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

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

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

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

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

Insight 1: The core idea is to iterate through the string `s` and greedily try to change each character to 'a' or as close to 'a' as possible to minimize lexicographical order while respecting the distance constraint. Insight 2: We need to calculate the minimum 'cyclic' distance between each character in `s` and 'a'. This involves considering both the distance in the forward direction (e.g., 'b' to 'a' is 1) and the distance in the reverse direction (e.g., 'z' to 'a' is also 1).

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Lexicographically Smallest String After Operations With Constraint - LeetCode 3106 Solution

Q: How to solve Lexicographically Smallest String After Operations With Constraint?

The solution iterates through the input string `s`, attempting to transform each character to 'a' if possible, considering the distance constraint `k`. For each character, the algorithm calculates the minimum cyclic distance to 'a'. If `k` is greater than or equal to the minimum distance, the character is changed to 'a', and `k` is reduced by the corresponding distance. Otherwise, the character is moved closer to 'a' by reducing its ASCII value by `k` amount, and `k` becomes zero. The process stops if `k` becomes zero at any point, as no further changes are needed.

Q: Which companies ask LeetCode 3106. Lexicographically Smallest String After Operations With Constraint in interviews?

This problem is frequently asked at ServiceNow.

Lexicographically Smallest String After Operations With Constraint - Complete Solution Guide

Lexicographically Smallest String After Operations With Constraint is LeetCode problem 3106, 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 and an integer k . Define a function distance(s 1 , s 2 ) between two strings s 1 and s 2 of the same length n as: The sum of the minimum distance between s 1 [i] and s 2 [i] when the characters from 'a' to 'z' are placed in a cyclic order, for all i in the range [0, n - 1] . For example, distance("ab", "cd") == 4 , and distance("a", "z") == 1 . You can change any letter of s to any other lowercase English letter, any number of times. Return a string denoting the lexicog

Detailed Explanation

The problem asks us to find the lexicographically smallest string `t` that can be obtained from a given string `s` by changing its characters, subject to a constraint on the 'distance' between `s` and `t`. The distance between two strings of the same length is defined as the sum of the minimum distances between corresponding characters in a cyclic fashion (a -> b -> c ... -> z -> a). We are given an integer `k`, which represents the maximum allowable distance between the original string `s` and the modified string `t`. The goal is to minimize `t` lexicographically while ensuring that `distance(s, t) <= k`.

Solution Approach

The solution iterates through the input string `s`, attempting to transform each character to 'a' if possible, considering the distance constraint `k`. For each character, the algorithm calculates the minimum cyclic distance to 'a'. If `k` is greater than or equal to the minimum distance, the character is changed to 'a', and `k` is reduced by the corresponding distance. Otherwise, the character is moved closer to 'a' by reducing its ASCII value by `k` amount, and `k` becomes zero. The process stops if `k` becomes zero at any point, as no further changes are needed.

Step-by-Step Algorithm

Step 1: Convert the input string `s` to a character array (or list) for easier modification.
Step 2: Initialize a loop to iterate through the character array.
Step 3: In each iteration, calculate the distance between the current character and 'a'. Calculate the cyclic distance `min(dist, 26 - dist)`. Dist is `ord(char) - ord('a')`.
Step 4: If the remaining distance `k` is greater than or equal to the calculated distance, change the current character to 'a' and reduce `k` by the distance.
Step 5: If `k` is less than the calculated distance, change the character by subtracting `k` from its ASCII value (effectively moving it closer to 'a') and set `k` to 0. No need to check reverse as the min distance logic already is in place.
Step 6: If `k` becomes 0 during any step, exit the loop as no more changes are allowed.
Step 7: Convert the modified character array back to a string and return it.

Key Insights

Insight 1: The core idea is to iterate through the string `s` and greedily try to change each character to 'a' or as close to 'a' as possible to minimize lexicographical order while respecting the distance constraint.
Insight 2: We need to calculate the minimum 'cyclic' distance between each character in `s` and 'a'. This involves considering both the distance in the forward direction (e.g., 'b' to 'a' is 1) and the distance in the reverse direction (e.g., 'z' to 'a' is also 1).
Insight 3: It's critical to update the remaining distance `k` after each character change and stop the process if `k` becomes zero.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

This problem involves: String, Greedy.

Companies

Asked at: ServiceNow.

