What is the time complexity of Pairs of Songs With Total Durations Divisible by 60?

The optimal solution has a time complexity of O(n).

What data structures are used to solve Pairs of Songs With Total Durations Divisible by 60?

This problem uses Array, Hash Table, Counting. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Pairs of Songs With Total Durations Divisible by 60?

Insight 1: The crucial insight is that we only need to consider the remainder of each song's duration when divided by 60. If `(time[i] + time[j]) % 60 == 0`, then `(time[i] % 60 + time[j] % 60) % 60 == 0`. This simplifies the problem. Insight 2: Instead of checking every pair, we can use a hash table (or in this case, an array of size 60) to store the frequency of each remainder. This allows us to efficiently find pairs that sum to a multiple of 60.

Which companies ask LeetCode 1010. Pairs of Songs With Total Durations Divisible by 60 in interviews?

This problem is frequently asked at Akamai, Atlassian, BlackRock, Citrix, Docusign and 2 more companies.

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Pairs of Songs With Total Durations Divisible by 60 - LeetCode 1010 Solution

Q: How to solve Pairs of Songs With Total Durations Divisible by 60?

The provided solution uses an array `remainders` of size 60 to store the count of each remainder. It iterates through the input array `time`, calculates the remainder of each song's duration when divided by 60, and updates the count. While iterating, for each song, it checks how many songs already exist that have a remainder that complements the current song's remainder (i.e., the sum of the remainders is a multiple of 60).

Pairs of Songs With Total Durations Divisible by 60 - Complete Solution Guide

Pairs of Songs With Total Durations Divisible by 60 is LeetCode problem 1010, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

You are given a list of songs where the i th 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

Detailed Explanation

The problem requires us to find the number of pairs of songs from a given list `time` whose total duration is divisible by 60. We need to iterate through all possible pairs of songs (i, j) where i < j and check if `(time[i] + time[j]) % 60 == 0`. The goal is to count how many such pairs exist.

Solution Approach

The provided solution uses an array `remainders` of size 60 to store the count of each remainder. It iterates through the input array `time`, calculates the remainder of each song's duration when divided by 60, and updates the count. While iterating, for each song, it checks how many songs already exist that have a remainder that complements the current song's remainder (i.e., the sum of the remainders is a multiple of 60).

Step-by-Step Algorithm

Step 1: Initialize an array `remainders` of size 60 with all elements set to 0. This array will store the counts of each remainder when the song durations are divided by 60.
Step 2: Initialize a variable `count` to 0. This variable will store the total number of pairs that satisfy the condition.
Step 3: Iterate through the input array `time`.
Step 4: For each song duration `t` in `time`, calculate its remainder `remainder = t % 60`.
Step 5: If the `remainder` is 0, increment `count` by the value of `remainders[0]`. Otherwise, increment `count` by the value of `remainders[60 - remainder]`. This step is finding pairs that sum up to a multiple of 60.
Step 6: Increment `remainders[remainder]` by 1. This updates the count of the current remainder.
Step 7: After iterating through all the song durations, return the final `count`.

Key Insights

Insight 1: The crucial insight is that we only need to consider the remainder of each song's duration when divided by 60. If `(time[i] + time[j]) % 60 == 0`, then `(time[i] % 60 + time[j] % 60) % 60 == 0`. This simplifies the problem.
Insight 2: Instead of checking every pair, we can use a hash table (or in this case, an array of size 60) to store the frequency of each remainder. This allows us to efficiently find pairs that sum to a multiple of 60.
Insight 3: We need to handle the case where both remainders are 0 separately. For a remainder r, we are looking for a remainder of 60-r. However, when r is 0, we are looking for another remainder of 0.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

This problem involves: Array, Hash Table, Counting.

Companies

Asked at: Akamai, Atlassian, BlackRock, Citrix, Docusign, PayPal, Salesforce.

Pairs of Songs With Total Durations Divisible by 60 | LeetCode 1010

Medium

Array Hash Table Counting

Time: O(n)Space: O(1)

Solution Code

Available in:

class Solution {
    public int numPairsDivisibleBy60(int[] time) {
        int[] remainders = new int[60];
        int count = 0;
        for (int t : time) {
            int remainder = t % 60;
            if (remainder == 0) {
                count += remainders[0];
            } else {
                count += remainders[60 - remainder];
            }
            remainders[remainder]++;
        }
        return count;
    }
}

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Problem Info

Description

Problem 1010: Pairs of Songs With Total Durations Divisible by 60 - Detailed explanation and solution approaches.

