How do you solve Count Pairs That Form a Complete Day II on LeetCode?

Count Pairs That Form a Complete Day II (LeetCode #3185) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Pairs that sum to a multiple of 24 can be efficiently found using remainders.

What is the time complexity of Count Pairs That Form a Complete Day II?

The optimal solution for Count Pairs That Form a Complete Day II runs in O(n) time and uses O(n) space. The time complexity is O(n) because we only traverse the array once. The space complexity is O(n) due to the hash map storing counts of remainders.

What is the brute force solution for Count Pairs That Form a Complete Day II?

The brute force approach for Count Pairs That Form a Complete Day II is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking every possible pair of indices in the array to see if their sum is a multiple of 24. This is straightforward but inefficient for large arrays. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Count Pairs That Form a Complete Day II?

Count Pairs That Form a Complete Day II uses the following concepts: Array, Hash Table, Counting. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Count Pairs That Form a Complete Day II?

Always consider edge cases, such as the smallest and largest values in the input. Explain your thought process clearly while coding — it shows your understanding. Practice optimizing your brute force solutions to get comfortable with different approaches.

What are common mistakes when solving Count Pairs That Form a Complete Day II?

Not considering the case where hours[i] + hours[j] is 0 modulo 24. Confusing the indices when checking pairs (i, j) — ensure i < j.

#3185

Count Pairs That Form a Complete Day II

Medium

ArrayHash TableCountingHash MapArray

LeetCode ↗

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)

The optimal solution uses a hash map to count occurrences of remainders when each hour is divided by 24. This allows us to efficiently find pairs that sum to a multiple of 24 without checking every combination.

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Algorithm

4 steps

1Step 1: Initialize a hash map to count occurrences of each remainder when hours[i] is divided by 24.
2Step 2: For each hour, calculate the required complement remainder (24 - (current remainder) % 24) to form a complete day.
3Step 3: For each hour, add the count of the complement remainder from the hash map to the result.
4Step 4: Update the hash map with the current hour's remainder.

solution.py11 lines

1# Full working Python code
2
3def countPairs(hours):
4    count = 0
5    remainder_count = {}
6    for hour in hours:
7        remainder = hour % 24
8        complement = (24 - remainder) % 24
9        count += remainder_count.get(complement, 0)
10        remainder_count[remainder] = remainder_count.get(remainder, 0) + 1
11    return count

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Complexity note: The time complexity is O(n) because we only traverse the array once. The space complexity is O(n) due to the hash map storing counts of remainders.

1Pairs that sum to a multiple of 24 can be efficiently found using remainders.
2Using a hash map allows us to count occurrences and find complements in constant time.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.