Lexicographically Smallest String After Operations With Constraint | LeetCode 3106

Medium

String Greedy

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

Solution Code

Available in:

class Solution {
    public String getSmallestString(String s, int k) {
        char[] res = s.toCharArray();

        for (int i = 0; i < res.length; i++) {
            if (k <= 0) {
                break;
            }

            int dist = res[i] - 'a';
            int cost_to_a = Math.min(dist, 26 - dist);

            if (k >= cost_to_a) {
                res[i] = 'a';
                k -= cost_to_a;
            } else {
                res[i] = (char) (res[i] - k);
                k = 0;
            }
        }

        return new String(res);
    }
}

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Problem Info

Description

Problem 3106: Lexicographically Smallest String After Operations With Constraint - Detailed explanation and solution approaches.

Topics

String Greedy

Companies

ServiceNow

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:3106

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: Convert the input string `s` to a character array (or list) for easier modification.
2
Step 2: Initialize a loop to iterate through the character array.
3
Step 3: In each iteration, calculate the distance between the current character and 'a'. Calculate the cyclic distance `min(dist, 26 - dist)`. Dist is `ord(char) - ord('a')`.
4
Step 4: If the remaining distance `k` is greater than or equal to the calculated distance, change the current character to 'a' and reduce `k` by the distance.
5
Step 5: If `k` is less than the calculated distance, change the character by subtracting `k` from its ASCII value (effectively moving it closer to 'a') and set `k` to 0. No need to check reverse as the min distance logic already is in place.
6
Step 6: If `k` becomes 0 during any step, exit the loop as no more changes are allowed.
7
Step 7: Convert the modified character array back to a 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)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The time complexity is directly proportional to the length of the string `s` because each character is visited and processed once. The space complexity arises from storing a copy of the string or character array, which takes space proportional to the length of `s`.

📖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

Dynamic Programming

Although this problem is solved using a greedy approach very efficiently, one could consider Dynamic Programming where dp[i][j] stores the smallest string possible up to index i using k = j. But, that would drastically change the TC.

Time: O(n*k*26) - Very inefficient!Space: O(n*k)

Trade-offs: DP will be much more complicated to implement. Would only be useful if we changed the problem to 'find the smallest string possible using exactly k distance' or something to this extent. Using DP here would be very inefficient.

✨

Key Insights

1Insight 1: The core idea is to iterate through the string `s` and greedily try to change each character to 'a' or as close to 'a' as possible to minimize lexicographical order while respecting the distance constraint.
2Insight 2: We need to calculate the minimum 'cyclic' distance between each character in `s` and 'a'. This involves considering both the distance in the forward direction (e.g., 'b' to 'a' is 1) and the distance in the reverse direction (e.g., 'z' to 'a' is also 1).
3Insight 3: It's critical to update the remaining distance `k` after each character change and stop the process if `k` becomes zero.

📚

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: `k` is 0. In this case, no changes are allowed, and the original string should be returned. The algorithm correctly handles this because the loop condition `if k <= 0` will immediately terminate the loop.
Edge case 2: The string is already lexicographically smallest (e.g., "aaa"). In this case, the algorithm will iterate through the string but not make any changes because the distance between each character and 'a' is 0.
Edge case 3: `k` is very large, much larger than the total possible distance. The algorithm correctly handles this by stopping when all possible reductions to `a` have been made or `k` is fully consumed.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not considering the cyclic distance between characters. For example, the distance between 'z' and 'a' is 1, not 25.

✗Common mistake 2: Not updating `k` correctly after each character change. It's crucial to subtract the distance from `k` to ensure the overall distance constraint is met.

✗Common mistake 3: Not stopping the algorithm when `k` becomes 0. Continuing to iterate after `k` is 0 will result in unnecessary calculations.

✗Common mistake 4: Using string concatenation within the loop to build the result string. This leads to O(n^2) time complexity, which exceeds requirements.

✅

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: This type of problem solving can be applied to data compression, where the goal is to minimize the amount of data needed to represent a piece of information while staying within a certain error tolerance.

🏦Real-world scenario 2: In network routing, algorithms are used to find the shortest path between two points subject to constraints such as bandwidth or latency, conceptually similar to minimizing the distance between strings subject to a distance constraint.

💡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

If stuck, try working through a small example by hand. The pattern often reveals itself.

⚠️ 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 17, 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

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

@AlgoMasterJul 13

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

@InterviewProJul 6

❓

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.

What techniques are commonly tested in medium-difficulty problems?

Medium problems often require combining multiple techniques: two-pointer with hash maps, BFS/DFS with state tracking, dynamic programming with optimization, or divide-and-conquer approaches. They test your ability to recognize patterns and choose the right combination of techniques.

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 String After Operations With Constraint"?

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.

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