problemmediumdatabaseleetcode-1699leetcode 1699leetcode1699

Number of Calls Between Two Persons

MediumUpdated: Aug 2, 2025
Practice on:
LeetCode

Problem

Table: Calls

+-------------+---------+
| Column Name | Type    |
+-------------+---------+
| from_id     | int     |
| to_id       | int     |
| duration    | int     |
+-------------+---------+
This table does not have a primary key (column with unique values), it may contain duplicates.
This table contains the duration of a phone call between from_id and to_id.
from_id != to_id

Write a solution to report the number of calls and the total call duration between each pair of distinct persons (person1, person2) where person1 < person2.

Return the result table in any order.

The result format is in the following example.

Examples

Example 1:

Input: 
Calls table:
+---------+-------+----------+
| from_id | to_id | duration |
+---------+-------+----------+
| 1       | 2     | 59       |
| 2       | 1     | 11       |
| 1       | 3     | 20       |
| 3       | 4     | 100      |
| 3       | 4     | 200      |
| 3       | 4     | 200      |
| 4       | 3     | 499      |
+---------+-------+----------+
Output: 
+---------+---------+------------+----------------+
| person1 | person2 | call_count | total_duration |
+---------+---------+------------+----------------+
| 1       | 2       | 2          | 70             |
| 1       | 3       | 1          | 20             |
| 3       | 4       | 4          | 999            |
+---------+---------+------------+----------------+
Explanation: 
Users 1 and 2 had 2 calls and the total duration is 70 (59 + 11).
Users 1 and 3 had 1 call and the total duration is 20.
Users 3 and 4 had 4 calls and the total duration is 999 (100 + 200 + 200 + 499).

Solution

Method 1 – SQL Group By Normalized Pair

Intuition

Normalize each call pair so that (person1, person2) is always (min(from_id, to_id), max(from_id, to_id)). Group by this pair and aggregate count and sum.

Approach

  1. For each row, set person1 = LEAST(from_id, to_id), person2 = GREATEST(from_id, to_id).
  2. Group by person1, person2 and count calls, sum durations.

Code

SQL
SELECT LEAST(from_id, to_id) AS person1,
       GREATEST(from_id, to_id) AS person2,
       COUNT(*) AS call_count,
       SUM(duration) AS total_duration
FROM Calls
GROUP BY person1, person2;

Complexity

  • ⏰ Time complexity: O(N)
  • 🧺 Space complexity: O(N)

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