LeetCode 3230: Customer Purchasing Behavior Analysis Solution

Master LeetCode problem 3230 (Customer Purchasing Behavior Analysis), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

3230. Customer Purchasing Behavior Analysis

Medium

Database

Problem Explanation

Explanation:

To solve this problem, we can use a hash map to keep track of each customer's purchasing behavior. We will iterate through the transactions and update the purchase count for each customer accordingly. After processing all transactions, we will identify the customers who made purchases on consecutive days and count them.

Algorithm:

Create a hash map to store the purchase behavior of each customer.
Iterate through the transactions and update the purchase count for each customer.
Iterate through the hash map and count the customers who made purchases on consecutive days.

Time Complexity: O(n), where n is the number of transactions. Space Complexity: O(n), for the hash map storing customer purchase behavior.

Solution Code

import java.util.*;

class Solution {
    public int analyzeCustomerPurchase(int[][] transactions) {
        Map<Integer, Set<Integer>> purchaseMap = new HashMap<>();
        
        for (int[] transaction : transactions) {
            int customer = transaction[0];
            int day = transaction[1];
            
            if (!purchaseMap.containsKey(customer)) {
                purchaseMap.put(customer, new HashSet<>());
            }
            purchaseMap.get(customer).add(day);
        }
        
        int count = 0;
        for (Set<Integer> days : purchaseMap.values()) {
            List<Integer> sortedDays = new ArrayList<>(days);
            Collections.sort(sortedDays);
            int consecutiveCount = 1;
            for (int i = 1; i < sortedDays.size(); i++) {
                if (sortedDays.get(i) - sortedDays.get(i - 1) == 1) {
                    consecutiveCount++;
                } else {
                    consecutiveCount = 1;
                }
                if (consecutiveCount >= 2) {
                    count++;
                    break;
                }
            }
        }
        
        return count;
    }
}

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