Topics

Array Hash Table Counting

Companies

AkamaiAtlassianBlackRockCitrixDocusignPayPalSalesforce

Stats

Difficulty:Medium

Languages:4

Companies:7

Problem #:1010

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

7 steps

1
Step 1: Initialize an array `remainders` of size 60 with all elements set to 0. This array will store the counts of each remainder when the song durations are divided by 60.
2
Step 2: Initialize a variable `count` to 0. This variable will store the total number of pairs that satisfy the condition.
3
Step 3: Iterate through the input array `time`.
4
Step 4: For each song duration `t` in `time`, calculate its remainder `remainder = t % 60`.
5
Step 5: If the `remainder` is 0, increment `count` by the value of `remainders[0]`. Otherwise, increment `count` by the value of `remainders[60 - remainder]`. This step is finding pairs that sum up to a multiple of 60.
6
Step 6: Increment `remainders[remainder]` by 1. This updates the count of the current remainder.
7
Step 7: After iterating through all the song durations, return the final `count`.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The time complexity is determined by the single loop iterating through the input array `time`. The space complexity is O(1) since the size of the `remainders` array is fixed and independent of the input size.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Brute Force

Iterate through all possible pairs of songs and check if their sum is divisible by 60. This is the naive approach.

Time: O(n^2), where n is the number of songs.Space: O(1)

Trade-offs: The brute force approach is simple to understand, but it is inefficient for large input sizes. The given solution is more efficient with O(n) time complexity.

✨

Key Insights

1Insight 1: The crucial insight is that we only need to consider the remainder of each song's duration when divided by 60. If `(time[i] + time[j]) % 60 == 0`, then `(time[i] % 60 + time[j] % 60) % 60 == 0`. This simplifies the problem.
2Insight 2: Instead of checking every pair, we can use a hash table (or in this case, an array of size 60) to store the frequency of each remainder. This allows us to efficiently find pairs that sum to a multiple of 60.
3Insight 3: We need to handle the case where both remainders are 0 separately. For a remainder r, we are looking for a remainder of 60-r. However, when r is 0, we are looking for another remainder of 0.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Hash Table

A data structure that maps keys to values for quick lookup. Like a dictionary!

Edge Cases

Edge case 1: When the input array is empty. The code correctly returns 0 since there are no pairs.
Edge case 2: When all songs have durations that are multiples of 60. The code correctly counts all possible pairs because the count is incremented by the previous count of songs with remainder 0.
Edge case 3: When there are songs with duration 0. Although constraint mentions duration >=1, it is good practice to ensure that 0 is handled correctly. The given approach treats 0 like any other number, therefore even if a duration were zero the program would correctly handle it and count pairs.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to handle the case where the remainder is 0. The complementary remainder to 0 is also 0, not 60.

✗Common mistake 2: Incorrectly calculating the index in the `remainders` array. It should be `60 - remainder` (or 0 when remainder is 0) to find the complementary remainder.

✗Common mistake 3: Counting a pair twice, which is avoided by the way remainders are counted during the iteration.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Consider scheduling tasks that have dependencies. If tasks i and j can only be done together every 60 minutes, this problem is similar.

🏦Real-world scenario 2: Imagine a music streaming service where you want to recommend pairs of songs that have a combined length that's a multiple of 60 to create a playlist with specific length requirements.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Consider multiple approaches before coding. Mentioning alternatives shows breadth of knowledge.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

👨‍💻 From Experience

This problem pattern appears frequently in real codebases. I've used similar logic when building production systems.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: September 19, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 14

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 7

❓

Frequently Asked Questions

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time complexity.

What are the trade-offs of using a hash table?

Hash tables provide O(1) average-case lookup, insertion, and deletion, making them ideal for frequency counting and duplicate detection. However, they use O(n) extra space and may degrade to O(n) time in worst-case scenarios with hash collisions. Always consider if the space trade-off is acceptable.

What techniques are commonly tested in medium-difficulty problems?

Medium problems often require combining multiple techniques: two-pointer with hash maps, BFS/DFS with state tracking, dynamic programming with optimization, or divide-and-conquer approaches. They test your ability to recognize patterns and choose the right combination of techniques.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Pairs of Songs With Total Durations Divisible by 60"?

For this medium-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether hash table techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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