# Solving Leetcode Interviews in Seconds with AI: Pairs of Songs With Total Durations Divisible by 60


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "1010" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

	# Problem Statement
	> You are given a list of songs where the ith song has a duration of time[i] seconds. Return the number of pairs of songs for which their total duration in seconds is divisible by 60. Formally, we want the number of indices i, j such that i < j with (time[i] + time[j]) % 60 == 0.   Example 1:  Input: time = [30,20,150,100,40] Output: 3 Explanation: Three pairs have a total duration divisible by 60: (time[0] = 30, time[2] = 150): total duration 180 (time[1] = 20, time[3] = 100): total duration 120 (time[1] = 20, time[4] = 40): total duration 60  Example 2:  Input: time = [60,60,60] Output: 3 Explanation: All three pairs have a total duration of 120, which is divisible by 60.    Constraints:  1 <= time.length <= 6 * 104 1 <= time[i] <= 500  

	# Explanation
	*   **Residue Counting:** Compute the remainders of each song duration when divided by 60.
*   **Pairing Remainders:** Count pairs of songs where the sum of their remainders is divisible by 60. This involves pairing a remainder `r` with `60 - r` (or `0` with `0`).
*   **Efficient Counting:** Use a frequency count (hash map or array) to efficiently count the occurrences of each remainder.

*   **Runtime Complexity:** O(n), where n is the number of songs. **Storage Complexity:** O(1), as we only store counts for remainders 0-59.

	
	# Code
	```python
	def numPairsDivisibleBy60(time):
    """
    Given a list of songs where the ith song has a duration of time[i] seconds.
    Return the number of pairs of songs for which their total duration in seconds is divisible by 60.
    """
    remainders = [0] * 60
    count = 0
    for t in time:
        remainder = t % 60
        complement = (60 - remainder) % 60  # Ensure complement is within [0, 59]
        count += remainders[complement]
        remainders[remainder] += 1
    return count
	```