LeetCode 3185: Count Pairs That Form a Complete Day II Solution

Master LeetCode problem 3185 (Count Pairs That Form a Complete Day II), 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.

3185. Count Pairs That Form a Complete Day II

Medium

Array Hash Table Counting

Problem Explanation

Explanation

To solve this problem, we can use a HashMap to store the frequencies of remainders when the sum of two hours is divided by 24. By iterating over the input array, we calculate the remainder for each element and check if the complement (24 - remainder) exists in the HashMap. If it does, we increment the count of valid pairs by the frequency of the complement in the HashMap. We also update the frequency of the current remainder in the HashMap.

Algorithm:

Initialize a HashMap to store the frequency of remainders.
Initialize a variable count to store the total count of valid pairs.
Iterate over the input array hours.
Calculate the remainder rem when hours[i] is divided by 24.
Check if the complement (24 - rem) exists in the HashMap.
- If it does, increment count by the frequency of the complement in the HashMap.
Update the frequency of the current remainder in the HashMap.
Return the total count of valid pairs.

Time Complexity: O(N), where N is the number of elements in the input array hours.

Space Complexity: O(N), where N is the number of elements in the input array hours.

Solution Code

import java.util.HashMap;

class Solution {
    public int countPairs(int[] hours) {
        HashMap<Integer, Integer> remainderFreq = new HashMap<>();
        int count = 0;
        
        for (int hour : hours) {
            int rem = hour % 24;
            if (remainderFreq.containsKey(24 - rem)) {
                count += remainderFreq.get(24 - rem);
            }
            remainderFreq.put(rem, remainderFreq.getOrDefault(rem, 0) + 1);
        }
        
        return count;
    }
}

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