LeetCode #3185 — MEDIUM

Count Pairs That Form a Complete Day II

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

Given an integer array hours representing times in hours, return an integer denoting the number of pairs i, j where i < j and hours[i] + hours[j] forms a complete day.

A complete day is defined as a time duration that is an exact multiple of 24 hours.

For example, 1 day is 24 hours, 2 days is 48 hours, 3 days is 72 hours, and so on.

Example 1:

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

Output: 2

Explanation: 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

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

Constraints:

1 <= hours.length <= 5 * 10⁵
1 <= hours[i] <= 10⁹

Step 01

Brute Force Baseline

Problem summary: Given an integer array hours representing times in hours, return an integer denoting the number of pairs i, j where i < j and hours[i] + hours[j] forms a complete day. A complete day is defined as a time duration that is an exact multiple of 24 hours. For example, 1 day is 24 hours, 2 days is 48 hours, 3 days is 72 hours, and so on.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Hash Map

Example 1

[12,12,30,24,24]

Example 2

[72,48,24,3]

Core Insight

What unlocks the optimal approach

A pair <code>(i, j)</code> forms a valid complete day if <code>(hours[i] + hours[j]) % 24 == 0</code>.
Using an array or a map, for each index <code>j</code> moving from left to right, increase the answer by the count of <code>(24 - hours[j]) % 24</code>, and then increase the count of <code>hours[j]</code>.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #3185: Count Pairs That Form a Complete Day II
class Solution {
    public long countCompleteDayPairs(int[] hours) {
        int[] cnt = new int[24];
        long ans = 0;
        for (int x : hours) {
            ans += cnt[(24 - x % 24) % 24];
            ++cnt[x % 24];
        }
        return ans;
    }
}

// Accepted solution for LeetCode #3185: Count Pairs That Form a Complete Day II
func countCompleteDayPairs(hours []int) (ans int64) {
	cnt := [24]int{}
	for _, x := range hours {
		ans += int64(cnt[(24-x%24)%24])
		cnt[x%24]++
	}
	return
}

# Accepted solution for LeetCode #3185: Count Pairs That Form a Complete Day II
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

// Accepted solution for LeetCode #3185: Count Pairs That Form a Complete Day II
/**
 * [3185] Count Pairs That Form a Complete Day II
 */
pub struct Solution {}

// submission codes start here

impl Solution {
    pub fn count_complete_day_pairs(hours: Vec<i32>) -> i64 {
        use std::collections::HashMap;

        let mut map = HashMap::new();
        let mut result = 0;

        for hour in hours {
            let hour = hour % 24;
            if let Some(&c) = map.get(&((24 - hour) % 24)) {
                result += c;
            }
            let entry = map.entry(hour).or_insert(0);
            *entry += 1;
        }

        result
    }
}

// submission codes end

#[cfg(test)]
mod tests {
    use super::*;

    #[test]
    fn test_3185() {
        assert_eq!(
            2,
            Solution::count_complete_day_pairs(vec![12, 12, 30, 24, 24])
        );
        assert_eq!(3, Solution::count_complete_day_pairs(vec![72, 48, 24, 3]));
    }
}

// Accepted solution for LeetCode #3185: Count Pairs That Form a Complete Day II
function countCompleteDayPairs(hours: number[]): number {
    const cnt: number[] = Array(24).fill(0);
    let ans: number = 0;
    for (const x of hours) {
        ans += cnt[(24 - (x % 24)) % 24];
        ++cnt[x % 24];
    }
    return ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(1)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.