#3184

Easy

auto_awesomeArray scanning plus hash lookup

LeetCode Problem Workspace

Count Pairs That Form a Complete Day I

Count all pairs in an array where their sum forms a complete day using hash-based counting for efficiency.

Problem Statement

Given an integer array hours, return the number of index pairs (i, j) where i < j and the sum hours[i] + hours[j] is a multiple of 24. Each pair represents a combination of work hours that completes a full day.

A complete day is any sum of hours divisible evenly by 24. For example, 24 hours, 48 hours, or 72 hours are considered complete days, and your task is to count all valid pairs forming these totals.

Examples

Example 1

Input: hours = [12,12,30,24,24]

Output: 2

The pairs of indices that form a complete day are (0, 1) and (3, 4) .

Example 2

Input: hours = [72,48,24,3]

Output: 3

The pairs of indices that form a complete day are (0, 1) , (0, 2) , and (1, 2) .

Constraints

1 <= hours.length <= 100
1 <= hours[i] <= 109

Solution Approach

Hash remainder counting

Use a hash map to count occurrences of hours modulo 24. For each number, calculate its remainder and find how many previous numbers have a complementary remainder to sum to 24.

Single pass accumulation

Iterate through the array once, updating the hash map of remainders as you go. Each time you encounter a number, add the count of matching remainders to the total pairs count.

Avoid double counting

Ensure that you only consider pairs where the first index is less than the second to prevent counting the same pair twice. Use the hash map to track prior numbers efficiently.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

Time complexity is O(n) due to a single pass over the array with constant-time hash operations. Space complexity is O(24) = O(1) for the remainder hash map since there are only 24 possible remainders.

What Interviewers Usually Probe

Asks about handling large numbers or values exceeding 24 hours.
Checks if you can optimize from brute force to a hash-based approach.
Wants clarity on handling modulo arithmetic and avoiding double counting.

Common Pitfalls or Variants

Common pitfalls

Brute force all pairs instead of using modulo hash counting.
Forgetting to handle the remainder 0 correctly, which forms complete days with itself.
Double counting pairs by not ensuring i < j.

Follow-up variants

Count pairs that sum to a multiple of k instead of 24, changing the modulo base.
Return the list of pairs themselves instead of just the count, requiring more storage.
Handle negative or zero hours while still using the modulo hash approach.

FAQ

How do I count pairs that form a complete day efficiently?

Track remainders modulo 24 in a hash map while scanning the array and count complementary remainders to form full days.

Why does the problem use hours[i] + hours[j] % 24?

Using modulo 24 identifies sums that are exact multiples of a day without computing large totals, keeping operations efficient.

Can GhostInterview handle arrays with more than 100 elements?

Yes, the hash map approach scales linearly with array size, so it works efficiently even near the upper limit of 100 elements.

What if the array has repeated values like multiple 24s?

The hash map automatically counts occurrences, so repeated values correctly contribute to multiple valid pairs without extra effort.

Is this problem an example of array scanning plus hash lookup?

Exactly, it demonstrates scanning the array while using a hash map to track and count complementary remainders for a target sum pattern.

terminal

Solution

Solution 1: Counting

We can use a hash table or an array $\textit{cnt}$ of length $24$ to record the occurrence count of each hour modulo $24$.

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class Solution:
    def countCompleteDayPairs(self, hours: List[int]) -> int:
        cnt = Counter()
        ans = 0
        for x in hours:
            ans += cnt[(24 - (x % 24)) % 24]
            cnt[x % 24] += 1
        return ans

Continue Practicing

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