What is the time complexity of Minimum Number of Operations to Make Word K-Periodic?

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

What is the space complexity of Minimum Number of Operations to Make Word K-Periodic?

The space complexity is O(n/k).

What data structures are used to solve Minimum Number of Operations to Make Word K-Periodic?

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

What are the key insights for solving Minimum Number of Operations to Make Word K-Periodic?

Insight 1: The core idea is to find the substring of length `k` that appears most frequently in the given string at k-intervals. The other substrings must be replaced with this most frequent substring. Insight 2: Using a hash map (or counter) to store the frequency of each k-length substring appearing at multiples of k indices is crucial for efficiently finding the most frequent substring.

Which companies ask LeetCode 3137. Minimum Number of Operations to Make Word K-Periodic in interviews?

This problem is frequently asked at Turing.

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Minimum Number of Operations to Make Word K-Periodic - LeetCode 3137 Solution

Minimum Number of Operations to Make Word K-Periodic - Complete Solution Guide

Minimum Number of Operations to Make Word K-Periodic is LeetCode problem 3137, 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 word of size n , and an integer k such that k divides n . In one operation, you can pick any two indices i and j , that are divisible by k , then replace the substring of length k starting at i with the substring of length k starting at j . That is, replace the substring word[i..i + k - 1] with the substring word[j..j + k - 1] . Return the minimum number of operations required to make word k-periodic . We say that word is k-periodic if there is some string s of length k su

Detailed Explanation

The problem asks us to find the minimum number of operations required to make a given string `word` k-periodic. A string is k-periodic if it can be formed by repeating a substring of length `k` multiple times. The allowed operation is to replace a substring of length `k` starting at index `i` (where `i` is a multiple of `k`) with a substring of length `k` starting at index `j` (where `j` is a multiple of `k`). The input is the string `word` and the integer `k`, and the output is the minimum number of operations.

Solution Approach

The solution involves counting the occurrences of each substring of length `k` that starts at an index divisible by `k`. After counting, the algorithm identifies the substring with the highest frequency. The number of operations is then calculated by subtracting the highest frequency from the total number of substrings of length `k` (which is `n/k`). This is because we want to find how many substrings are *not* equal to the most frequent substring and therefore need to be replaced.

Step-by-Step Algorithm

Step 1: Initialize a hash map (or counter) to store the frequencies of substrings of length `k`.
Step 2: Iterate through the string `word` with a step size of `k`. This ensures we only consider substrings starting at indices that are multiples of `k`.
Step 3: For each such substring, extract it from the word and update its count in the hash map.
Step 4: Iterate through the hash map to find the substring with the maximum frequency.
Step 5: Calculate the total number of k-length substrings by dividing the length of the word `n` by `k`.
Step 6: Subtract the maximum frequency from the total number of substrings to find the minimum number of operations required. Return this value.

Key Insights

Insight 1: The core idea is to find the substring of length `k` that appears most frequently in the given string at k-intervals. The other substrings must be replaced with this most frequent substring.
Insight 2: Using a hash map (or counter) to store the frequency of each k-length substring appearing at multiples of k indices is crucial for efficiently finding the most frequent substring.
Insight 3: The number of operations needed is simply the total number of k-length substrings (n/k) minus the frequency of the most frequent substring.

Complexity Analysis

Time Complexity: O(n/k)

Space Complexity: O(n/k)

Topics

This problem involves: Hash Table, String, Counting.

Companies

Asked at: Turing.

