What is the time complexity of Can Convert String in K Moves?

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

What data structures are used to solve Can Convert String in K Moves?

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

What are the key insights for solving Can Convert String in K Moves?

Insight 1: The core idea is to calculate the shift difference for each corresponding character pair in `s` and `t`. If the shift difference is greater than 0, it represents the number of moves needed to convert the character. Insight 2: The same shift difference can appear multiple times. We need to ensure that we have enough moves `k` to perform each of these shifts, taking into account that each move number can only be used once.

Which companies ask LeetCode 1540. Can Convert String in K Moves in interviews?

This problem is frequently asked at Infosys.

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Can Convert String in K Moves - LeetCode 1540 Solution

Q: How to solve Can Convert String in K Moves?

The solution calculates the required shift difference for each corresponding character in `s` and `t`. It then counts how many times each shift difference (1-25) is needed. Finally, it iterates through each possible shift difference and checks if the maximum move needed for that shift difference and its duplicates exceeds `k`. If any exceedance is found, it returns `false`. Otherwise, returns `true`.

Can Convert String in K Moves - Complete Solution Guide

Can Convert String in K Moves is LeetCode problem 1540, 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

Given two strings s and t , your goal is to convert s into t in k moves or less. During the i th ( 1 <= i <= k ) move you can: Choose any index j (1-indexed) from s , such that 1 <= j <= s.length and j has not been chosen in any previous move, and shift the character at that index i times. Do nothing. Shifting a character means replacing it by the next letter in the alphabet (wrapping around so that 'z' becomes 'a' ). Shifting a character by i means applying the shift operations i times. Remembe

Detailed Explanation

The problem asks whether string `s` can be converted to string `t` in at most `k` moves. Each move allows shifting a character in `s` by a certain number of times (1 to k) to match the character in `t` at the same index. Shifting wraps around the alphabet ('z' becomes 'a'). Each index can be shifted only once. The inputs are the two strings `s` and `t` and the maximum number of moves `k`. The output is a boolean indicating whether the conversion is possible.

Solution Approach

The solution calculates the required shift difference for each corresponding character in `s` and `t`. It then counts how many times each shift difference (1-25) is needed. Finally, it iterates through each possible shift difference and checks if the maximum move needed for that shift difference and its duplicates exceeds `k`. If any exceedance is found, it returns `false`. Otherwise, returns `true`.

Step-by-Step Algorithm

Step 1: Check if the lengths of `s` and `t` are equal. If not, return `false` because conversion is impossible.
Step 2: Initialize an array `counts` of size 26 to store the frequency of each shift difference (1 to 25).
Step 3: Iterate through the strings `s` and `t`. For each index `i`, calculate the shift difference between `t[i]` and `s[i]` using `(ord(t[i]) - ord(s[i]) + 26) % 26`.
Step 4: If the shift difference is greater than 0, increment the corresponding count in the `counts` array.
Step 5: Iterate through the `counts` array from shift 1 to 25. For each shift value, check if its occurrences can be handled within `k` moves. The maximum number of moves needed is `shift + (count - 1) * 26`. For example, if a shift value of `1` occurs 3 times. The moves needed could be 1, 27 and 53.
Step 6: If the `max_move_needed` is greater than `k` for any shift value, return `false`.
Step 7: If all shift values can be handled within `k` moves, return `true`.

Key Insights

Insight 1: The core idea is to calculate the shift difference for each corresponding character pair in `s` and `t`. If the shift difference is greater than 0, it represents the number of moves needed to convert the character.
Insight 2: The same shift difference can appear multiple times. We need to ensure that we have enough moves `k` to perform each of these shifts, taking into account that each move number can only be used once.
Insight 3: Optimizing the shift numbers required. Since each move number can be used once, we need to maximize the total number of available move numbers. For a shift count `x`, we can also use `x + 26`, `x + 52` and so on, up to `k`.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

This problem involves: Hash Table, String.

Companies

Asked at: Infosys.

Can Convert String in K Moves | LeetCode 1540

Medium

Hash Table String

Time: O(n)Space: O(1)

Solution Code

Available in:

class Solution {
    public boolean canConvertString(String s, String t, int k) {
        if (s.length() != t.length()) {
            return false;
        }

        int[] counts = new int[26];

        for (int i = 0; i < s.length(); i++) {
            int diff = (t.charAt(i) - s.charAt(i) + 26) % 26;
            if (diff > 0) {
                counts[diff]++;
            }
        }

        for (int shift = 1; shift < 26; shift++) {
            int count = counts[shift];
            if (count > 0) {
                int max_move_needed = shift + (count - 1) * 26;
                if (max_move_needed > k) {
                    return false;
                }
            }
        }

        return true;
    }
}

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Problem Info

Description

Problem 1540: Can Convert String in K Moves - Detailed explanation and solution approaches.

