);
            
            if (!visited.contains(add_res)) {
                visited.add(add_res);
                q.offer(add_res);
            }
            
            // Operation 2: Rotate the string to the right by 'b' positions.
            String rotate_res = current_s.substring(n-b) + current_s.substring(0, n-b);
            
            if (!visited.contains(rotate_res)) {
                visited.add(rotate_res);
                q.offer(rotate_res);
            }

        }

        return min_s;
    }
}

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Problem Info

Description

Problem 1625: Lexicographically Smallest String After Applying Operations - Detailed explanation and solution approaches.

Topics

String Depth-First Search Breadth-First Search Enumeration

Companies

J.P. Morgan

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:1625

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

10 steps

1
Step 1: Initialize a queue `q` with the starting string `s` and a set `visited` to keep track of visited strings. Add `s` to `visited`.
2
Step 2: Initialize `min_s` with the starting string `s` as the current lexicographically smallest string.
3
Step 3: While the queue `q` is not empty:
4
Step 4: Dequeue a string `current_s` from the queue.
5
Step 5: Compare `current_s` with `min_s`. If `current_s` is lexicographically smaller than `min_s`, update `min_s`.
6
Step 6: Apply the 'add' operation to `current_s` to obtain `add_res`. If `add_res` has not been visited:
7
Step 7: Add `add_res` to `visited` and enqueue it into `q`.
8
Step 8: Apply the 'rotate' operation to `current_s` to obtain `rotate_res`. If `rotate_res` has not been visited:
9
Step 9: Add `rotate_res` to `visited` and enqueue it into `q`.
10
Step 10: After the queue is empty, return `min_s`.

💡

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^2)

🐢 Quadratic - Gets slow with large inputs. Consider optimization!

💾

Space Complexity

O(n^2)

📚 Quadratic space - Uses lots of memory!

📝

Why?

The time complexity depends on the number of reachable states. Each string comparison takes O(n) time. The space complexity is proportional to the number of reachable states, as these states are stored in the `visited` set and queue. The actual number of reachable states is difficult to determine without more analysis of `a` and `b`.

📖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

Depth-First Search (DFS)

DFS could be used instead of BFS. However, it might not find the lexicographically smallest string as quickly as BFS, as it doesn't explore the graph level by level. You'd need to explore all possible paths and then compare, or perform a min-update during the DFS.

Time: O(n^2 * 10^(n/2)) in worst-case, similar to BFSSpace: O(n * depth of recursion tree). Potentially O(n^2) in the worst case.

Trade-offs: DFS might be more space-efficient if the depth of the recursion tree is smaller than the number of nodes explored by BFS, but it's generally less efficient for finding the lexicographically smallest string, as it doesn't guarantee finding it first.

✨

Key Insights

1Insight 1: The problem can be modeled as a graph traversal, where each string is a node, and the operations represent edges to other nodes (strings).
2Insight 2: Breadth-First Search (BFS) is a suitable algorithm to explore all reachable states (strings) because it systematically explores the graph level by level, guaranteeing that the lexicographically smallest string is found first.
3Insight 3: Using a `visited` set is crucial to prevent cycles and redundant computations, avoiding infinite loops and improving efficiency.

📚

Key Terms Explained

For Students

String

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

Edge Cases

Edge case 1: When `a` is 0 and `b` is a multiple of the string length, the string remains unchanged. The algorithm correctly handles this because it adds the initial string to the `visited` set and compares it with the current minimum.
Edge case 2: When the input string is already lexicographically the smallest possible, the algorithm still works correctly, as it explores all reachable states and eventually finds that the initial string is indeed the smallest.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not using a `visited` set, which leads to infinite loops and Time Limit Exceeded (TLE) errors.

✗Common mistake 2: Not handling the wraparound correctly when adding `a` to digits at odd indices (i.e., not using the modulo operator `%`).

✗Common mistake 3: Incorrect string manipulation (e.g., index out of bounds errors, memory leaks in C).

✗Common mistake 4: Prematurely returning from the function before exploring all reachable states.

✅

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: Optimization problems where a sequence of operations needs to be applied to find the optimal state (e.g., resource allocation, scheduling).

🏦Real-world scenario 2: Cryptography, where a string transformation algorithm needs to be applied repeatedly, and you want to find the 'smallest' encrypted version (though the definition of 'smallest' would need to be tailored to the application).

💡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: June 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 14

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

@AlgoMasterJul 7

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

@InterviewProJun 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.

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 "Lexicographically Smallest String After Applying Operations"?

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 depth-first search techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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