Minimum Number of Operations to Make Word K-Periodic | LeetCode 3137

Medium

Hash Table String Counting

Time: O(n/k)Space: O(n/k)

Solution Code

Available in:

import java.util.HashMap;
import java.util.Map;

class Solution {
    public int minimumOperationsToMakeKPeriodic(String word, int k) {
        int n = word.length();
        Map<String, Integer> counts = new HashMap<>();

        for (int i = 0; i < n; i += k) {
            String substring = word.substring(i, Math.min(i + k, n));
            counts.put(substring, counts.getOrDefault(substring, 0) + 1);
        }

        int maxFreq = 0;
        if (!counts.isEmpty()) {
            for (int freq : counts.values()) {
                maxFreq = Math.max(maxFreq, freq);
            }
        }

        int totalSubstrings = n / k;

        return totalSubstrings - maxFreq;
    }
}

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Problem Info

Description

Problem 3137: Minimum Number of Operations to Make Word K-Periodic - Detailed explanation and solution approaches.

Topics

Hash Table String Counting

Companies

Turing

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:3137

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

6 steps

1
Step 1: Initialize a hash map (or counter) to store the frequencies of substrings of length `k`.
2
Step 2: Iterate through the string `word` with a step size of `k`. This ensures we only consider substrings starting at indices that are multiples of `k`.
3
Step 3: For each such substring, extract it from the word and update its count in the hash map.
4
Step 4: Iterate through the hash map to find the substring with the maximum frequency.
5
Step 5: Calculate the total number of k-length substrings by dividing the length of the word `n` by `k`.
6
Step 6: Subtract the maximum frequency from the total number of substrings to find the minimum number of operations required. Return this value.

💡

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/k)

💾

Space Complexity

O(n/k)

📝

Why?

The time complexity is dominated by the loop that iterates `n/k` times. The space complexity is determined by the maximum number of unique k-length substrings, which is `n/k` in the worst case.

📖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

Brute-force comparison

Compare all substrings of length `k` with each other directly. This involves nested loops.

Time: O(n^2)Space: O(1)

Trade-offs: This approach is less efficient than using a hash map, especially for large input strings, because it involves multiple comparisons of the same substring. It avoids using extra space but it sacrifices computational efficiency. Only suitable for extremely small strings.

✨

Key Insights

1Insight 1: The core idea is to find the substring of length `k` that appears most frequently in the given string at k-intervals. The other substrings must be replaced with this most frequent substring.
2Insight 2: Using a hash map (or counter) to store the frequency of each k-length substring appearing at multiples of k indices is crucial for efficiently finding the most frequent substring.
3Insight 3: The number of operations needed is simply the total number of k-length substrings (n/k) minus the frequency of the most frequent substring.

📚

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: Empty string or k = n. If the string is empty, the result should be 0. If k=n, then n/k = 1. Only one possible substring which covers the entire string. the code handles this case correctly, returning 0 if all substrings are the same, and zero if there is only one such substring
Edge case 2: All substrings are the same. In this case, the frequency of the most frequent substring will be `n/k`, so `n/k - n/k = 0` operations will be needed, which is the correct answer. The code correctly manages this case. The HashMap will just hold a count of n/k for the same substring, and the calculation n/k - (n/k) results in 0.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not considering all substrings that start at indices divisible by `k`. It's important to increment the index by `k` in each iteration.

✗Common mistake 2: Incorrectly calculating the number of operations. It's important to subtract the frequency of the most frequent substring from the *total* number of k-length substrings (n/k).

✅

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: Data compression algorithms. Identifying frequently occurring patterns is crucial for efficient compression. This pattern identifies repetitive blocks.

🏦Real-world scenario 2: DNA sequencing. Analyzing the frequency of certain DNA sequences can reveal insights into genetic variations and diseases. Finding repetitive sequences helps in genome analysis.

💡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

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

⚠️ Watch Out

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

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: August 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

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

@InterviewProJul 14

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 7

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJun 30

❓

Frequently Asked Questions

How do I choose between a hash map and a hash set?

Use a hash set when you only need to track presence/absence of elements (like detecting duplicates). Use a hash map when you need to associate values with keys (like counting frequencies or storing indices). Hash sets use less memory since they only store keys.

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 "Minimum Number of Operations to Make Word K-Periodic"?

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.

🤖 Need help with this problem?