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, 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

    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