What is the time complexity of Number of Subarrays With LCM Equal to K?

The optimal solution has a time complexity of O(n^2 * log(max(nums))).

What data structures are used to solve Number of Subarrays With LCM Equal to K?

This problem uses Array, Math, Number Theory. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Number of Subarrays With LCM Equal to K?

Insight 1: We can iterate through all possible subarrays of `nums` and calculate the LCM of each subarray to check if it equals `k`. Insight 2: To efficiently calculate the LCM of a subarray, we can use the formula `lcm(a, b) = (a * b) / gcd(a, b)`, where `gcd(a, b)` is the greatest common divisor of `a` and `b`.

Which companies ask LeetCode 2470. Number of Subarrays With LCM Equal to K in interviews?

This problem is frequently asked at Unity.

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Number of Subarrays With LCM Equal to K - LeetCode 2470 Solution

Number of Subarrays With LCM Equal to K - Complete Solution Guide

Number of Subarrays With LCM Equal to K is LeetCode problem 2470, 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

Given an integer array nums and an integer k , return the number of subarrays of nums where the least common multiple of the subarray's elements is k . A subarray is a contiguous non-empty sequence of elements within an array. The least common multiple of an array is the smallest positive integer that is divisible by all the array elements. Example 1: Input: nums = [3,6,2,7,1], k = 6 Output: 4 Explanation: The subarrays of nums where 6 is the least common multiple of all the subarray's elements

Detailed Explanation

The problem asks us to find the number of subarrays within a given array `nums` whose least common multiple (LCM) is equal to a given integer `k`. A subarray is a contiguous sequence of elements within the array. The LCM of an array is the smallest positive integer that is divisible by all elements of the array. The input consists of an integer array `nums` and an integer `k`, and the output is the number of subarrays of `nums` that satisfy the LCM condition. The constraints are that the length of `nums` is between 1 and 1000, and each element in `nums` and `k` are between 1 and 1000.

Solution Approach

The provided solution iterates through all possible subarrays using nested loops. For each subarray, it checks if all elements of the subarray divide `k`. If any element doesn't divide `k`, the inner loop breaks because the LCM cannot be `k`. If all elements divide `k`, it calculates the LCM of the elements in the subarray using the formula `lcm(a, b) = (a * b) / gcd(a, b)` iteratively. If the LCM equals `k`, the count is incremented. Finally, the total count of subarrays with LCM equal to `k` is returned.

Step-by-Step Algorithm

Step 1: Initialize a counter `count` to 0. This variable will store the number of subarrays with LCM equal to `k`.
Step 2: Iterate through the input array `nums` using a nested loop structure. The outer loop iterates from `i = 0` to `n - 1`, where `n` is the length of `nums`. This loop represents the starting index of the subarray.
Step 3: The inner loop iterates from `j = i` to `n - 1`. This loop represents the ending index of the subarray.
Step 4: Inside the inner loop, initialize `current_lcm` to 1. This variable will store the LCM of the current subarray.
Step 5: Check if `k` is divisible by `nums[j]`. If it's not, break the inner loop because the LCM of the subarray cannot be `k` if any element in the subarray doesn't divide `k`.
Step 6: If `k` is divisible by `nums[j]`, update `current_lcm` using the formula `current_lcm = (current_lcm * nums[j]) / gcd(current_lcm, nums[j])`.
Step 7: After updating `current_lcm`, check if it is equal to `k`. If it is, increment `count`.
Step 8: After the inner loop completes, the outer loop continues to the next starting index.
Step 9: After the outer loop completes, return `count`.

Key Insights

Insight 1: We can iterate through all possible subarrays of `nums` and calculate the LCM of each subarray to check if it equals `k`.
Insight 2: To efficiently calculate the LCM of a subarray, we can use the formula `lcm(a, b) = (a * b) / gcd(a, b)`, where `gcd(a, b)` is the greatest common divisor of `a` and `b`.
Insight 3: If any element in the subarray does not divide `k` evenly, then it is impossible for the LCM of that subarray to be equal to `k`. This optimization avoids unnecessary LCM calculations.

Complexity Analysis

Time Complexity: O(n^2 * log(max(nums)))

Space Complexity: O(1)

Topics

This problem involves: Array, Math, Number Theory.

Companies

Asked at: Unity.

Number of Subarrays With LCM Equal to K | LeetCode 2470

Medium

Array Math Number Theory

Time: O(n^2 * log(max(nums)))Space: O(1)

Solution Code

Available in:

class Solution {
    public int subarrayLCM(int[] nums, int k) {
        int n = nums.length;
        int count = 0;
        for (int i = 0; i < n; i++) {
            int current_lcm = 1;
            for (int j = i; j < n; j++) {
                if (k % nums[j] != 0) {
                    break;
                }
                
                // In Java, we use the same lcm(a, b) = (a * b) / gcd(a, b) formula.
                current_lcm = (current_lcm * nums[j]) / gcd(current_lcm, nums[j]);
                
                if (current_lcm == k) {
                    count++;
                }
            }
        }
        return count;
    }
    
    // Helper function to calculate the greatest common divisor (GCD).
    private int gcd(int a, int b) {
        while (b != 0) {
            int temp = b;
            b = a % b;
            a = temp;
        }
        return a;
    }
}

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Problem Info

Description

Problem 2470: Number of Subarrays With LCM Equal to K - Detailed explanation and solution approaches.

