# LC 3185. Count Pairs That Form a Complete Day II

### [Problem link](https://leetcode.com/problems/count-pairs-that-form-a-complete-day-ii/)

###### tags: `leedcode` `medium` `c++`

Given an integer array <code>hours</code> representing times in **hours** , return an integer denoting the number of pairs <code>i</code>, <code>j</code> where <code>i < j</code> and <code>hours[i] + hours[j]</code> 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:**

<div class="example-block">
Input: <span class="example-io">hours = [12,12,30,24,24]

Output: <span class="example-io">2

Explanation: The pairs of indices that form a complete day are <code>(0, 1)</code> and <code>(3, 4)</code>.

**Example 2:**

<div class="example-block">
Input: <span class="example-io">hours = [72,48,24,3]

Output: <span class="example-io">3

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

**Constraints:**

- <code>1 <= hours.length <= 5 * 10⁵</code>
- <code>1 <= hours[i] <= 10⁹</code>

## Solution 1
#### C++

class Solution {
public:
    long long countCompleteDayPairs(vector<int>& hours) {
        long long res = 0;
        unordered_map<int, int> umap;
        for (int& hour: hours) {
            int tmp = hour % 24;
            if (tmp == 0) {
                res += umap[0];
            } else {
                res += umap[24 - tmp];
            }
            umap[tmp]++;
        }
        return res;
    }
};

>### Complexity
>| | Time Complexity | Space Complexity |
>| ----------- | --------------- | ---------------- |
>| Solution 1 | O(n) | O(n) |

## Note
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