Topics

Hash Table String

Companies

Infosys

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:1540

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: Check if the lengths of `s` and `t` are equal. If not, return `false` because conversion is impossible.
2
Step 2: Initialize an array `counts` of size 26 to store the frequency of each shift difference (1 to 25).
3
Step 3: Iterate through the strings `s` and `t`. For each index `i`, calculate the shift difference between `t[i]` and `s[i]` using `(ord(t[i]) - ord(s[i]) + 26) % 26`.
4
Step 4: If the shift difference is greater than 0, increment the corresponding count in the `counts` array.
5
Step 5: Iterate through the `counts` array from shift 1 to 25. For each shift value, check if its occurrences can be handled within `k` moves. The maximum number of moves needed is `shift + (count - 1) * 26`. For example, if a shift value of `1` occurs 3 times. The moves needed could be 1, 27 and 53.
6
Step 6: If the `max_move_needed` is greater than `k` for any shift value, return `false`.
7
Step 7: If all shift values can be handled within `k` moves, return `true`.

💡

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(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The dominant factor in time complexity is the loop through the string which takes O(n) time. The space is constant because it uses an array of size 26.

📖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 a HashMap

Instead of an array `counts` of size 26, we could use a HashMap to store the count of each required shift value. This would be useful if the range of possible shift values was very large, but the actual number of different shift values was small.

Time: O(n)Space: O(m), where m is the number of distinct shift values.

Trade-offs: Using a HashMap would have a space complexity dependent on the number of distinct shift values. If the shift values are always distributed in a limited range (as they are here, from 1 to 25), the array approach is more efficient because of its constant space complexity. This approach is better when the range of possible shift values is very high and sparse, because in the original case all shift differences could potentially occur making space complexity O(1) better. When `m << 26`, HashMap becomes more space efficient.

✨

Key Insights

1Insight 1: The core idea is to calculate the shift difference for each corresponding character pair in `s` and `t`. If the shift difference is greater than 0, it represents the number of moves needed to convert the character.
2Insight 2: The same shift difference can appear multiple times. We need to ensure that we have enough moves `k` to perform each of these shifts, taking into account that each move number can only be used once.
3Insight 3: Optimizing the shift numbers required. Since each move number can be used once, we need to maximize the total number of available move numbers. For a shift count `x`, we can also use `x + 26`, `x + 52` and so on, up to `k`.

📚

Key Terms Explained

For Students

Hash Table

A data structure that maps keys to values for quick lookup. Like a dictionary!

String

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

Edge Cases

Edge case 1: `s` and `t` have different lengths. The solution handles this by immediately returning `false`. This is correct as we can only convert strings of same lengths by the nature of the problem.
Edge case 2: `s` and `t` are identical. The solution correctly computes all shift differences as 0, not incrementing the counts array. Hence no move will be attempted so it will return true.
Edge case 3: k is very large (approaching 10^9). The solution uses an `int` or `long long` type (in cpp) to store the `max_move_needed` to prevent overflow. Also, the counts array is iterated only 26 times, preventing time limit issues.
Edge case 4: All characters are the same except for one character where a large shift is needed. For instance, 'aaaaa' and 'aaaaz' where k = 25 will pass because shift is 25 and counts[25] is 1, so max needed move is 25 <= 25.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Integer overflow when calculating the maximum moves needed. This can happen when `shift + (count - 1) * 26` exceeds the maximum value of an `int`. Use `long long` if necessary to avoid overflow.

✗Common mistake 2: Incorrectly calculating the shift difference. Remember to wrap around the alphabet by using the modulo operator `% 26` and add 26 before the mod to handle cases where the character in `t` comes before the character in `s` in the alphabet.

✗Common mistake 3: Not considering that you can use a shift greater than needed. If we require shift 1 for two indices, we could use moves 1 and 27.

✅

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 problem is related to data encryption and decryption, where you might need to transform one string to another using a series of shifts or operations with constraints.

🏦Real-world scenario 2: Code optimization where transforming one piece of code to another version with minimal operations and specific constrains on sequence of operations.

💡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

Hash tables trade space for time. Before using one, consider if the problem constraints allow the extra memory. In interviews, always mention this trade-off.

🎯 Interview Tip

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

⚠️ Watch Out

Don't forget to handle key collisions and null/undefined keys. Consider what happens with duplicate values.

👨‍💻 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: September 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

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

❓

Frequently Asked Questions

What are the trade-offs of using a hash table?

Hash tables provide O(1) average-case lookup, insertion, and deletion, making them ideal for frequency counting and duplicate detection. However, they use O(n) extra space and may degrade to O(n) time in worst-case scenarios with hash collisions. Always consider if the space trade-off is acceptable.

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.

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 "Can Convert String in K Moves"?

For this medium-level Hash Table problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether string techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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