Topics

Array Math Number Theory

Companies

Unity

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:2470

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

9 steps

1
Step 1: Initialize a counter `count` to 0. This variable will store the number of subarrays with LCM equal to `k`.
2
Step 2: Iterate through the input array `nums` using a nested loop structure. The outer loop iterates from `i = 0` to `n - 1`, where `n` is the length of `nums`. This loop represents the starting index of the subarray.
3
Step 3: The inner loop iterates from `j = i` to `n - 1`. This loop represents the ending index of the subarray.
4
Step 4: Inside the inner loop, initialize `current_lcm` to 1. This variable will store the LCM of the current subarray.
5
Step 5: Check if `k` is divisible by `nums[j]`. If it's not, break the inner loop because the LCM of the subarray cannot be `k` if any element in the subarray doesn't divide `k`.
6
Step 6: If `k` is divisible by `nums[j]`, update `current_lcm` using the formula `current_lcm = (current_lcm * nums[j]) / gcd(current_lcm, nums[j])`.
7
Step 7: After updating `current_lcm`, check if it is equal to `k`. If it is, increment `count`.
8
Step 8: After the inner loop completes, the outer loop continues to the next starting index.
9
Step 9: After the outer loop completes, return `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^2 * log(max(nums)))

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The time complexity is O(n^2 * log(max(nums))) because of the nested loops which take O(n^2) time and the gcd calculation inside the inner loop takes O(log(max(nums))) time. The space complexity is O(1) because we are only using a few constant space variables.

📖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

Optimized GCD/LCM computation

Precompute GCD of elements for each subarray to improve performance in scenarios where the same subarrays are evaluated repeatedly (although this does not change the overall O(n^2) complexity). Also, memoize calculated LCM values for improved calculation speed if there are repetitive number subarrays. However, the problem constraints are small, so this does not offer meaningful improvements.

Time: O(n^2 * log(max(nums)))Space: O(n^2) in the worst case

Trade-offs: This approach might slightly improve performance if many subarrays share common elements, but it requires additional space to store the precomputed GCDs/LCMs.

✨

Key Insights

1Insight 1: We can iterate through all possible subarrays of `nums` and calculate the LCM of each subarray to check if it equals `k`.
2Insight 2: To efficiently calculate the LCM of a subarray, we can use the formula `lcm(a, b) = (a * b) / gcd(a, b)`, where `gcd(a, b)` is the greatest common divisor of `a` and `b`.
3Insight 3: If any element in the subarray does not divide `k` evenly, then it is impossible for the LCM of that subarray to be equal to `k`. This optimization avoids unnecessary LCM calculations.

📚

Key Terms Explained

For Students

Array

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

Edge Cases

Edge case 1: `nums` is empty. The code handles this case correctly as the outer loop won't execute, and 0 will be returned.
Edge case 2: `k` is 1. The code correctly handles this case by calculating the LCM of each subarray and comparing it with `k`.
Edge case 3: `nums` contains a zero. This is not explicitly handled but since the problem states `1 <= nums[i] <= 1000`, it is not a valid test case. The gcd calculation will throw an error if any element is 0.
Edge case 4: If `k` is a large prime number and no element in `nums` divides `k`, the inner loop breaks immediately, resulting in faster execution. The code does handle this efficiently by checking if `k % nums[j] != 0` before computing the LCM.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to check if the current element divides `k` before calculating the LCM. This can lead to unnecessary calculations and incorrect results.

✗Common mistake 2: Using integer division in the LCM calculation, which can cause precision issues. The provided solutions use appropriate integer division to handle this.

✗Common mistake 3: Not handling the edge case where the input array is empty. The given solution implicitly handles the case when nums is empty by returning 0, which is correct.

✗Common mistake 4: Integer overflow when calculating the product of numbers in the LCM formula. The CPP solution casts the values to `long long` to avoid the overflow issue.

✅

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: Optimizing resource allocation. For example, you might have a set of tasks with different time requirements, and you want to find combinations of tasks that fit within a specific timeframe determined by an LCM.

🏦Real-world scenario 2: Analyzing periodic events. If you have a set of events that occur at different intervals, finding the LCM of those intervals can help you determine when they will all occur simultaneously.

💡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

If stuck, try working through a small example by hand. The pattern often reveals itself.

⚠️ 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: October 9, 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 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 "Number of Subarrays With LCM Equal to K"?

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 math techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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