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Leetcode 3019. Number of Changing Keys

Problem Statement

You are given a list of keys where each key is represented as a list of characteristics. Two keys are considered identical if they have exactly the same characteristics in any order. The task is to determine how many keys need to be changed to make all keys identical.

Clarifying Questions

  1. Characteristics Uniqueness: Are the characteristics unique within a single key?
    • Assume yes.
  2. Characteristics Order: Does the order of characteristics in each key matter?
    • No, order does not matter.
  3. Input Format: What is the format of the input?
    • A list of lists, where each sub-list contains characteristics of a key.
  4. Output Format: What should be the format of the output?
    • An integer indicating the number of keys that need to be changed.

Strategy

To determine the number of keys which need to be changed to make all keys identical, we can approach the problem as follows:

  1. Normalize Keys: Convert each key to a tuple of sorted characteristics to ensure identical keys have the same representation.
  2. Count Occurrences: Use a dictionary to count how many times each normalized key appears.
  3. Find Most Frequent Key: Identify the normalized key with the maximum number of occurrences.
  4. Calculate Changes: The number of keys to change is then the total number of keys minus the count of the most frequent key.

Code

import java.util.*;

public class ChangingKeys {

    public static int numberOfChangingKeys(List<List<String>> keys) {
        if (keys == null || keys.isEmpty()) return 0;

        // Map to store the frequency of each normalized key
        Map<List<String>, Integer> keyCountMap = new HashMap<>();
        
        for (List<String> key : keys) {
            // Sort characteristics to standardize the representation
            List<String> sortedKey = new ArrayList<>(key);
            Collections.sort(sortedKey);
            
            keyCountMap.put(sortedKey, keyCountMap.getOrDefault(sortedKey, 0) + 1);
        }
        
        // Find the maximum frequency
        int maxFrequency = 0;
        for (int count : keyCountMap.values()) {
            maxFrequency = Math.max(maxFrequency, count);
        }
        
        // Number of keys to change is total keys minus the most frequent key count
        return keys.size() - maxFrequency;
    }

    public static void main(String[] args) {
        List<List<String>> keys = Arrays.asList(
            Arrays.asList("A", "B", "C"),
            Arrays.asList("C", "B", "A"),
            Arrays.asList("A", "B"),
            Arrays.asList("B", "C", "B")
        );
        
        System.out.println(numberOfChangingKeys(keys));  // Example usage
    }
}

Time Complexity

  1. Normalization: Sorting each key takes (O(M \log M)) time where (M) is the number of characteristics in a key.
  2. Total Sorting: If the total number of keys is (N), then the time complexity for normalization for all keys will be (O(N \cdot M \log M)).
  3. Counting: Counting occurrences will take (O(N)).
  4. Finding Maximum Frequency: This again is (O(N)).

Combining these, the overall time complexity is (O(N \cdot M \log M)).

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