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

The optimal solution for Minimum Number of Operations to Make Word K-Periodic runs in O(n) time and uses O(n) space. This complexity is linear because we only traverse the string once to build the frequency map and then once more to find the maximum frequency.

What is the brute force solution for Minimum Number of Operations to Make Word K-Periodic?

The brute force approach for Minimum Number of Operations to Make Word K-Periodic is Brute Force, with O(n²) time complexity and O(1) space complexity. In the brute-force approach, we try to generate all possible k-length substrings from the string and compare them. This helps us understand how many operations are needed to make the string k-periodic by counting mismatches. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Minimum Number of Operations to Make Word K-Periodic?

Minimum Number of Operations to Make Word K-Periodic uses the following concepts: Hash Table, String, Counting. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Minimum Number of Operations to Make Word K-Periodic?

Always clarify the problem constraints and definitions before jumping to solutions. Think about edge cases, such as strings that are already k-periodic. Practice explaining your thought process clearly, as communication is key in interviews.

What are common mistakes when solving Minimum Number of Operations to Make Word K-Periodic?

Not considering all k-length substrings starting from indices divisible by k. Overlooking the need to count the maximum frequency correctly.

#3137

Minimum Number of Operations to Make Word K-Periodic

Medium

Hash TableStringCountingHash MapArray

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Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

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Intuition

Time O(n)Space O(n)

In the optimal approach, we count the frequency of each k-length substring that starts at indices divisible by k. The minimum operations needed will be the total number of substrings minus the frequency of the most common substring.

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Algorithm

3 steps

1Step 1: Create a frequency map to count occurrences of each k-length substring starting at indices divisible by k.
2Step 2: Find the maximum frequency from the frequency map.
3Step 3: Calculate the minimum operations as the total number of k-length substrings minus the maximum frequency.

solution.py10 lines

1from collections import defaultdict
2
3def min_operations(word, k):
4    freq = defaultdict(int)
5    n = len(word)
6    for i in range(0, n, k):
7        freq[word[i:i + k]] += 1
8    max_freq = max(freq.values())
9    total_substrings = n // k
10    return total_substrings - max_freq

ℹ

Complexity note: This complexity is linear because we only traverse the string once to build the frequency map and then once more to find the maximum frequency.

1Understanding the frequency of substrings helps minimize operations.
2The most common substring dictates how many changes are needed.

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